# Bias-variance derivation with the cross term visible

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This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented.

derivation-mko-bias-variance-decomposition-01 · ocme-new142-reader-adoption-20261021



## Goal and scope

Prove the decomposition symbol by symbol in one law and check both theta values.

Finite scalar laws, kept separate at each theta.





## Givens and symbols


- work


- G1 — Theta={1/4,3/4}, Omega={00,01,10,11}; P_(1/4)=(9,3,3,1)/16 and P_(3/4)=(1,3,3,9)/16.
- G2 — U(x)=(x1+x2)/4 has total table(0,1/4,1/4,1/2) in Omega order, with target g(theta)=theta.


- θ — fixed unknown parameter
- U — one total statistic
- m,b — same-law mean and bias





## Rules and sources


- mko_definition · cited — For every theta supply a complete normalized finite sampling law. Under that same law separately compute the mean, bias=E_theta[T]−theta, variance=E_theta[(T−E_theta[T])^2] and MSE=E_theta[(T−theta)^2]. Expanding T−theta=(T−E_theta[T])+bias gives a cross term that vanishes because E_theta[T−E_theta[T]]=0. Thus MSE=variance+bias^2 pointwise in theta. Do not pool different theta laws. Unbiasedness gives MSE=variance, not zero MSE or consistency. One observed squared error is not the definition of expected loss, even if its numerical value coincides with MSE. · mko-bias-variance-decomposition
- work · prerequisite — Linearity, constant expectations under normalized laws and square expansion; at fixed theta, m=EthetaU, bias=m-theta, variance=Etheta[(U-m)²] and MSE=Etheta[(U-theta)²]. The decomposition itself is not assumed.





## Main derivation

S01 · impliesBefore

\theta\in\Theta\text{ fixed},\quad m=E_\theta U

After

b=m-\theta,\quad U-\theta=(U-m)+b

Conditions: none

Rule: work · Uses: G1, G2, work

Leaf basis: work

Fix one theta and split error into centered error and constant bias.



S02 · impliesBefore

U-\theta=(U-m)+b

After

(U-\theta)^2=(U-m)^2+2b(U-m)+b^2

Conditions: none

Rule: work · Uses: G1, G2, S01, work

Leaf basis: work

Expand the square fully, retaining the cross term.



S03 · impliesBefore

E_\theta[U-m]=E_\theta U-m

After

E_\theta[U-m]=0

Conditions: none

Rule: work · Uses: G1, G2, S02, work

Leaf basis: work

Linearity and the definition of m give the centering identity.



S04 · impliesBefore

E_\theta[2b(U-m)]

After

2bE_\theta[U-m]=0

Conditions: none

Rule: work · Uses: G1, G2, S03, work

Leaf basis: work

At fixed theta b is constant, so factor it out and use centering.



S05 · impliesBefore

E_\theta[(U-\theta)^2]

After

E_\theta[(U-m)^2]+0+b^2

Conditions: none

Rule: work · Uses: G1, G2, S04, work

Leaf basis: work

Take expectation of the complete expansion under the same law; b² remains itself.



S06 · impliesBefore

\operatorname{MSE}_\theta(U)=E_\theta[(U-m)^2]+b^2

After

\operatorname{MSE}_\theta(U)=\operatorname{Var}_\theta(U)+b(\theta)^2

Conditions: none

Rule: work · Uses: G1, G2, S05, work

Leaf basis: work

Identify variance and bias by definition, completing the finite-law argument.



S07 · impliesBefore

\theta=1/4,\quad U\in\{0,1/4,1/2\},\quad P_{1/4}^U=(9,6,1)/16

After

m=0(9/16)+(1/4)(6/16)+(1/2)(1/16)=1/8,\quad b=1/8-1/4=-1/8

Conditions: none

Rule: work · Uses: G1, G2, S06, work

Leaf basis: work

Relist the complete normalized sampling law in this guide and directly compute its mean and signed bias.



S08 · impliesBefore

v=\sum_uP_{1/4}^U(u)(u-1/8)^2,\quad M=\sum_uP_{1/4}^U(u)(u-1/4)^2

After

v=\frac{9(1/8)^2+6(1/8)^2+(3/8)^2}{16}=3/128;\quad M=\frac{9(1/4)^2+6(0)^2+(1/4)^2}{16}=5/128

Conditions: none

Rule: work · Uses: G1, G2, S07, work

Leaf basis: work

Weight the centered variance and target losses independently, without using the proved decomposition to obtain either value.



S09 · impliesBefore

\theta=1/4:\quad M=5/128,\ b=-1/8,\ v=3/128

After

5/128=3/128+(-1/8)^2

Conditions: none

Rule: work · Uses: G1, G2, S08, work

Leaf basis: work

Cross-check the two direct-sum results against the general decomposition.



S10 · impliesBefore

\theta=3/4,\quad P_{3/4}^U=(1,6,9)/16

After

m=0(1/16)+(1/4)(6/16)+(1/2)(9/16)=3/8,\quad b=3/8-3/4=-3/8

Conditions: none

Rule: work · Uses: G1, G2, S09, work

Leaf basis: work

Independently compute the second law's mean and bias without reusing first-law weights.



S11 · impliesBefore

v=\sum_uP_{3/4}^U(u)(u-3/8)^2,\quad M=\sum_uP_{3/4}^U(u)(u-3/4)^2

After

v=\frac{(3/8)^2+6(1/8)^2+9(1/8)^2}{16}=3/128;\quad M=\frac{(3/4)^2+6(1/2)^2+9(1/4)^2}{16}=21/128

Conditions: none

Rule: work · Uses: G1, G2, S10, work

Leaf basis: work

Evaluate the second law's centered-square sum and target-loss sum separately.



S12 · impliesBefore

\theta=3/4:\quad M=21/128,\ b=-3/8,\ v=3/128

After

21/128=3/128+(-3/8)^2

Conditions: none

Rule: work · Uses: G1, G2, S11, work

Leaf basis: work

The second law's direct values verify the decomposition without mixing parameters.







## Substeps and lemmas

No substeps in this guide.





## Reverse audits


- S12 · premise_audit · G1, G2, S11, work — Unbiasedness only removes squared bias, not variance.
- S11 · premise_audit · G1, G2, S10, work — The three direct MSE contributions are 9/256,24/256,9/256.
- S10 · premise_audit · G1, G2, S09, work — The final support mass under this law is 9/16.
- S09 · premise_audit · G1, G2, S08, work — Square bias first; negative bias is not subtracted.
- S08 · premise_audit · G1, G2, S07, work — The variance center is 1/8 while the MSE target is 1/4; they differ.
- S07 · premise_audit · G1, G2, S06, work — Pair the three support values with their three masses in the same order.
- S06 · premise_audit · G1, G2, S05, work — Theta was arbitrary, so the identity holds separately at every model parameter.
- S05 · premise_audit · G1, G2, S04, work — Normalization gives Etheta[b²]=b².
- S04 · premise_audit · G1, G2, S03, work — No independence is required here.
- S03 · premise_audit · G1, G2, S02, work — Signs alone do not give mean zero; the definition of m does.
- S02 · premise_audit · G1, G2, S01, work — The factor2 is required; b is not a random observed value.
- S01 · premise_audit · G1, G2, work — Both m and b belong to this Ptheta, not a mixture.





## Error controls and negative controls

Visible cross termRetain2b(U-m) before eliminating it by centered expectation zero.

Direct MSE auditCompare both direct loss sums with 3/128 plus squared bias.


- The cross term vanishes only by independence. → reject — b is constant and Etheta[U-m]=0 suffices.
- Unbiasedness makes MSE zero. → reject — It only gives MSE=variance; an unbiased Bernoulli observation has positive variance.
- MSE=variance+bias. → reject — Expansion gives squared bias; unsquared bias would even give a negative value in the first law.





## Result and limitations

MSEtheta=Vartheta+bias(theta)²; direct values are 5/128 and 21/128 at the two theta values.







## Verification and source binding

6 declared checks for this guide; only the listed expressions are covered.


- source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
- adopted source SHA-256: 684c90c6e4cc67e67bac21d84d6391f4ac7df58ae8a555e711b2c67724a97dc5
- original source SHA-256: 8b7541177d92f5388c355193b85c8de3a5067e1f823493728c074e77a687ec37
- Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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