In one of the slides from my class related to bayesian linear regression, I have the following scenario.

Under g-prior, the shrinkage estimator induced by the prior is $$\hat{\beta_{\alpha}} = \alpha\hat{\beta}$$

now it states that the value of $\alpha$ that minimizes the mean squared error (conditioned on $\beta$ and $\sigma^2$) for this estimator is

$$\alpha^* = \frac{1}{\frac{\sigma^2tr((X'X)^{-1})}{\beta'\beta}+1}$$

It also says that the proof uses the result: if $Z$ is a random vector with mean $\mu$ and covariance $\Sigma$ then with matrix $\Delta$ $$E[Z'\Delta Z] = tr(\Delta\Sigma) + \mu'\Delta\mu$$

Could anyone help deriving this $\alpha^*$? I am approaching by trying to differentiate $E[\| Y - X\hat{\beta_{\alpha}}\|^2 | \beta, \sigma^2]$ not really going anywhere.


1 Answer 1


It comes as \begin{align*} \mathbb{E}[\| Y - X\hat{\beta_{\alpha}}\|^2]&= \mathbb{E}[\| Y - X\beta\|^2] \overbrace{+}^{\text{Pythagore's}} \mathbb{E}[\| X\beta - X\hat{\beta_{\alpha}}\|^2]\\ &= \underbrace{C}_\text{independent of $\alpha$} + \mathbb{E}[(\beta - \hat{\beta_{\alpha}})'X'X (\beta - \hat{\beta_{\alpha}})]\\ &= C + \mathbb{E}[(\beta - \alpha\hat{\beta})'X'X (\hat\beta - \hat{\beta})]\\ &= C + \alpha^2\sigma^2\text{tr}\{X'X(X'X)^{-1}\}+(\beta - \alpha{\beta})'X'X (\beta - \alpha\beta)\\ &= C + \alpha^2\sigma^2\,p+(\alpha-1)^2{\beta}'X'X \beta \end{align*} Hence there seems to be a mistake in the formula as provided.


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