# Theta-Specific Bias-Variance Identities and Centering

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-12, complete estimator values=[-12, -11, -10] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-12, [[-12, 1/4], [-11, 1/2], [-10, 1/4]], -11, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-12, [[-12, 1/4], [-11, 1/2], [-10, 1/4]], -11, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0001/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 2. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-11, complete estimator values=[-11, -9, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-11, [[-11, 1/4], [-9, 1/2], [-7, 1/4]], -9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-11, [[-11, 1/4], [-9, 1/2], [-7, 1/4]], -9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0002/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 3. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-10, complete estimator values=[-10, -7, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-10, [[-10, 1/4], [-7, 1/2], [-4, 1/4]], -7, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-10, [[-10, 1/4], [-7, 1/2], [-4, 1/4]], -7, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0003/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 4. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-9, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-9, [[-9, 1/4], [-8, 1/2], [-7, 1/4]], -8, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-9, [[-9, 1/4], [-8, 1/2], [-7, 1/4]], -8, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0004/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 5. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-8, complete estimator values=[-8, -6, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-8, [[-8, 1/4], [-6, 1/2], [-4, 1/4]], -6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-8, [[-8, 1/4], [-6, 1/2], [-4, 1/4]], -6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0005/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 6. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-7, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-7, [[-7, 1/4], [-4, 1/2], [-1, 1/4]], -4, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-7, [[-7, 1/4], [-4, 1/2], [-1, 1/4]], -4, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0006/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 7. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-6, complete estimator values=[-6, -5, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-6, [[-6, 1/4], [-5, 1/2], [-4, 1/4]], -5, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-6, [[-6, 1/4], [-5, 1/2], [-4, 1/4]], -5, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0007/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 8. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-5, complete estimator values=[-5, -3, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-5, [[-5, 1/4], [-3, 1/2], [-1, 1/4]], -3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-5, [[-5, 1/4], [-3, 1/2], [-1, 1/4]], -3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0008/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 9. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-4, complete estimator values=[-4, -1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-4, [[-4, 1/4], [-1, 1/2], [2, 1/4]], -1, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-4, [[-4, 1/4], [-1, 1/2], [2, 1/4]], -1, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0009/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 10. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-3, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-3, [[-3, 1/4], [-2, 1/2], [-1, 1/4]], -2, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-3, [[-3, 1/4], [-2, 1/2], [-1, 1/4]], -2, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0010/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 11. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-2, complete estimator values=[-2, 0, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-2, [[-2, 1/4], [0, 1/2], [2, 1/4]], 0, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-2, [[-2, 1/4], [0, 1/2], [2, 1/4]], 0, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0011/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 12. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=-1, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(-1, [[-1, 1/4], [2, 1/2], [5, 1/4]], 2, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (-1, [[-1, 1/4], [2, 1/2], [5, 1/4]], 2, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0012/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 13. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=0, complete estimator values=[0, 1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(0, [[0, 1/4], [1, 1/2], [2, 1/4]], 1, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (0, [[0, 1/4], [1, 1/2], [2, 1/4]], 1, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0013/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 14. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=1, complete estimator values=[1, 3, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(1, [[1, 1/4], [3, 1/2], [5, 1/4]], 3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (1, [[1, 1/4], [3, 1/2], [5, 1/4]], 3, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0014/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 15. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=2, complete estimator values=[2, 5, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(2, [[2, 1/4], [5, 1/2], [8, 1/4]], 5, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (2, [[2, 1/4], [5, 1/2], [8, 1/4]], 5, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0015/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 16. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=3, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(3, [[3, 1/4], [4, 1/2], [5, 1/4]], 4, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (3, [[3, 1/4], [4, 1/2], [5, 1/4]], 4, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0016/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 17. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=4, complete estimator values=[4, 6, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(4, [[4, 1/4], [6, 1/2], [8, 1/4]], 6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (4, [[4, 1/4], [6, 1/2], [8, 1/4]], 6, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0017/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 18. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=5, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(5, [[5, 1/4], [8, 1/2], [11, 1/4]], 8, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (5, [[5, 1/4], [8, 1/2], [11, 1/4]], 8, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0018/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 19. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=6, complete estimator values=[6, 7, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(6, [[6, 1/4], [7, 1/2], [8, 1/4]], 7, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (6, [[6, 1/4], [7, 1/2], [8, 1/4]], 7, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0019/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 20. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=7, complete estimator values=[7, 9, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(7, [[7, 1/4], [9, 1/2], [11, 1/4]], 9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (7, [[7, 1/4], [9, 1/2], [11, 1/4]], 9, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0020/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 21. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=8, complete estimator values=[8, 11, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(8, [[8, 1/4], [11, 1/2], [14, 1/4]], 11, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (8, [[8, 1/4], [11, 1/2], [14, 1/4]], 11, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0021/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 22. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=9, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(9, [[9, 1/4], [10, 1/2], [11, 1/4]], 10, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (9, [[9, 1/4], [10, 1/2], [11, 1/4]], 10, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0022/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 23. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=10, complete estimator values=[10, 12, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(10, [[10, 1/4], [12, 1/2], [14, 1/4]], 12, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (10, [[10, 1/4], [12, 1/2], [14, 1/4]], 12, 2, 2, 6, 6, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0023/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 24. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=11, complete estimator values=[11, 14, 17] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(11, [[11, 1/4], [14, 1/2], [17, 1/4]], 14, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (11, [[11, 1/4], [14, 1/2], [17, 1/4]], 14, 3, 9/2, 27/2, 27/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0024/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 25. MSE Equals Variance plus Squared Bias under One Complete Theta Law

Use scalar theta=12, complete estimator values=[12, 13, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (theta, law_rows, mean, bias, variance, MSE, variance_plus_bias_squared, identity_holds, certificate).

View answer and explanation(12, [[12, 1/4], [13, 1/2], [14, 1/4]], 13, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law)

Compute mean, bias, variance and MSE separately under the same law, then compare the direct MSE with variance+bias^2. The identity is a finite scalar-law certificate, not consistency or an asymptotic claim. The exact result, retaining the requested order, is (12, [[12, 1/4], [13, 1/2], [14, 1/4]], 13, 1, 1/2, 3/2, 3/2, Yes, MSE_equals_variance_plus_squared_bias_under_one_complete_sampling_law).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0025/)

Bias-variance decomposition · qf-mod46-biasvar-single-theta-identity · candidate





## 26. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-12, complete estimator values=[-11, -10, -9] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0026/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 27. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-11, complete estimator values=[-10, -9, -8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0027/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 28. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-10, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0028/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 29. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-9, complete estimator values=[-8, -7, -6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0029/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 30. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-8, complete estimator values=[-7, -6, -5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0030/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 31. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-7, complete estimator values=[-6, -5, -4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0031/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 32. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-6, complete estimator values=[-5, -4, -3] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0032/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 33. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-5, complete estimator values=[-4, -3, -2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0033/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 34. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-4, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0034/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 35. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-3, complete estimator values=[-2, -1, 0] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(-1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (-1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0035/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 36. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-2, complete estimator values=[-1, 0, 1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(0, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (0, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0036/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 37. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=-1, complete estimator values=[0, 1, 2] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (1, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0037/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 38. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=0, complete estimator values=[1, 2, 3] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (2, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0038/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 39. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=1, complete estimator values=[2, 3, 4] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (3, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0039/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 40. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=2, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (4, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0040/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 41. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=3, complete estimator values=[4, 5, 6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (5, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0041/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 42. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=4, complete estimator values=[5, 6, 7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (6, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0042/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 43. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=5, complete estimator values=[6, 7, 8] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (7, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0043/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 44. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=6, complete estimator values=[7, 8, 9] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (8, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0044/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 45. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=7, complete estimator values=[8, 9, 10] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (9, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0045/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 46. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=8, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (10, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0046/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 47. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=9, complete estimator values=[10, 11, 12] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(11, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (11, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0047/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 48. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=10, complete estimator values=[11, 12, 13] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(12, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (12, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0048/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 49. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=11, complete estimator values=[12, 13, 14] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(13, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (13, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0049/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 50. The Cross Term Vanishes Because the Centered Estimator Has Mean Zero

Use scalar theta=12, complete estimator values=[13, 14, 15] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Give (mean, centered_mean, bias, cross_term, cross_term_zero, certificate).

View answer and explanation(14, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes)

Compute E[T−E[T]], bias and twice their product. The cross term vanishes by the centering identity, not because every observed deviation or the bias itself must be zero. The exact result, retaining the requested order, is (14, 0, 2, 0, Yes, E_of_T_minus_E_T_is_zero_so_the_cross_term_vanishes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0050/)

Bias-variance decomposition · qf-mod46-biasvar-cross-term-cancellation · candidate





## 51. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-12, [-12, -11, -10], [[1, 4], [1, 2], [1, 4]]], [-11, [-12, -11, -10], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-11, -11, 0, 1/2, 1/2, 1/2, Yes], [-12, -11, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-11, -11, 0, 1/2, 1/2, 1/2, Yes], [-12, -11, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0051/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 52. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-11, [-11, -9, -7], [[1, 4], [1, 2], [1, 4]]], [-10, [-11, -10, -9], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-10, -10, 0, 1/2, 1/2, 1/2, Yes], [-11, -9, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-10, -10, 0, 1/2, 1/2, 1/2, Yes], [-11, -9, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0052/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 53. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-10, [-10, -7, -4], [[1, 4], [1, 2], [1, 4]]], [-9, [-10, -9, -8], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-10, -7, 3, 9/2, 27/2, 27/2, Yes], [-9, -9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-10, -7, 3, 9/2, 27/2, 27/2, Yes], [-9, -9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0053/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 54. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-9, [-9, -8, -7], [[1, 4], [1, 2], [1, 4]]], [-8, [-9, -8, -7], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-8, -8, 0, 1/2, 1/2, 1/2, Yes], [-9, -8, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-8, -8, 0, 1/2, 1/2, 1/2, Yes], [-9, -8, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0054/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 55. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-8, [-8, -6, -4], [[1, 4], [1, 2], [1, 4]]], [-7, [-8, -7, -6], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-7, -7, 0, 1/2, 1/2, 1/2, Yes], [-8, -6, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-7, -7, 0, 1/2, 1/2, 1/2, Yes], [-8, -6, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0055/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 56. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-7, [-7, -4, -1], [[1, 4], [1, 2], [1, 4]]], [-6, [-7, -6, -5], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-6, -6, 0, 1/2, 1/2, 1/2, Yes], [-7, -4, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-6, -6, 0, 1/2, 1/2, 1/2, Yes], [-7, -4, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0056/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 57. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-6, [-6, -5, -4], [[1, 4], [1, 2], [1, 4]]], [-5, [-6, -5, -4], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-5, -5, 0, 1/2, 1/2, 1/2, Yes], [-6, -5, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-5, -5, 0, 1/2, 1/2, 1/2, Yes], [-6, -5, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0057/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 58. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-5, [-5, -3, -1], [[1, 4], [1, 2], [1, 4]]], [-4, [-5, -4, -3], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-4, -4, 0, 1/2, 1/2, 1/2, Yes], [-5, -3, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-4, -4, 0, 1/2, 1/2, 1/2, Yes], [-5, -3, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0058/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 59. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-4, [-4, -1, 2], [[1, 4], [1, 2], [1, 4]]], [-3, [-4, -3, -2], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-3, -3, 0, 1/2, 1/2, 1/2, Yes], [-4, -1, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-3, -3, 0, 1/2, 1/2, 1/2, Yes], [-4, -1, 3, 9/2, 27/2, 27/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0059/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 60. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-3, [-3, -2, -1], [[1, 4], [1, 2], [1, 4]]], [-2, [-3, -2, -1], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-2, -2, 0, 1/2, 1/2, 1/2, Yes], [-3, -2, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-2, -2, 0, 1/2, 1/2, 1/2, Yes], [-3, -2, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0060/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 61. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-2, [-2, 0, 2], [[1, 4], [1, 2], [1, 4]]], [-1, [-2, -1, 0], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-1, -1, 0, 1/2, 1/2, 1/2, Yes], [-2, 0, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-1, -1, 0, 1/2, 1/2, 1/2, Yes], [-2, 0, 2, 2, 6, 6, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0061/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 62. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[-1, [-1, 2, 5], [[1, 4], [1, 2], [1, 4]]], [0, [-1, 0, 1], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[-1, 2, 3, 9/2, 27/2, 27/2, Yes], [0, 0, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[-1, 2, 3, 9/2, 27/2, 27/2, Yes], [0, 0, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0062/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 63. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[0, [0, 1, 2], [[1, 4], [1, 2], [1, 4]]], [1, [0, 1, 2], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[0, 1, 1, 1/2, 3/2, 3/2, Yes], [1, 1, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[0, 1, 1, 1/2, 3/2, 3/2, Yes], [1, 1, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0063/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 64. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[1, [1, 3, 5], [[1, 4], [1, 2], [1, 4]]], [2, [1, 2, 3], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[1, 3, 2, 2, 6, 6, Yes], [2, 2, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[1, 3, 2, 2, 6, 6, Yes], [2, 2, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0064/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 65. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[2, [2, 5, 8], [[1, 4], [1, 2], [1, 4]]], [3, [2, 3, 4], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[2, 5, 3, 9/2, 27/2, 27/2, Yes], [3, 3, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[2, 5, 3, 9/2, 27/2, 27/2, Yes], [3, 3, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0065/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 66. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[3, [3, 4, 5], [[1, 4], [1, 2], [1, 4]]], [4, [3, 4, 5], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[3, 4, 1, 1/2, 3/2, 3/2, Yes], [4, 4, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[3, 4, 1, 1/2, 3/2, 3/2, Yes], [4, 4, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0066/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 67. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[4, [4, 6, 8], [[1, 4], [1, 2], [1, 4]]], [5, [4, 5, 6], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[4, 6, 2, 2, 6, 6, Yes], [5, 5, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[4, 6, 2, 2, 6, 6, Yes], [5, 5, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0067/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 68. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[5, [5, 8, 11], [[1, 4], [1, 2], [1, 4]]], [6, [5, 6, 7], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[5, 8, 3, 9/2, 27/2, 27/2, Yes], [6, 6, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[5, 8, 3, 9/2, 27/2, 27/2, Yes], [6, 6, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0068/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 69. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[6, [6, 7, 8], [[1, 4], [1, 2], [1, 4]]], [7, [6, 7, 8], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[6, 7, 1, 1/2, 3/2, 3/2, Yes], [7, 7, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[6, 7, 1, 1/2, 3/2, 3/2, Yes], [7, 7, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0069/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 70. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[7, [7, 9, 11], [[1, 4], [1, 2], [1, 4]]], [8, [7, 8, 9], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[7, 9, 2, 2, 6, 6, Yes], [8, 8, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[7, 9, 2, 2, 6, 6, Yes], [8, 8, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0070/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 71. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[8, [8, 11, 14], [[1, 4], [1, 2], [1, 4]]], [9, [8, 9, 10], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[8, 11, 3, 9/2, 27/2, 27/2, Yes], [9, 9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[8, 11, 3, 9/2, 27/2, 27/2, Yes], [9, 9, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0071/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 72. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[9, [9, 10, 11], [[1, 4], [1, 2], [1, 4]]], [10, [9, 10, 11], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[10, 10, 0, 1/2, 1/2, 1/2, Yes], [9, 10, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[10, 10, 0, 1/2, 1/2, 1/2, Yes], [9, 10, 1, 1/2, 3/2, 3/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0072/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 73. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[10, [10, 12, 14], [[1, 4], [1, 2], [1, 4]]], [11, [10, 11, 12], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[10, 12, 2, 2, 6, 6, Yes], [11, 11, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[10, 12, 2, 2, 6, 6, Yes], [11, 11, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0073/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 74. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[11, [11, 14, 17], [[1, 4], [1, 2], [1, 4]]], [12, [11, 12, 13], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[11, 14, 3, 9/2, 27/2, 27/2, Yes], [12, 12, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[11, 14, 3, 9/2, 27/2, 27/2, Yes], [12, 12, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0074/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 75. Bias-Variance Decomposition Is Checked Pointwise under Each Theta Law

Use complete laws=[[12, [12, 13, 14], [[1, 4], [1, 2], [1, 4]]], [13, [12, 13, 14], [[1, 4], [1, 2], [1, 4]]]], each [theta,estimator values,aligned probabilities]. Every exact law is separately normalized and theta labels are unique. Give (theta_identity_rows, all_identities_hold, certificate).

View answer and explanation([[12, 13, 1, 1/2, 3/2, 3/2, Yes], [13, 13, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim)

For each law separately retain [theta,mean,bias,variance,MSE,variance+bias^2,identity]. Preserve the source canonical law order rather than numeric theta order. Do not pool laws or infer asymptotic behavior from a finite table. The exact result, retaining the requested order, is ([[12, 13, 1, 1/2, 3/2, 3/2, Yes], [13, 13, 0, 1/2, 1/2, 1/2, Yes]], Yes, identity_is_pointwise_in_theta_not_an_asymptotic_claim).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0075/)

Bias-variance decomposition · qf-mod46-biasvar-multi-theta-table · candidate





## 76. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-12, complete estimator values=[-13, -12, -11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-13. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-12, -13, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-12, -13, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0076/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 77. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-11, complete estimator values=[-11, -9, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-11, -11, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-11, -11, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0077/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 78. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-10, complete estimator values=[-13, -10, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-13. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-10, -13, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-10, -13, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0078/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 79. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-9, complete estimator values=[-9, -8, -7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-9. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-9, -9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-9, -9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0079/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 80. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-8, complete estimator values=[-10, -8, -6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-10. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-8, -10, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-8, -10, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0080/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 81. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-7, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-7, -7, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-7, -7, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0081/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 82. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-6, complete estimator values=[-7, -6, -5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-6, -7, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-6, -7, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0082/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 83. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-5, complete estimator values=[-5, -3, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-5, -5, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-5, -5, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0083/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 84. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-4, complete estimator values=[-7, -4, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-4, -7, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-4, -7, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0084/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 85. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-3, complete estimator values=[-3, -2, -1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-3. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-3, -3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-3, -3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0085/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 86. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-2, complete estimator values=[-4, -2, 0] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-4. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-2, -4, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-2, -4, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0086/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 87. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=-1, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(-1, -1, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (-1, -1, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0087/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 88. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=0, complete estimator values=[-1, 0, 1] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(0, -1, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (0, -1, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0088/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 89. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=1, complete estimator values=[1, 3, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(1, 1, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (1, 1, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0089/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 90. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=2, complete estimator values=[-1, 2, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=-1. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(2, -1, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (2, -1, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0090/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 91. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=3, complete estimator values=[3, 4, 5] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=3. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(3, 3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (3, 3, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0091/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 92. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=4, complete estimator values=[2, 4, 6] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=2. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(4, 2, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (4, 2, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0092/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 93. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=5, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(5, 5, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (5, 5, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0093/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 94. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=6, complete estimator values=[5, 6, 7] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(6, 5, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (6, 5, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0094/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 95. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=7, complete estimator values=[7, 9, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=7. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(7, 7, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (7, 7, 0, 2, No, 2, 6, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0095/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 96. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=8, complete estimator values=[5, 8, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=5. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(8, 5, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (8, 5, 9, 0, Yes, 9/2, 9/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0096/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 97. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=9, complete estimator values=[9, 10, 11] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=9. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(9, 9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (9, 9, 0, 1, No, 1/2, 3/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0097/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 98. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=10, complete estimator values=[8, 10, 12] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=8. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(10, 8, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (10, 8, 4, 0, Yes, 2, 2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0098/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 99. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=11, complete estimator values=[11, 14, 17] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(11, 11, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (11, 11, 0, 3, No, 9/2, 27/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0099/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate





## 100. Unbiasedness Gives MSE Equal to Variance, Not Zero Error or Consistency

Use scalar theta=12, complete estimator values=[11, 12, 13] and aligned exact probabilities=[1/4, 1/2, 1/4]. Under this same P_theta the nonnegative probabilities sum to1. Retain zero-mass listed rows and sort values with their own probabilities. Given listed observed estimate=11. Give (theta, observed, observed_squared_error, bias, unbiased, variance, MSE, variance_comparison_holds, unbiased_and_zero_MSE, consistency_claim, boundary).

variance_comparison_holds checks MSE=variance when unbiased, otherwise MSE≥variance. One observed loss may coincide with MSE; the final No does not deny coincidence or establish consistency.

View answer and explanation(12, 11, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE)

Compute the observed squared error separately from expected loss. If unbiased, verify MSE=variance; otherwise verify MSE≥variance. The last No withholds consistency, not possible coincidence between one observed loss and MSE. Unbiasedness may coexist with positive variance and MSE. The exact result, retaining the requested order, is (12, 11, 1, 0, Yes, 1/2, 1/2, Yes, No, No, unbiased_only_gives_MSE_equals_variance_not_zero_MSE_or_consistency_and_observed_error_is_not_MSE).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261115-r03-bias-variance-decomposition-0100/)

Bias-variance decomposition · qf-mod46-biasvar-biased-unbiased-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261115-r03-bias-variance-decomposition-foundations/validation-report.json)