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How can I show that the in-sample bias error in Ridge regression can be decomposed into model bias plus estimation bias? I.e., if $Avg$ takes the average over all the input variables $x$ in the sample, and $E$ denotes the in-sample expectation value over the noise component in the underlying modeled function $f$, then:

$$ Avg[(f(x)-E[x'\beta_{Ridge}])^2] = Avg[(f(x)-x'\beta_{OLS})^2] + Avg[(x'\beta_{OLS}-E[x'\beta_{Ridge}])^2] $$ $$ = Squared Model Bias + Squared Estimation Bias $$ This is stated in Elements of Statistical Learning, Hastie et al, p.224, result 7.14

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It turns out this can be resolved by rewriting as matrices and using the optimality condition for $\beta_{OLS}$. Rewriting as matrices, taking the expectation value over $x'\beta_{Ridge}$ (this just results in dropping the error term $\epsilon$ in $y+\epsilon$), dropping the subscript $OLS$, and using subscript $R$ instead of $Ridge$, the above expression can be re-written:

$$ \frac{1}{N}[(y-X\beta)^2+(X\beta-X\beta_R)^2] $$

Dropping $N$ and using the optimality condition for $\beta$, i.e. $X'X\beta=X'y$, the first term can be re-written:

$$ y'y-y'X\beta $$

Using the "hat" matrix $H = X(X'X)^{-1}X'$, this can be written:

$$ y'y-y'Hy \quad \text{(eq. 1)}\\ $$

The $SquaredEstimationBias$ can be further simplified using the $H$ matrix and the corresponding matrix for Ridge regression, $X(X'X+\lambda I)^{-1}X'$, let us call it $H_R$. One can check that these matrices have the following nice properties, which have to do with the fact that $H$ is a projection matrix:

$$ H'H = H $$ $$ H'H_R = H_R'H= H_R $$

Now, if we write out the $Squared Estimation Bias$ term using the "hat" matrices and their properties, we arrive at:

$$ y'Hy-2y'H_Ry+y'H_R'H_Ry $$

The first of the terms directly above cancels the second term in eq. 1 for the $Squared Model Bias$, $(-y'Hy)$, leaving a total sum for $Squared Model Bias + Squared Estimation Bias$ of:

$$ y'y-2y'H_Ry+y'H_R'H_Ry $$

Equating $y$ with $f(x)$ (this is implied by the taking of expectations) this can be re-written to match the total bias error, as required:

$$ (y-X\beta_R)^2 = N*Avg(f(x)-x'\beta_R)^2=N*Avg(f(x)-E[x'\beta_R])^2 $$

Given the properties that we have used here, I believe this result holds for any model that can be written as $X\hat{\beta}$, a linear combination of the input variables, not just Ridge regression.

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