{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261031-r07-martingale-0001","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3257f2c0f579304ce0304be5cd5820bf59e1a30de4db30127e84ef458d7fd587","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317001,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":1}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=1>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [1, -1], [-1, 1], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [1, -1], [-1, 1], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0002","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"18863d5e52c21392572b5f287412bff171b086ab4f5c04d3722ff16c54a3125e","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317002,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":2}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=2>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [2, -2], [-2, 2], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [2, -2], [-2, 2], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0003","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"272dbb966a536b22552c74f03e654485a547a8e9a2f34e1deac5e35abd90e950","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317003,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":3}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=3>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [3, -3], [-3, 3], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [3, -3], [-3, 3], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0004","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a15738d3bf68e7b9cede2e8ebf91f2bf5c165b0485a9a89fe8aed8b45e5d10f9","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317004,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":4}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=4>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [4, -4], [-4, 4], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [4, -4], [-4, 4], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0005","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"724e49dfa2d6432b7537eb6d2578bb7a81eb6c33e25a7e57bd041c6c2af772e4","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317005,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":5}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=5>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [5, -5], [-5, 5], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [5, -5], [-5, 5], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0006","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6901019f01079ae81b4b8036d604a8dedc5d56dce600f400e1eb9aa9b8cf1fdd","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317006,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":6}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=6>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [6, -6], [-6, 6], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [6, -6], [-6, 6], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0007","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1c1b492f65af20afcf4a1e02b251584ec948504aa5b00a2923257539fd2c5f0","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317007,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":7}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=7>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [7, -7], [-7, 7], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [7, -7], [-7, 7], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0008","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54fc0828af094649b027ba44df0c15a956227a04e3bce21421853a46c32dd82b","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317008,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":8}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=8>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [8, -8], [-8, 8], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [8, -8], [-8, 8], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0009","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b358204dc6eb3e3ffe3c4d695f66b8e72dbad67aaa30a1ad08c02b6af3587770","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317009,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":9}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=9>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [9, -9], [-9, 9], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [9, -9], [-9, 9], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0010","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d55d496f381f06157ea565b260b4da31391c6a0be4b6d5d95c63a90d81cda024","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317010,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":10}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=10>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [10, -10], [-10, 10], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [10, -10], [-10, 10], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0011","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b72d4e1859b72fc5accc1817c23ede4abd950dc154122443c2bb5f3ccc451362","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317011,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":11}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=11>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [11, -11], [-11, 11], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [11, -11], [-11, 11], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0012","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b5594a60f3a9152b3e8cd508e00d6091ec35306d5e9621ceedc86085a14fb39","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317012,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":12}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=12>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [12, -12], [-12, 12], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [12, -12], [-12, 12], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0013","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"456fb3580bc3963a65abd40217f337283115b5ec10458374c2f1b66ebee2a6d4","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317013,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":13}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=13>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [13, -13], [-13, 13], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [13, -13], [-13, 13], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0014","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"600bf7532ee12bd4fc5dfb0088e8bfebbe7de5822d6014e3413f92e3e218de94","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317014,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":14}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=14>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [14, -14], [-14, 14], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [14, -14], [-14, 14], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0015","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ab697abd799b28c6bf6ef4c8b7ff20ff08ae87d49ee4937979b1346e48db7f2","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317015,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":15}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=15>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [15, -15], [-15, 15], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [15, -15], [-15, 15], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0016","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"775322bf9a8790757bbb23c01100f0eb008394dad987e0b842e712e24feca049","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317016,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":16}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=16>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [16, -16], [-16, 16], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [16, -16], [-16, 16], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0017","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e70b751b1ee9689fe771b8dec0089458df16fd1509477f3fbcb8f41cc598a0a5","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317017,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":17}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=17>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [17, -17], [-17, 17], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [17, -17], [-17, 17], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0018","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6321aa323bcc55a6991d02cd23deb8c73573c1004fcdba277ffba476f05be186","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317018,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":18}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=18>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [18, -18], [-18, 18], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [18, -18], [-18, 18], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0019","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e8bef52cf889602b8cbefbfd4d727d172b671c68033296314ecf28fcc51f3dd","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317019,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":19}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=19>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [19, -19], [-19, 19], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [19, -19], [-19, 19], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0020","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acb9d53455d0267a1df57d202b45310896b885d5aa582c1eae3d55652e726e20","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317020,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":20}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=20>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [20, -20], [-20, 20], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [20, -20], [-20, 20], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0021","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04d91f472dc1681576c80b50b429216ca2d7e7bcea97c1f33f3314684ade7a7c","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317021,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":21}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=21>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [21, -21], [-21, 21], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [21, -21], [-21, 21], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0022","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99c75288670186f02951c3a7e87bcb4290a6c50e67744ea76c36a2ea8a8b9a96","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317022,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":22}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=22>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [22, -22], [-22, 22], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [22, -22], [-22, 22], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0023","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ffc1c42572d4b78cfc88d9526f9498e1d2b2b1410fd7c89cc0c81f73aa60de1","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317023,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":23}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=23>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [23, -23], [-23, 23], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [23, -23], [-23, 23], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0024","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b44d7bef47868705d8deb5207432741e23403397aa5ec443f167b352011f4b7c","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317024,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":24}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=24>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [24, -24], [-24, 24], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [24, -24], [-24, 24], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0025","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a63ecfd02304e710d6777edc5a3ce711ea338e8e92d5b956324b358328166548","family_id":"qf-mod31-martdeep-constant-expectation-counterexample","seed":202610317025,"semantic_template":"qf-mod31-martdeep-constant-expectation-counterexample/{\"amplitude\":25}"},"fields":{"title":"Constant Total Expectations Can Fail Conditional Martingale Steps","stem":"Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=25>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).","answer":{"kind":"expression","value":"(0, [25, -25], [-25, 25], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)","explanation":"All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [25, -25], [-25, 25], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)."},"skills":["Constant Total Expectations Can Fail Conditional Martingale Steps"],"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-20261014-advance-20261031-r07-martingale-0026","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"929516810743c691b155983df457117134699c3635f0c98e371e48905aae9189","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317026,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[1,-1,2,-2]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[1, -1, 2, -2] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [1, -1], 0], [1, [2, -2], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [1, -1], 0], [1, [2, -2], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0027","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0048de19ad9c4855aa190a08761e0cd5632db5aeec5aba52cdfc23d3f4f69e5","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317027,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[2,-2,3,-3]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[2, -2, 3, -3] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [2, -2], 0], [1, [3, -3], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [2, -2], 0], [1, [3, -3], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0028","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cda53374d004053e4f06ad5a83d25f4998b221bf889f7ff650edd3b812ebf249","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317028,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[3,-3,4,-4]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[3, -3, 4, -4] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [3, -3], 0], [1, [4, -4], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [3, -3], 0], [1, [4, -4], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0029","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bd8fcfd4b4aaf23f31f2cb1c7634a2f325a05ed98ee797f06a38393d3750d47","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317029,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[4,-4,5,-5]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[4, -4, 5, -5] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [4, -4], 0], [1, [5, -5], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [4, -4], 0], [1, [5, -5], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0030","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36d79f02e7659a6c27ba718cb2d0d74890e1848e0b0f8efedf7c03cbd350a36a","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317030,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[5,-5,6,-6]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[5, -5, 6, -6] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [5, -5], 0], [1, [6, -6], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [5, -5], 0], [1, [6, -6], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0031","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b2a7e63eb0a578a2403b3a7bfe16392d4331945396df6c3b187ef02a7057266","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317031,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[6,-6,7,-7]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[6, -6, 7, -7] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [6, -6], 0], [1, [7, -7], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [6, -6], 0], [1, [7, -7], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0032","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96fd3119912927803fdf69c4aa570760bbd764259aec09e5f48f1bae0c255e69","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317032,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[7,-7,8,-8]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[7, -7, 8, -8] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [7, -7], 0], [1, [8, -8], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [7, -7], 0], [1, [8, -8], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0033","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94661bea3c257fc168e4c8a877f9005d874d35b7ee0ae81083ee2bfa3c71ac4c","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317033,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[8,-8,9,-9]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[8, -8, 9, -9] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [8, -8], 0], [1, [9, -9], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [8, -8], 0], [1, [9, -9], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0034","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3775ba77a0120cc7fcfe3fc76bfb654b342a9f2ece309b46c5d376d530a46f17","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317034,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[9,-9,10,-10]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[9, -9, 10, -10] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [9, -9], 0], [1, [10, -10], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [9, -9], 0], [1, [10, -10], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0035","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f4416bce9e8e32036e983e58fc8c5e58be00aa2ebd6b6113409264a4683a8fa","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317035,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[10,-10,11,-11]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[10, -10, 11, -11] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [10, -10], 0], [1, [11, -11], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [10, -10], 0], [1, [11, -11], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0036","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6382d01b23147f4b0d5c2324ae3f134566a3381549e97838df367cd051f1e46","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317036,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[11,-11,12,-12]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[11, -11, 12, -12] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [11, -11], 0], [1, [12, -12], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [11, -11], 0], [1, [12, -12], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0037","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce1e07bcf6067144633f69650663398bbd8063dd5b916c205796fc45e0da270f","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317037,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[12,-12,13,-13]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[12, -12, 13, -13] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [12, -12], 0], [1, [13, -13], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [12, -12], 0], [1, [13, -13], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0038","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13c93ae083db764771be12c5ab7904581597c172b126ceba8a63eaebfba6fcd4","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317038,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[13,-13,14,-14]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[13, -13, 14, -14] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [13, -13], 0], [1, [14, -14], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [13, -13], 0], [1, [14, -14], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0039","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d871569b9304c3fcffe1baad5e7ec9bcd6afff7ef528ad1254a8fd5f179885e3","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317039,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[14,-14,15,-15]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[14, -14, 15, -15] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [14, -14], 0], [1, [15, -15], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [14, -14], 0], [1, [15, -15], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0040","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1143b980762b44f18233e1b0cb8773dc52085390a152a89ee9c5fb0285e26420","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317040,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[15,-15,16,-16]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[15, -15, 16, -16] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [15, -15], 0], [1, [16, -16], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [15, -15], 0], [1, [16, -16], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0041","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b84498ad80a814fd17d31d9271d70f8dc724dcd1c4487d321687fd8b3f931012","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317041,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[16,-16,17,-17]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[16, -16, 17, -17] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [16, -16], 0], [1, [17, -17], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [16, -16], 0], [1, [17, -17], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0042","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01fcfd61b952f6c6f78db417aad89f8b15eb0ff42651edea82c60fc82f1479f8","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317042,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[17,-17,18,-18]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[17, -17, 18, -18] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [17, -17], 0], [1, [18, -18], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [17, -17], 0], [1, [18, -18], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0043","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"330f10f1f9e53a6f97e44f54d04c718daf04c9550e5d8901059b3d1bc39c0336","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317043,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[18,-18,19,-19]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[18, -18, 19, -19] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [18, -18], 0], [1, [19, -19], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [18, -18], 0], [1, [19, -19], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0044","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4741159aa26bac8f8a0307accc92ac7d7436349462c8bd4bfa5524df4b64b69a","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317044,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[19,-19,20,-20]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[19, -19, 20, -20] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [19, -19], 0], [1, [20, -20], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [19, -19], 0], [1, [20, -20], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0045","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86b3e703e2d7530e49e6714fa5fda6e1aca0b1f4aaa3f5255d19073494ee53fe","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317045,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[20,-20,21,-21]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[20, -20, 21, -21] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [20, -20], 0], [1, [21, -21], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [20, -20], 0], [1, [21, -21], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0046","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7f4d5fb040d5b6327eb5e4c64b5ce13b4df2d41e4341351d9c727530a02a584","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317046,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[21,-21,22,-22]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[21, -21, 22, -22] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [21, -21], 0], [1, [22, -22], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [21, -21], 0], [1, [22, -22], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0047","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3853fe419f942a35213081f94bd4f1ff82a83752ed3fd4a7a30938fd1a18eca0","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317047,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[22,-22,23,-23]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[22, -22, 23, -23] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [22, -22], 0], [1, [23, -23], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [22, -22], 0], [1, [23, -23], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0048","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2de04abda63dd1a5d9e8990b5a9fd30e476bdf312e8f247b0f9187d2f1b79e62","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317048,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[23,-23,24,-24]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[23, -23, 24, -24] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [23, -23], 0], [1, [24, -24], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [23, -23], 0], [1, [24, -24], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0049","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45a71611e82e0be38ed752241d61e0a35216cf3540fa67b0a43919c96462b8c5","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317049,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[24,-24,25,-25]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[24, -24, 25, -25] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [24, -24], 0], [1, [25, -25], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [24, -24], 0], [1, [25, -25], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0050","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce39dfc10588f85e55f7c917f53191c24e13529d4ed0910c8b9b72859e427ebd","family_id":"qf-mod31-martdeep-difference-mean","seed":202610317050,"semantic_template":"qf-mod31-martdeep-difference-mean/{\"blocks\":[0,0,1,1],\"increments\":[25,-25,26,-26]}"},"fields":{"title":"Zero Total Mean Is Not Enough: Check Each Information Block","stem":"Use four equiprobable outcomes with aligned increments=[25, -25, 26, -26] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.","answer":{"kind":"expression","value":"([[0, [25, -25], 0], [1, [26, -26], 0]], Yes, conditional_mean_zero_difference)","explanation":"Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [25, -25], 0], [1, [26, -26], 0]], Yes, conditional_mean_zero_difference)."},"skills":["Zero Total Mean Is Not Enough: Check Each Information Block"],"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-20261014-advance-20261031-r07-martingale-0051","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"148fda23ad19b1d77f1ebd4e090f3220be9ef04f9d2c35fde00bed9b132058e7","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317051,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-12,\"scale\":-2,\"shift\":3,\"terminal\":[-13,-11,-12,-12]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-12 and complete terminal values=[-13, -11, -12, -12]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=3. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-12, [-13, -11, -12, -12], -2, 3, 27, [29, 25, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-12, [-13, -11, -12, -12], -2, 3, 27, [29, 25, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0052","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58577405eac87e394d99591370e2897ba3ec74a5a1d3be7fa7f12332351d1add","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317052,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-11,\"scale\":-1,\"shift\":4,\"terminal\":[-12,-10,-11,-11]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-11 and complete terminal values=[-12, -10, -11, -11]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=4. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-11, [-12, -10, -11, -11], -1, 4, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-11, [-12, -10, -11, -11], -1, 4, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0053","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"308d2a2e2cbdb58113392123c97d9578344d00b7550d041df4c626343c2dbf8c","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317053,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-10,\"scale\":0,\"shift\":5,\"terminal\":[-11,-9,-10,-10]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-10 and complete terminal values=[-11, -9, -10, -10]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=5. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-10, [-11, -9, -10, -10], 0, 5, 5, [5, 5, 5, 5], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-10, [-11, -9, -10, -10], 0, 5, 5, [5, 5, 5, 5], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0054","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b03d1ed5d3fb4ba7dc8c489f029363b7f3ed0ab7aecc75f5b0e59070ec0cdd92","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317054,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-9,\"scale\":1,\"shift\":6,\"terminal\":[-10,-8,-9,-9]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-9 and complete terminal values=[-10, -8, -9, -9]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=6. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-9, [-10, -8, -9, -9], 1, 6, -3, [-4, -2, -3, -3], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-9, [-10, -8, -9, -9], 1, 6, -3, [-4, -2, -3, -3], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0055","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72270b3cd8167b9cafa234f07e7861c9144f21ec937869ae69ee4fc4d60576da","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317055,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-8,\"scale\":2,\"shift\":7,\"terminal\":[-9,-7,-8,-8]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-8 and complete terminal values=[-9, -7, -8, -8]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=7. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-8, [-9, -7, -8, -8], 2, 7, -9, [-11, -7, -9, -9], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-8, [-9, -7, -8, -8], 2, 7, -9, [-11, -7, -9, -9], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0056","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fadd3b650b0776baf5692c858c5e739b709403a4fe70b89691d812e2f33fea4","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317056,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-7,\"scale\":-2,\"shift\":8,\"terminal\":[-8,-6,-7,-7]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-7 and complete terminal values=[-8, -6, -7, -7]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=8. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-7, [-8, -6, -7, -7], -2, 8, 22, [24, 20, 22, 22], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-7, [-8, -6, -7, -7], -2, 8, 22, [24, 20, 22, 22], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0057","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3c492f15359faab6d208e1a7035949fb8ead0f953b65dde2282f888f745348d","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317057,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-6,\"scale\":-1,\"shift\":9,\"terminal\":[-7,-5,-6,-6]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-6 and complete terminal values=[-7, -5, -6, -6]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=9. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-6, [-7, -5, -6, -6], -1, 9, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-6, [-7, -5, -6, -6], -1, 9, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0058","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f036519a680063da6536053705507474fe077e2bb3d85a16501b35bdb398c012","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317058,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-5,\"scale\":0,\"shift\":10,\"terminal\":[-6,-4,-5,-5]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-5 and complete terminal values=[-6, -4, -5, -5]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=10. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-5, [-6, -4, -5, -5], 0, 10, 10, [10, 10, 10, 10], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-5, [-6, -4, -5, -5], 0, 10, 10, [10, 10, 10, 10], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0059","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d9b80bbf7ea6859cb666c07c77d4d887d845a531af2795e46fc4bedf6fb1a4f","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317059,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-4,\"scale\":1,\"shift\":11,\"terminal\":[-5,-3,-4,-4]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-4 and complete terminal values=[-5, -3, -4, -4]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=11. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-4, [-5, -3, -4, -4], 1, 11, 7, [6, 8, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-4, [-5, -3, -4, -4], 1, 11, 7, [6, 8, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0060","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"daa31b901ec107302f14521dbb226999b5cb774084ec8414f4bebbfca7ce9401","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317060,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-3,\"scale\":2,\"shift\":12,\"terminal\":[-4,-2,-3,-3]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-3 and complete terminal values=[-4, -2, -3, -3]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=12. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-3, [-4, -2, -3, -3], 2, 12, 6, [4, 8, 6, 6], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-3, [-4, -2, -3, -3], 2, 12, 6, [4, 8, 6, 6], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0061","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf9e306e6223132b99962d8b93bdaa1570c3b374c837473060fc67e5cf9840c6","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317061,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-2,\"scale\":-2,\"shift\":13,\"terminal\":[-3,-1,-2,-2]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-2 and complete terminal values=[-3, -1, -2, -2]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=13. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-2, [-3, -1, -2, -2], -2, 13, 17, [19, 15, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-2, [-3, -1, -2, -2], -2, 13, 17, [19, 15, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0062","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f75c3b425947ebbe34e6978ccf175aca30cf9d84650eb37791f94464fdce8947","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317062,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":-1,\"scale\":-1,\"shift\":14,\"terminal\":[-2,0,-1,-1]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=-1 and complete terminal values=[-2, 0, -1, -1]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=14. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(-1, [-2, 0, -1, -1], -1, 14, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-1, [-2, 0, -1, -1], -1, 14, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0063","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b3636231081aee236779e7ea8731ab7cccc95eb570176e76b7178a020b505f1","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317063,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":0,\"scale\":0,\"shift\":15,\"terminal\":[-1,1,0,0]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=0 and complete terminal values=[-1, 1, 0, 0]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=15. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(0, [-1, 1, 0, 0], 0, 15, 15, [15, 15, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (0, [-1, 1, 0, 0], 0, 15, 15, [15, 15, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0064","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"758df3a4851a436e7db9a6889482b419227fc7c007d9a8d7d9fc36d8b81026fb","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317064,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":1,\"scale\":1,\"shift\":16,\"terminal\":[0,2,1,1]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=1 and complete terminal values=[0, 2, 1, 1]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=16. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(1, [0, 2, 1, 1], 1, 16, 17, [16, 18, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (1, [0, 2, 1, 1], 1, 16, 17, [16, 18, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0065","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11ec7522b3daf882d19aa95ac9220f4a98daf89cdb2322fc3638b6252bb41756","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317065,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":2,\"scale\":2,\"shift\":17,\"terminal\":[1,3,2,2]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=2 and complete terminal values=[1, 3, 2, 2]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=17. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(2, [1, 3, 2, 2], 2, 17, 21, [19, 23, 21, 21], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (2, [1, 3, 2, 2], 2, 17, 21, [19, 23, 21, 21], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0066","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7ac24b9f1ce7b84714fe5f93c36302350801be739debf82025400354a99b7c1","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317066,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":3,\"scale\":-2,\"shift\":18,\"terminal\":[2,4,3,3]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=3 and complete terminal values=[2, 4, 3, 3]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=18. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(3, [2, 4, 3, 3], -2, 18, 12, [14, 10, 12, 12], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (3, [2, 4, 3, 3], -2, 18, 12, [14, 10, 12, 12], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0067","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9ec2339b4a0923837492eb6f79e92e314065460a2a3b7edcca1ae5ceed2ec2c","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317067,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":4,\"scale\":-1,\"shift\":19,\"terminal\":[3,5,4,4]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=4 and complete terminal values=[3, 5, 4, 4]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=19. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(4, [3, 5, 4, 4], -1, 19, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (4, [3, 5, 4, 4], -1, 19, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0068","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23ef6acb6404de7018c4466f091ee77e7d48ee1f2426c7154df886e959d0fabb","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317068,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":5,\"scale\":0,\"shift\":20,\"terminal\":[4,6,5,5]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=5 and complete terminal values=[4, 6, 5, 5]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=20. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(5, [4, 6, 5, 5], 0, 20, 20, [20, 20, 20, 20], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (5, [4, 6, 5, 5], 0, 20, 20, [20, 20, 20, 20], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0069","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40c279efabcaeefb21066192ff03af7858d676c5264edff429b3386307b5b5dd","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317069,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":6,\"scale\":1,\"shift\":21,\"terminal\":[5,7,6,6]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=6 and complete terminal values=[5, 7, 6, 6]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=21. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(6, [5, 7, 6, 6], 1, 21, 27, [26, 28, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (6, [5, 7, 6, 6], 1, 21, 27, [26, 28, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0070","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"940554aca2088a6919ef2a205311f69949374345288ed6622114b1f0400f76b0","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317070,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":7,\"scale\":2,\"shift\":22,\"terminal\":[6,8,7,7]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=7 and complete terminal values=[6, 8, 7, 7]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=22. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(7, [6, 8, 7, 7], 2, 22, 36, [34, 38, 36, 36], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (7, [6, 8, 7, 7], 2, 22, 36, [34, 38, 36, 36], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0071","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"900cee38f93b6ae01f776a5826605081bc5890933c23e3b316b58ffd6a66e94b","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317071,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":8,\"scale\":-2,\"shift\":23,\"terminal\":[7,9,8,8]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=8 and complete terminal values=[7, 9, 8, 8]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=23. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(8, [7, 9, 8, 8], -2, 23, 7, [9, 5, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (8, [7, 9, 8, 8], -2, 23, 7, [9, 5, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0072","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73d0f6594fd4e7538439615e83d3e2e2852407ef580d6aaf8bd38e5d4d5c630d","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317072,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":9,\"scale\":-1,\"shift\":24,\"terminal\":[8,10,9,9]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=9 and complete terminal values=[8, 10, 9, 9]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=24. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(9, [8, 10, 9, 9], -1, 24, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (9, [8, 10, 9, 9], -1, 24, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0073","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7b50981298374a90ce47ac10c0d60db05fbdc455726e76b21bc39aa95f6222d","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317073,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":10,\"scale\":0,\"shift\":25,\"terminal\":[9,11,10,10]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=10 and complete terminal values=[9, 11, 10, 10]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=25. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(10, [9, 11, 10, 10], 0, 25, 25, [25, 25, 25, 25], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (10, [9, 11, 10, 10], 0, 25, 25, [25, 25, 25, 25], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0074","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"542cbdb8152908a18d650cd35238bfa33aa8c8cc2917f886b1bc669b0e438814","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317074,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":11,\"scale\":1,\"shift\":26,\"terminal\":[10,12,11,11]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=11 and complete terminal values=[10, 12, 11, 11]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=26. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(11, [10, 12, 11, 11], 1, 26, 37, [36, 38, 37, 37], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (11, [10, 12, 11, 11], 1, 26, 37, [36, 38, 37, 37], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0075","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96daa2f25c222315200105d3aeca5aa555a45cded3f54f962302b5c2953c774e","family_id":"qf-mod31-martdeep-affine-stability","seed":202610317075,"semantic_template":"qf-mod31-martdeep-affine-stability/{\"initial\":12,\"scale\":2,\"shift\":27,\"terminal\":[11,13,12,12]}"},"fields":{"title":"One Fixed Affine Transformation Preserves the Same-Filtration Martingale","stem":"Use the declared equiprobable one-step martingale with constant initial=12 and complete terminal values=[11, 13, 12, 12]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=27. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).","answer":{"kind":"expression","value":"(12, [11, 13, 12, 12], 2, 27, 51, [49, 53, 51, 51], Yes, affine_transform_preserves_same_filtration_martingale)","explanation":"First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (12, [11, 13, 12, 12], 2, 27, 51, [49, 53, 51, 51], Yes, affine_transform_preserves_same_filtration_martingale)."},"skills":["One Fixed Affine Transformation Preserves the Same-Filtration Martingale"],"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-20261014-advance-20261031-r07-martingale-0076","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"468e2ded99910ed491664c71a6a0b0a1f1e1bfc6bb41ebf785756df64db901ed","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317076,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":1}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=1 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(1, 1, [-1, 1], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 1, [-1, 1], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0077","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f11b245b4024646a17b8ef20a628a2752a3516e472e68c78c4bf9e4f099b1408","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317077,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":2}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=2 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(2, 1, [-2, 0, 2], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (2, 1, [-2, 0, 2], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0078","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8eceba26463b55b9aa35689e5623c8fd0e37e3bb50a6bc455e6b4c29c9cbbaa5","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317078,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":3}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=3 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(3, 1, [-3, -1, 1, 3], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (3, 1, [-3, -1, 1, 3], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0079","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb950c1aa40543ea9ebffb84e32f13f665f812d3035993f67e02cfd8e71dd8dd","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317079,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":4}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=4 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(4, 1, [-4, -2, 0, 2, 4], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (4, 1, [-4, -2, 0, 2, 4], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0080","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab0f57ffe2a3753dc8bed1687b696565793c68113bc02f554764deb285ebbd9f","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317080,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":5}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=5 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(5, 1, [-5, -3, -1, 1, 3, 5], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (5, 1, [-5, -3, -1, 1, 3, 5], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0081","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a5caafe18a2a62d68ed6aaa6a4c77112ac9a33567a64687fce65aa1b007e51d","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317081,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":6}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=6 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(6, 1, [-6, -4, -2, 0, 2, 4, 6], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (6, 1, [-6, -4, -2, 0, 2, 4, 6], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0082","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d735a732ccc651c35bc9b6ddf11a86fd28949c151f6637a246572b2f84241fe1","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317082,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":7}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=7 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(7, 1, [-7, -5, -3, -1, 1, 3, 5, 7], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (7, 1, [-7, -5, -3, -1, 1, 3, 5, 7], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0083","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eae13f9288ad1e76a10f19d223ea4c522dec68456d1a1a2493de428b7963a7d6","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317083,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":8}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=8 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(8, 1, [-8, -6, -4, -2, 0, 2, 4, 6, 8], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (8, 1, [-8, -6, -4, -2, 0, 2, 4, 6, 8], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0084","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f543e9b75ee806a029d9afb186d5761ca674f4933f7af005131c5c487457df08","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317084,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":9}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=9 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(9, 1, [-9, -7, -5, -3, -1, 1, 3, 5, 7, 9], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (9, 1, [-9, -7, -5, -3, -1, 1, 3, 5, 7, 9], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0085","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae42c6f4bd97de08698c3f1144d1f03381409f59050b0e8fe3f4928c421330aa","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317085,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":10}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=10 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(10, 1, [-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (10, 1, [-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0086","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71b741020a60ab522009f7822bb72a926f63e450980bcec6a60f04d8e2361052","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317086,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":11}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=11 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(11, 1, [-11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (11, 1, [-11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0087","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21af56e963a6e55d92b880ab6077bbc357523a8a23b6ea6aec2bc0ff5ae5e1f6","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317087,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":1,\"steps\":12}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=12 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(12, 1, [-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (12, 1, [-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0088","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d438fea46514a6bbc932fb3c1eea6b260a5750c558a5bd184e70c78392b85f6d","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317088,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":1}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=1 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(1, 2, [-2, 2], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 2, [-2, 2], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0089","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe3afe22ec967742cf30065cedf9512eab7dbb7666aaf9cdf7ac6f9280902024","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317089,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":2}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=2 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(2, 2, [-4, 0, 4], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (2, 2, [-4, 0, 4], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0090","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9146450856af9b0198e83d1f530cf362d70d818cefc55f943f00a274a72d5740","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317090,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":3}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=3 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(3, 2, [-6, -2, 2, 6], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (3, 2, [-6, -2, 2, 6], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0091","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de868d8ab0d0245c9b4ab91bb4493da81bb32d6cd21bcedde2cea3f4cae2657d","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317091,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":4}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=4 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(4, 2, [-8, -4, 0, 4, 8], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (4, 2, [-8, -4, 0, 4, 8], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0092","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a205b2d2a4b5520e2d7a29be49cbb05628302178d7b87b7092b02efeb20bdd28","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317092,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":5}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=5 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(5, 2, [-10, -6, -2, 2, 6, 10], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (5, 2, [-10, -6, -2, 2, 6, 10], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0093","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7a923607b4512089849053406e6e8cd9dd9c88723d7ad22fb1bb9ecda87ccf8","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317093,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":6}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=6 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(6, 2, [-12, -8, -4, 0, 4, 8, 12], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (6, 2, [-12, -8, -4, 0, 4, 8, 12], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0094","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"76448634c02d428105040c91c09e6ee19418b1b36a101d866799e6a72afd4876","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317094,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":7}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=7 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(7, 2, [-14, -10, -6, -2, 2, 6, 10, 14], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (7, 2, [-14, -10, -6, -2, 2, 6, 10, 14], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0095","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ad423145f5950e70f37c725a68ee7f2f0fd0b61582fab6d96ed317e14906839","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317095,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":8}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=8 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(8, 2, [-16, -12, -8, -4, 0, 4, 8, 12, 16], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (8, 2, [-16, -12, -8, -4, 0, 4, 8, 12, 16], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0096","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8e4870f69fa920a981e1376c0ac73490d06a39fb61bbf8e351687101c83865a","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317096,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":9}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=9 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(9, 2, [-18, -14, -10, -6, -2, 2, 6, 10, 14, 18], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (9, 2, [-18, -14, -10, -6, -2, 2, 6, 10, 14, 18], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0097","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1724ed349bfb1d447270fb38971be4ab287069680dd90c492448d09e6cbff74d","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317097,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":10}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=10 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(10, 2, [-20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (10, 2, [-20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0098","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"223c4f64568318d0733d8fc5b164832326b30758819dc4ef7cc486985ff6f8c2","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317098,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":11}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=11 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(11, 2, [-22, -18, -14, -10, -6, -2, 2, 6, 10, 14, 18, 22], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (11, 2, [-22, -18, -14, -10, -6, -2, 2, 6, 10, 14, 18, 22], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0099","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f99fe38380e0c82d259217e691e248b345cb6450b722a8b367fb9172e0a86de","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317099,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":2,\"steps\":12}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=12 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(12, 2, [-24, -20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20, 24], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (12, 2, [-24, -20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20, 24], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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-20261014-advance-20261031-r07-martingale-0100","batch_id":"batch-20261014-advance-20261031-r07-martingale-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34bbc57e575fa5ea356c60dfe9522f919539b572700870483dd186961656766d","family_id":"qf-mod31-martdeep-fair-partial-sum","seed":202610317100,"semantic_template":"qf-mod31-martdeep-fair-partial-sum/{\"scale\":3,\"steps\":1}"},"fields":{"title":"A Declared Fair Independent-Increment Model Supplies a Specific Martingale","stem":"Use steps=1 independent equiprobable signs, each increment ±scale with scale=3>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.","answer":{"kind":"expression","value":"(1, 3, [-3, 3], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)","explanation":"The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 3, [-3, 3], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)."},"skills":["A Declared Fair Independent-Increment Model Supplies a Specific Martingale"],"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."}}