{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0001","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4d5be0a5e23a0468881e07a56383ebb91e59bf8c2b828bad33dfc372c0127ab","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153001,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-10,1],[-11,1],[-12,1]],\"theta\":[-12,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-12, complete estimator values=[-12, -11, -10] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-12, [[-12, 1/4], [-11, 1/2], [-10, 1/4]], -11, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-12, [[-12, 1/4], [-11, 1/2], [-10, 1/4]], -11, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0002","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"970306176662449e649c2d83cd8bd8830baadafa8ae550798926712e4e726545","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153002,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[-11,1],[-7,1],[-9,1]],\"theta\":[-11,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-11, complete estimator values=[-11, -9, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-11, [[-11, 1/4], [-9, 1/2], [-7, 1/4]], -9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-11, [[-11, 1/4], [-9, 1/2], [-7, 1/4]], -9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0003","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d9b84280d82b19f5da151bacacaf8e788a0cece1bdd546bd3277fd62921e182","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153003,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[-10,1],[-4,1],[-7,1]],\"theta\":[-10,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-10, complete estimator values=[-10, -7, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-10, [[-10, 1/4], [-7, 1/2], [-4, 1/4]], -7, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-10, [[-10, 1/4], [-7, 1/2], [-4, 1/4]], -7, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0004","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9156ad49d2e94f74452399fa25995386b360d23a498db20aee0430ae7fa89066","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153004,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-7,1],[-8,1],[-9,1]],\"theta\":[-9,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-9, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-9, [[-9, 1/4], [-8, 1/2], [-7, 1/4]], -8, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-9, [[-9, 1/4], [-8, 1/2], [-7, 1/4]], -8, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0005","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ab880a173dfbe5b5bb77495aa97ff616c0a2290cfe1c02fb35340cd6b5d109f","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153005,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-4,1],[-6,1],[-8,1]],\"theta\":[-8,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-8, complete estimator values=[-8, -6, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-8, [[-8, 1/4], [-6, 1/2], [-4, 1/4]], -6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-8, [[-8, 1/4], [-6, 1/2], [-4, 1/4]], -6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0006","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c2717a82acf77c1db94ac0a2f5ae088280396b132121fed2ff6c07888479f38","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153006,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-4,1],[-7,1]],\"theta\":[-7,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-7, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-7, [[-7, 1/4], [-4, 1/2], [-1, 1/4]], -4, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-7, [[-7, 1/4], [-4, 1/2], [-1, 1/4]], -4, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0007","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"949732321f656c785f554a03f905651af508e0ab370e02247b4eed10e91707c4","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153007,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-4,1],[-5,1],[-6,1]],\"theta\":[-6,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-6, complete estimator values=[-6, -5, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-6, [[-6, 1/4], [-5, 1/2], [-4, 1/4]], -5, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-6, [[-6, 1/4], [-5, 1/2], [-4, 1/4]], -5, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0008","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1db81b6d40205765f358d8488d2ad640e73413a5a381aad949bd11b8bbc2d910","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153008,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-3,1],[-5,1]],\"theta\":[-5,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-5, complete estimator values=[-5, -3, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-5, [[-5, 1/4], [-3, 1/2], [-1, 1/4]], -3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-5, [[-5, 1/4], [-3, 1/2], [-1, 1/4]], -3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0009","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05ece0c095a278b368aca57c88fa1f4229b1fcb97f14277b4c9482ec923e393b","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153009,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[-1,1],[-4,1],[2,1]],\"theta\":[-4,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-4, complete estimator values=[-4, -1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-4, [[-4, 1/4], [-1, 1/2], [2, 1/4]], -1, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-4, [[-4, 1/4], [-1, 1/2], [2, 1/4]], -1, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0010","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d89d6d51b91ac27c84e2bf90df05ca48f166a31f24338a23203b10028706420","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153010,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-2,1],[-3,1]],\"theta\":[-3,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-3, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-3, [[-3, 1/4], [-2, 1/2], [-1, 1/4]], -2, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-3, [[-3, 1/4], [-2, 1/2], [-1, 1/4]], -2, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0011","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d77e25c9e73cb9e818fa41528bac21f0a121f7c4172ab625683f45c8141d5737","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153011,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-2,1],[0,1],[2,1]],\"theta\":[-2,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-2, complete estimator values=[-2, 0, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-2, [[-2, 1/4], [0, 1/2], [2, 1/4]], 0, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-2, [[-2, 1/4], [0, 1/2], [2, 1/4]], 0, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0012","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1f9006ba4576b6e36aafbb0374c6d1316cc6dc0787894457b1ad56a802216a0","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153012,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[2,1],[5,1]],\"theta\":[-1,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=-1, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(-1, [[-1, 1/4], [2, 1/2], [5, 1/4]], 2, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-1, [[-1, 1/4], [2, 1/2], [5, 1/4]], 2, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0013","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e0eebbd08da18659b5b6346059c5bda5a8717b26343b50726dff72f0d71e50b","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153013,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[0,1],[1,1],[2,1]],\"theta\":[0,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=0, complete estimator values=[0, 1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(0, [[0, 1/4], [1, 1/2], [2, 1/4]], 1, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (0, [[0, 1/4], [1, 1/2], [2, 1/4]], 1, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0014","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0507152a79fa328aaba61805d30509102abd3c77a06ff02d8cd67794769084f8","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153014,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[1,1],[3,1],[5,1]],\"theta\":[1,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=1, complete estimator values=[1, 3, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(1, [[1, 1/4], [3, 1/2], [5, 1/4]], 3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (1, [[1, 1/4], [3, 1/2], [5, 1/4]], 3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0015","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"170f6ecf7874fdcc21566e52c8134983d782757afc3eb144060ec36b0afb4948","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153015,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[2,1],[5,1],[8,1]],\"theta\":[2,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=2, complete estimator values=[2, 5, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(2, [[2, 1/4], [5, 1/2], [8, 1/4]], 5, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (2, [[2, 1/4], [5, 1/2], [8, 1/4]], 5, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0016","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9fd5f6209f2a9ce07328a589870631f7201a9a48e4be6544d84751935225376","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153016,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[3,1],[4,1],[5,1]],\"theta\":[3,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=3, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(3, [[3, 1/4], [4, 1/2], [5, 1/4]], 4, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (3, [[3, 1/4], [4, 1/2], [5, 1/4]], 4, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0017","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7e720845abc26e80d9ed70f30a2ccb3b5f5c9ea0ff948948f3e8b96fd18b1ea","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153017,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[4,1],[6,1],[8,1]],\"theta\":[4,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=4, complete estimator values=[4, 6, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(4, [[4, 1/4], [6, 1/2], [8, 1/4]], 6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (4, [[4, 1/4], [6, 1/2], [8, 1/4]], 6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0018","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55031e46325e4dd4c4fb0b890aca2f94c0a82434a50a3c1678a651d0b69a4763","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153018,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[11,1],[5,1],[8,1]],\"theta\":[5,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=5, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(5, [[5, 1/4], [8, 1/2], [11, 1/4]], 8, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (5, [[5, 1/4], [8, 1/2], [11, 1/4]], 8, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0019","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a8b09be49cc54a6deb8f44993025ad75f61e3ed6616cc19ba628dab98acc47a","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153019,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[6,1],[7,1],[8,1]],\"theta\":[6,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=6, complete estimator values=[6, 7, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(6, [[6, 1/4], [7, 1/2], [8, 1/4]], 7, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (6, [[6, 1/4], [7, 1/2], [8, 1/4]], 7, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0020","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab15d63449031f1a54de50bfc6a58e5abd33b714e8c6d15701a7eaec6c930c41","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153020,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[11,1],[7,1],[9,1]],\"theta\":[7,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=7, complete estimator values=[7, 9, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(7, [[7, 1/4], [9, 1/2], [11, 1/4]], 9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (7, [[7, 1/4], [9, 1/2], [11, 1/4]], 9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0021","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27f293995131bd993bd87eff88544d1f4bc72125d2c0bfe4150f496a692dba7f","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153021,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[11,1],[14,1],[8,1]],\"theta\":[8,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=8, complete estimator values=[8, 11, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(8, [[8, 1/4], [11, 1/2], [14, 1/4]], 11, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (8, [[8, 1/4], [11, 1/2], [14, 1/4]], 11, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0022","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"674c5b4e66c1bbd78b1a5acf715177e7eb8d705c8aaee417022daae060801f8b","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153022,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[10,1],[11,1],[9,1]],\"theta\":[9,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=9, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(9, [[9, 1/4], [10, 1/2], [11, 1/4]], 10, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (9, [[9, 1/4], [10, 1/2], [11, 1/4]], 10, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0023","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c7de4c2f75d88305599a095ab8b24dffda394bfd36522ccf42e5f4d3d483e72","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153023,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[10,1],[12,1],[14,1]],\"theta\":[10,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=10, complete estimator values=[10, 12, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(10, [[10, 1/4], [12, 1/2], [14, 1/4]], 12, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (10, [[10, 1/4], [12, 1/2], [14, 1/4]], 12, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0024","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"989ba40f4415ef8b14b9b3e3dcdcb4447cbd0b63a135596b8a6a3a066478413d","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153024,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[11,1],[14,1],[17,1]],\"theta\":[11,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=11, complete estimator values=[11, 14, 17] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(11, [[11, 1/4], [14, 1/2], [17, 1/4]], 14, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (11, [[11, 1/4], [14, 1/2], [17, 1/4]], 14, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0025","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8155d9254b2f3e5e8e001d3f970f2f198a8455fa10626d85c2ec6677dffbe1d","family_id":"qf-mod46-biasvar-single-theta-identity","seed":202611153025,"semantic_template":"qf-mod46-biasvar-single-theta-identity/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[12,1],[13,1],[14,1]],\"theta\":[12,1]}"},"fields":{"title":"MSE Equals Variance plus Squared Bias under One Complete Theta Law","stem":"Use scalar theta=12, complete estimator values=[12, 13, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).","answer":{"kind":"expression","value":"(12, [[12, 1/4], [13, 1/2], [14, 1/4]], 13, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)","explanation":"Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (12, [[12, 1/4], [13, 1/2], [14, 1/4]], 13, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)."},"skills":["MSE Equals Variance plus Squared Bias under One Complete Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0026","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7082bf8d538daf82d3576453340ea92e49ec6bfec7e9c56d1835196dfecd5d54","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153026,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[-10,1],[-11,1],[-9,1]],\"theta\":[-12,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-12, complete estimator values=[-11, -10, -9] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0027","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"711417f3aea771dd425df10a2846890851b88d07799aa9b1573171ad2adce33f","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153027,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[-10,1],[-8,1],[-9,1]],\"theta\":[-11,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-11, complete estimator values=[-10, -9, -8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0028","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aac4aaca913c3adc7a3b39d8bb463756edff5d57c07cf10076d4fcbad23b6f8a","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153028,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-7,1],[-8,1],[-9,1]],\"theta\":[-10,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-10, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0029","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3b193e06a3cd31d0dfbb4c147f15dfc69a1143cdfa85fe1afbee2c0200eaf1a","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153029,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-6,1],[-7,1],[-8,1]],\"theta\":[-9,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-9, complete estimator values=[-8, -7, -6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0030","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f6efd1ec810eabdbfb63aef098c4a387fc6cec1e9257c6a4a9f84a030510d95","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153030,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-5,1],[-6,1],[-7,1]],\"theta\":[-8,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-8, complete estimator values=[-7, -6, -5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0031","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35d20d4bd9c939d6a78758a24ab4fec02f37042c60e71078c6ec8552ccef9e79","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153031,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-4,1],[-5,1],[-6,1]],\"theta\":[-7,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-7, complete estimator values=[-6, -5, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0032","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df85a0f93526a7f449330a5eb0dc4ae056606dafd64cbf32b482bfc01e062ab7","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153032,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-3,1],[-4,1],[-5,1]],\"theta\":[-6,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-6, complete estimator values=[-5, -4, -3] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0033","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c00e38f15d323ef682b388830cac56f6092a9abf7d53a84433b302910b7348a","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153033,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-2,1],[-3,1],[-4,1]],\"theta\":[-5,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-5, complete estimator values=[-4, -3, -2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0034","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9b71b83744d39aef5f4f83ad4aaeb15d3306df761761be1de40b3f5199fa575","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153034,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-2,1],[-3,1]],\"theta\":[-4,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-4, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0035","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40a617d8b9cd00168693e2ef05bf9d4f1a85b2e95d80924ef060ff8fd8d96b9c","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153035,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[-1,1],[-2,1],[0,1]],\"theta\":[-3,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-3, complete estimator values=[-2, -1, 0] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(-1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0036","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56ffb104b2b7ffc0a805f989ac0643cddcccb3e6de1c0fc4359275b073fd074a","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153036,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[0,1],[1,1]],\"theta\":[-2,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-2, complete estimator values=[-1, 0, 1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(0, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (0, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0037","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3c08f35b4f5ecfce37aff494751c4106243cad387abe5ec7d388aaf9dbc3ed7","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153037,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[0,1],[1,1],[2,1]],\"theta\":[-1,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=-1, complete estimator values=[0, 1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0038","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa6a4b57bd6ac8744ba15c191c192a8968e895181f944839a9867b20565aff2c","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153038,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[1,1],[2,1],[3,1]],\"theta\":[0,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=0, complete estimator values=[1, 2, 3] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0039","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39565c73778e04b99a5c056aab2a62f33b1d45925d357ef677c696dbd0e17091","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153039,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[2,1],[3,1],[4,1]],\"theta\":[1,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=1, complete estimator values=[2, 3, 4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0040","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3845666594780273f9c52ea85655eb036819f37b1d6934332a470e18089bc2f4","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153040,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[3,1],[4,1],[5,1]],\"theta\":[2,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=2, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0041","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e92644bfffbd532dc79a5ab5ac00953073c3231a8048972fea195499b49cffac","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153041,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[4,1],[5,1],[6,1]],\"theta\":[3,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=3, complete estimator values=[4, 5, 6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0042","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc14622c10231f002f7aefbb59294b6bc268aeb6f16452fb4ec328d48d0074f1","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153042,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[5,1],[6,1],[7,1]],\"theta\":[4,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=4, complete estimator values=[5, 6, 7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0043","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"472d37be88db21d6fae3ff8b343e156f5059d9f2d3e47c31f7dc12cd90ee07ce","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153043,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[6,1],[7,1],[8,1]],\"theta\":[5,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=5, complete estimator values=[6, 7, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0044","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dc0b853f11ab8dd865bfd02993bba96e8a462335679d678b686ce0a94871fd8","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153044,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[7,1],[8,1],[9,1]],\"theta\":[6,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=6, complete estimator values=[7, 8, 9] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0045","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8807f1d3f72a4213ff86b68b59de1639ab5dec4feca859ef4f7b797fb661539e","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153045,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[10,1],[8,1],[9,1]],\"theta\":[7,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=7, complete estimator values=[8, 9, 10] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0046","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbeb44b2b0497cf2fc0cb119b1ec4fc85018c91c7446bc3d249febfc4d02288d","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153046,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[10,1],[11,1],[9,1]],\"theta\":[8,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=8, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0047","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9043c08b95ec8c19b1c8f196df34ec1a0b28a0af607aaac86c5280b1ffaaf5f","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153047,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[10,1],[11,1],[12,1]],\"theta\":[9,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=9, complete estimator values=[10, 11, 12] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(11, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (11, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0048","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec1cd179ca62aaf0ca96ffc9c8a2ed5cdf7983ac2e1a00783c207e9323a16207","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153048,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[11,1],[12,1],[13,1]],\"theta\":[10,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=10, complete estimator values=[11, 12, 13] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(12, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (12, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0049","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7682e28c68ea09fd54d6b44d96c7cf14a8628d66d4503e9ba8e98c0e20159ffe","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153049,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[12,1],[13,1],[14,1]],\"theta\":[11,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=11, complete estimator values=[12, 13, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(13, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (13, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0050","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c14598c969819be4c0b5cbe5f289727e103eccc5cfd8566cf02e0201c56107f0","family_id":"qf-mod46-biasvar-cross-term-cancellation","seed":202611153050,"semantic_template":"qf-mod46-biasvar-cross-term-cancellation/{\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[13,1],[14,1],[15,1]],\"theta\":[12,1]}"},"fields":{"title":"The Cross Term Vanishes Because the Centered Estimator Has Mean Zero","stem":"Use scalar theta=12, complete estimator values=[13, 14, 15] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).","answer":{"kind":"expression","value":"(14, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)","explanation":"Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (14, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)."},"skills":["The Cross Term Vanishes Because the Centered Estimator Has Mean Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0051","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79d596525fdc4b19eee18029e6a39cc06834aafdb4a745594c36545f3cf171c8","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153051,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-11,1],[[-10,1],[-11,1],[-12,1]],[[1,4],[1,2],[1,4]]],[[-12,1],[[-10,1],[-11,1],[-12,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-12, [-12, -11, -10], [[1, 4], [1, 2], [1, 4]]], [-11, [-12, -11, -10], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-11, -11, 0, 1/2, 1/2, 1/2, Yes], [-12, -11, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-11, -11, 0, 1/2, 1/2, 1/2, Yes], [-12, -11, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0052","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b247ca992e60161d0c6c6a25b8d0b7da5b84f448efe57c472518722171ec34f0","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153052,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-10,1],[[-10,1],[-11,1],[-9,1]],[[1,2],[1,4],[1,4]]],[[-11,1],[[-11,1],[-7,1],[-9,1]],[[1,4],[1,4],[1,2]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-11, [-11, -9, -7], [[1, 4], [1, 2], [1, 4]]], [-10, [-11, -10, -9], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-10, -10, 0, 1/2, 1/2, 1/2, Yes], [-11, -9, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-10, -10, 0, 1/2, 1/2, 1/2, Yes], [-11, -9, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0053","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"352c59d865d457a06107215f4875dabb8c7561340f25759ba263462576d1b183","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153053,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-10,1],[[-10,1],[-4,1],[-7,1]],[[1,4],[1,4],[1,2]]],[[-9,1],[[-10,1],[-8,1],[-9,1]],[[1,4],[1,4],[1,2]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-10, [-10, -7, -4], [[1, 4], [1, 2], [1, 4]]], [-9, [-10, -9, -8], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-10, -7, 3, 9/2, 27/2, 27/2, Yes], [-9, -9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-10, -7, 3, 9/2, 27/2, 27/2, Yes], [-9, -9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0054","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f00ec7c78e5fc5704aabec16696b0687ed2f472df838b4710f9a6ad42feeb358","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153054,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-8,1],[[-7,1],[-8,1],[-9,1]],[[1,4],[1,2],[1,4]]],[[-9,1],[[-7,1],[-8,1],[-9,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-9, [-9, -8, -7], [[1, 4], [1, 2], [1, 4]]], [-8, [-9, -8, -7], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-8, -8, 0, 1/2, 1/2, 1/2, Yes], [-9, -8, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-8, -8, 0, 1/2, 1/2, 1/2, Yes], [-9, -8, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0055","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54517456c4996b73bf69c7c10ba9c5ed95fd03ff6ec815ca4f1799bd9232e9a6","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153055,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-7,1],[[-6,1],[-7,1],[-8,1]],[[1,4],[1,2],[1,4]]],[[-8,1],[[-4,1],[-6,1],[-8,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-8, [-8, -6, -4], [[1, 4], [1, 2], [1, 4]]], [-7, [-8, -7, -6], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-7, -7, 0, 1/2, 1/2, 1/2, Yes], [-8, -6, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-7, -7, 0, 1/2, 1/2, 1/2, Yes], [-8, -6, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0056","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b57fd8081280c2fe4bb4e6bf9d1df556ce990713c6d60325d26df816b9874aa4","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153056,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-6,1],[[-5,1],[-6,1],[-7,1]],[[1,4],[1,2],[1,4]]],[[-7,1],[[-1,1],[-4,1],[-7,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-7, [-7, -4, -1], [[1, 4], [1, 2], [1, 4]]], [-6, [-7, -6, -5], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-6, -6, 0, 1/2, 1/2, 1/2, Yes], [-7, -4, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-6, -6, 0, 1/2, 1/2, 1/2, Yes], [-7, -4, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0057","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e25e84697d78e6bbcb22a40018687014abad524fa70cc9e54376cecf91cab2c","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153057,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-5,1],[[-4,1],[-5,1],[-6,1]],[[1,4],[1,2],[1,4]]],[[-6,1],[[-4,1],[-5,1],[-6,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-6, [-6, -5, -4], [[1, 4], [1, 2], [1, 4]]], [-5, [-6, -5, -4], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-5, -5, 0, 1/2, 1/2, 1/2, Yes], [-6, -5, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-5, -5, 0, 1/2, 1/2, 1/2, Yes], [-6, -5, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0058","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53a35ea0d856db062b03c0a42a4e7f6fa8dcaedca877a52defb204b62abefa99","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153058,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-4,1],[[-3,1],[-4,1],[-5,1]],[[1,4],[1,2],[1,4]]],[[-5,1],[[-1,1],[-3,1],[-5,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-5, [-5, -3, -1], [[1, 4], [1, 2], [1, 4]]], [-4, [-5, -4, -3], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-4, -4, 0, 1/2, 1/2, 1/2, Yes], [-5, -3, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-4, -4, 0, 1/2, 1/2, 1/2, Yes], [-5, -3, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0059","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70b414628fd55b84c82bc64db81beadcf8f46fb84c6c3de2b8e7a4e248538182","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153059,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-3,1],[[-2,1],[-3,1],[-4,1]],[[1,4],[1,2],[1,4]]],[[-4,1],[[-1,1],[-4,1],[2,1]],[[1,2],[1,4],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-4, [-4, -1, 2], [[1, 4], [1, 2], [1, 4]]], [-3, [-4, -3, -2], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-3, -3, 0, 1/2, 1/2, 1/2, Yes], [-4, -1, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-3, -3, 0, 1/2, 1/2, 1/2, Yes], [-4, -1, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0060","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a39e66748c16af56e0b3db0c6a5edcc2e9f7150675669716c668828f26493f0","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153060,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-2,1],[[-1,1],[-2,1],[-3,1]],[[1,4],[1,2],[1,4]]],[[-3,1],[[-1,1],[-2,1],[-3,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-3, [-3, -2, -1], [[1, 4], [1, 2], [1, 4]]], [-2, [-3, -2, -1], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-2, -2, 0, 1/2, 1/2, 1/2, Yes], [-3, -2, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-2, -2, 0, 1/2, 1/2, 1/2, Yes], [-3, -2, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0061","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8a838cf4c373facd965293f00893576acd06394eeec70e43ddbce063df6d653","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153061,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-1,1],[[-1,1],[-2,1],[0,1]],[[1,2],[1,4],[1,4]]],[[-2,1],[[-2,1],[0,1],[2,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-2, [-2, 0, 2], [[1, 4], [1, 2], [1, 4]]], [-1, [-2, -1, 0], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-1, -1, 0, 1/2, 1/2, 1/2, Yes], [-2, 0, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-1, -1, 0, 1/2, 1/2, 1/2, Yes], [-2, 0, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0062","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4148a43907a32420483c74886685e874553ad89b6799bb9ffbb1e951196942d","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153062,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[-1,1],[[-1,1],[2,1],[5,1]],[[1,4],[1,2],[1,4]]],[[0,1],[[-1,1],[0,1],[1,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[-1, [-1, 2, 5], [[1, 4], [1, 2], [1, 4]]], [0, [-1, 0, 1], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[-1, 2, 3, 9/2, 27/2, 27/2, Yes], [0, 0, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-1, 2, 3, 9/2, 27/2, 27/2, Yes], [0, 0, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0063","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0843a974c3f04fb52cdb90a62e061f6a336c82b7f6e69e5249292615a3f027a","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153063,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[0,1],[[0,1],[1,1],[2,1]],[[1,4],[1,2],[1,4]]],[[1,1],[[0,1],[1,1],[2,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[0, [0, 1, 2], [[1, 4], [1, 2], [1, 4]]], [1, [0, 1, 2], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[0, 1, 1, 1/2, 3/2, 3/2, Yes], [1, 1, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[0, 1, 1, 1/2, 3/2, 3/2, Yes], [1, 1, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0064","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20d2d9b656c0c1849279e2c1e1bb755b7c65d19193d51010cf6f69421f1ede26","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153064,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[1,1],[[1,1],[3,1],[5,1]],[[1,4],[1,2],[1,4]]],[[2,1],[[1,1],[2,1],[3,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[1, [1, 3, 5], [[1, 4], [1, 2], [1, 4]]], [2, [1, 2, 3], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[1, 3, 2, 2, 6, 6, Yes], [2, 2, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[1, 3, 2, 2, 6, 6, Yes], [2, 2, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0065","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d1d069b9b7975bf41798d0fbd5c68f34f6a241b811138763860f2fa954d7dfc","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153065,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[2,1],[[2,1],[5,1],[8,1]],[[1,4],[1,2],[1,4]]],[[3,1],[[2,1],[3,1],[4,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[2, [2, 5, 8], [[1, 4], [1, 2], [1, 4]]], [3, [2, 3, 4], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[2, 5, 3, 9/2, 27/2, 27/2, Yes], [3, 3, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[2, 5, 3, 9/2, 27/2, 27/2, Yes], [3, 3, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0066","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"889f5867e77f67ad9e251a87521573ba020a11db630bc764e4a878d2f8ddadd7","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153066,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[3,1],[[3,1],[4,1],[5,1]],[[1,4],[1,2],[1,4]]],[[4,1],[[3,1],[4,1],[5,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[3, [3, 4, 5], [[1, 4], [1, 2], [1, 4]]], [4, [3, 4, 5], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[3, 4, 1, 1/2, 3/2, 3/2, Yes], [4, 4, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[3, 4, 1, 1/2, 3/2, 3/2, Yes], [4, 4, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0067","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d461e3e809673f17d7c3d0032b90690e389ae2fcf11ece3bd9f3cd66fd0cda05","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153067,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[4,1],[[4,1],[6,1],[8,1]],[[1,4],[1,2],[1,4]]],[[5,1],[[4,1],[5,1],[6,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[4, [4, 6, 8], [[1, 4], [1, 2], [1, 4]]], [5, [4, 5, 6], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[4, 6, 2, 2, 6, 6, Yes], [5, 5, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[4, 6, 2, 2, 6, 6, Yes], [5, 5, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0068","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a61c7cfdecba9a7277ed9bb6251a6b1447c1b18f2db0a970f05af2761f5501f2","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153068,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[5,1],[[11,1],[5,1],[8,1]],[[1,4],[1,4],[1,2]]],[[6,1],[[5,1],[6,1],[7,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[5, [5, 8, 11], [[1, 4], [1, 2], [1, 4]]], [6, [5, 6, 7], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[5, 8, 3, 9/2, 27/2, 27/2, Yes], [6, 6, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[5, 8, 3, 9/2, 27/2, 27/2, Yes], [6, 6, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0069","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77ab29522a1871da266afe861c46ca8431d7412ac03acf67e37763a1b62bd139","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153069,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[6,1],[[6,1],[7,1],[8,1]],[[1,4],[1,2],[1,4]]],[[7,1],[[6,1],[7,1],[8,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[6, [6, 7, 8], [[1, 4], [1, 2], [1, 4]]], [7, [6, 7, 8], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[6, 7, 1, 1/2, 3/2, 3/2, Yes], [7, 7, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[6, 7, 1, 1/2, 3/2, 3/2, Yes], [7, 7, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0070","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f53bda9cdf9c9ebc44df53b4d922b1790525dbf9b2054bb42ebf98d23bedf1e","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153070,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[7,1],[[11,1],[7,1],[9,1]],[[1,4],[1,4],[1,2]]],[[8,1],[[7,1],[8,1],[9,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[7, [7, 9, 11], [[1, 4], [1, 2], [1, 4]]], [8, [7, 8, 9], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[7, 9, 2, 2, 6, 6, Yes], [8, 8, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[7, 9, 2, 2, 6, 6, Yes], [8, 8, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0071","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce73fd5f0b345fc1130915c2487cfd5a5da7af9428bc58bf799597a62394d557","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153071,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[8,1],[[11,1],[14,1],[8,1]],[[1,2],[1,4],[1,4]]],[[9,1],[[10,1],[8,1],[9,1]],[[1,4],[1,4],[1,2]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[8, [8, 11, 14], [[1, 4], [1, 2], [1, 4]]], [9, [8, 9, 10], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[8, 11, 3, 9/2, 27/2, 27/2, Yes], [9, 9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[8, 11, 3, 9/2, 27/2, 27/2, Yes], [9, 9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0072","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5be344bb0577c72f68ad0c2fb25c958de03e8b8690dc33061e72d4e0f395c43","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153072,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[10,1],[[10,1],[11,1],[9,1]],[[1,2],[1,4],[1,4]]],[[9,1],[[10,1],[11,1],[9,1]],[[1,2],[1,4],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[9, [9, 10, 11], [[1, 4], [1, 2], [1, 4]]], [10, [9, 10, 11], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[10, 10, 0, 1/2, 1/2, 1/2, Yes], [9, 10, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[10, 10, 0, 1/2, 1/2, 1/2, Yes], [9, 10, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0073","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e744ad683ed911b9062c9f8fe5f4bf302e92133afb5f441b4613c0c152d1edd8","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153073,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[10,1],[[10,1],[12,1],[14,1]],[[1,4],[1,2],[1,4]]],[[11,1],[[10,1],[11,1],[12,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[10, [10, 12, 14], [[1, 4], [1, 2], [1, 4]]], [11, [10, 11, 12], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[10, 12, 2, 2, 6, 6, Yes], [11, 11, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[10, 12, 2, 2, 6, 6, Yes], [11, 11, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0074","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34c103bf133557e7d31d2a2721585cb6c36b061eab342a79650b8ca67675d4bf","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153074,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[11,1],[[11,1],[14,1],[17,1]],[[1,4],[1,2],[1,4]]],[[12,1],[[11,1],[12,1],[13,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[11, [11, 14, 17], [[1, 4], [1, 2], [1, 4]]], [12, [11, 12, 13], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[11, 14, 3, 9/2, 27/2, 27/2, Yes], [12, 12, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[11, 14, 3, 9/2, 27/2, 27/2, Yes], [12, 12, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0075","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"479a1e7f642008414e868ba26503de574a64d17260e57f36af2f2de55f594a53","family_id":"qf-mod46-biasvar-multi-theta-table","seed":202611153075,"semantic_template":"qf-mod46-biasvar-multi-theta-table/{\"laws\":[[[12,1],[[12,1],[13,1],[14,1]],[[1,4],[1,2],[1,4]]],[[13,1],[[12,1],[13,1],[14,1]],[[1,4],[1,2],[1,4]]]]}"},"fields":{"title":"Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law","stem":"Use complete laws=[[12, [12, 13, 14], [[1, 4], [1, 2], [1, 4]]], [13, [12, 13, 14], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).","answer":{"kind":"expression","value":"([[12, 13, 1, 1/2, 3/2, 3/2, Yes], [13, 13, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)","explanation":"For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[12, 13, 1, 1/2, 3/2, 3/2, Yes], [13, 13, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)."},"skills":["Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0076","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13b2ffac0b05d0c30e1122b368776de8fe1f9aa8af63de6a65ecad535197576c","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153076,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-13,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-11,1],[-12,1],[-13,1]],\"theta\":[-12,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-12, complete estimator values=[-13, -12, -11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-13. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-12, -13, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-12, -13, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0077","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"492dcc53f17930a338a2cda0bf593399a22ad1ec8798e26ca9d778009858a3ba","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153077,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-11,1],\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[-11,1],[-7,1],[-9,1]],\"theta\":[-11,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-11, complete estimator values=[-11, -9, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-11, -11, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-11, -11, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0078","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0824d5f132340a974d0299127ef63eacf3d0a2d9e11c0c69a393a08b8d1ab9cd","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153078,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-13,1],\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[-10,1],[-13,1],[-7,1]],\"theta\":[-10,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-10, complete estimator values=[-13, -10, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-13. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-10, -13, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-10, -13, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0079","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fca85cf9a65027269a7bad25d210937aedea5e5aff7d8366085dec1f5e73fd15","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153079,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-9,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-7,1],[-8,1],[-9,1]],\"theta\":[-9,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-9, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-9. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-9, -9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-9, -9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0080","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3333f80d92309b31653e4d3916206bb5e8eb4e4ca4ea24999b37ab458043c6c","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153080,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-10,1],\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[-10,1],[-6,1],[-8,1]],\"theta\":[-8,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-8, complete estimator values=[-10, -8, -6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-10. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-8, -10, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-8, -10, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0081","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dbb393f07f9db50dbf18351891c8782322d1d323b734064f51cf35a21a0753e2","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153081,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-7,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-4,1],[-7,1]],\"theta\":[-7,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-7, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-7, -7, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-7, -7, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0082","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b37596de7be0ab246841017332e4417b0642350dcbff8fbaab62133d5778739a","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153082,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-7,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-5,1],[-6,1],[-7,1]],\"theta\":[-6,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-6, complete estimator values=[-7, -6, -5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-6, -7, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-6, -7, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0083","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25ae72d642b753f01176c1ec6d2bae9b5a81b6e7a7f4e5d68bad12fc60594844","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153083,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-5,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-3,1],[-5,1]],\"theta\":[-5,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-5, complete estimator values=[-5, -3, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-5, -5, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-5, -5, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0084","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"861b82953c96717817e175d180501345f01c557b9bcce31c62deaa54df4cd88f","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153084,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-7,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-4,1],[-7,1]],\"theta\":[-4,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-4, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-4, -7, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-4, -7, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0085","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b716ed76187c0ee51e0421f3971ab34f3e44c426272dcecf413d868911c95eb0","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153085,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-3,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[-2,1],[-3,1]],\"theta\":[-3,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-3, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-3. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-3, -3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-3, -3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0086","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b939aa6f6d10fb43c6f7ee040187b5218e62939f0439fff4db7da21d6de33ab","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153086,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-4,1],\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[-2,1],[-4,1],[0,1]],\"theta\":[-2,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-2, complete estimator values=[-4, -2, 0] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-4. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-2, -4, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-2, -4, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0087","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b930b226c9a1cf9426ce6a4c336419e0e497a56db064db6ff53e25d76d219b9","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153087,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-1,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[2,1],[5,1]],\"theta\":[-1,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=-1, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(-1, -1, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-1, -1, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0088","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98f84e47a65375aaf90b84f47ee9a17819679dcd04bfac21a45bc2d496280b61","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153088,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-1,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[0,1],[1,1]],\"theta\":[0,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=0, complete estimator values=[-1, 0, 1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(0, -1, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (0, -1, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0089","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f721037ca4e17611ee4cc3ca8fe73053b9b45a581fd04bce1dde9f3876de90f","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153089,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[1,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[1,1],[3,1],[5,1]],\"theta\":[1,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=1, complete estimator values=[1, 3, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(1, 1, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (1, 1, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0090","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"07af16b853974b858140abe0389802b7a5b5293d3126cd407814f0ef7cefc426","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153090,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[-1,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[-1,1],[2,1],[5,1]],\"theta\":[2,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=2, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(2, -1, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (2, -1, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0091","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03205425826542fe517ea78d379ee8e90c936312416619d9642a8097d08e432c","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153091,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[3,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[3,1],[4,1],[5,1]],\"theta\":[3,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=3, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=3. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(3, 3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (3, 3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0092","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5839ef449bdac1f89e99ddc944aae5dca128a6c754e879da35ec6aed84d02f6f","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153092,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[2,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[2,1],[4,1],[6,1]],\"theta\":[4,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=4, complete estimator values=[2, 4, 6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=2. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(4, 2, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (4, 2, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0093","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cb73f5ac54a188af237823edb65f29829fbf21a3fbd885038d3121ca3e1968a","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153093,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[5,1],\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[11,1],[5,1],[8,1]],\"theta\":[5,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=5, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(5, 5, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (5, 5, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0094","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a93b58db9904a54b91a83df3407952cffb3885f0d1db3679430dc1cd45254f19","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153094,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[5,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[5,1],[6,1],[7,1]],\"theta\":[6,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=6, complete estimator values=[5, 6, 7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(6, 5, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (6, 5, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0095","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdf1696e85c22c7a49a110c287112401164f2c0c5ce74ce847bb2fb9a5a35d86","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153095,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[7,1],\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[11,1],[7,1],[9,1]],\"theta\":[7,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=7, complete estimator values=[7, 9, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(7, 7, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (7, 7, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0096","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f9881f97b8b35d0392e706d1dd7160c99bd4f2f58866cd4da1bde0dffd36d01","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153096,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[5,1],\"probabilities\":[[1,4],[1,4],[1,2]],\"support\":[[11,1],[5,1],[8,1]],\"theta\":[8,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=8, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(8, 5, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (8, 5, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0097","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13ad10e382a5b88ed57b154ed9e2f3044525471e6ed136af6404a2f8c4f1e4c8","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153097,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[9,1],\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[10,1],[11,1],[9,1]],\"theta\":[9,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=9, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=9. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(9, 9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (9, 9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0098","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff274ec3bc988ca387258c77793f6924083de9838bc0f492be68617a8820653c","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153098,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[8,1],\"probabilities\":[[1,2],[1,4],[1,4]],\"support\":[[10,1],[12,1],[8,1]],\"theta\":[10,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=10, complete estimator values=[8, 10, 12] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=8. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(10, 8, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (10, 8, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0099","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b353b3decedcd7c36d6dbd366c2ca7270732975ca8404d238edac7de0a6e8738","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153099,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[11,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[11,1],[14,1],[17,1]],\"theta\":[11,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=11, complete estimator values=[11, 14, 17] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(11, 11, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (11, 11, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261115-r03-bias-variance-decomposition-0100","batch_id":"batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14acf1f074a721dcfe4e65fa6de21ee50976c5b811e4fbb60e228365adffe51b","family_id":"qf-mod46-biasvar-biased-unbiased-boundary","seed":202611153100,"semantic_template":"qf-mod46-biasvar-biased-unbiased-boundary/{\"observed\":[11,1],\"probabilities\":[[1,4],[1,2],[1,4]],\"support\":[[11,1],[12,1],[13,1]],\"theta\":[12,1]}"},"fields":{"title":"Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency","stem":"Use scalar theta=12, complete estimator values=[11, 12, 13] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).","answer":{"kind":"expression","value":"(12, 11, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)","explanation":"Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (12, 11, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)."},"skills":["Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}