 Solution

Put $f=(f(x_1),\dots ,f(x_n){)}^T$, so $Y=f+\epsilon$ and $Y^{\ast }=f+{\epsilon }^{\ast }$ with independent noise vectors having covariance matrix $I_n$. For any deterministic linear smoother $H$,whereas independence givesExpanding the first trace shows that the second expression exceeds the first by $2\mathrm{tr}(H)$, proving the identity.

For ridge regression,Its effective degrees of freedom are $\mathrm{tr}(H_{\lambda })$. Thus training error is optimistically biased for independent-copy prediction error by $2\mathrm{tr}(H_{\lambda })/n$. The graph exhibits exactly this effect: the dashed training curve keeps falling as $\lambda$ decreases, while the solid test curve eventually rises through overfitting.