# Show that $\operatorname{E}[\beta_i\mid Y_i,\tau_i] = (1-u_i)Y_i$.

Let\begin{align} Y\mid\beta,\sigma^2&\sim N(X\beta, \sigma^2I)\\ \beta\mid\sigma^2,\tau_1^2,\ldots,\tau_p^2&\sim N(0,\sigma^2D), \,\,D = diag(\tau_1^2, \ldots, \tau_p^2)\\ \tau_j^2\mid\lambda&\stackrel{iid}{\sim}\operatorname{Exp}(\lambda^2/2) \end{align}

and suppose that $X = I, \sigma^2 = 1$ and $\lambda = 1$. I want to show that $$E[\beta_i\mid Y_i, \tau_i] = (1-u_i)Y_i$$ with $u_i = (1+\tau_i^2)^{-1}$. I think I need to compute the posterior density of $\beta_i$ by multiplying the prior of $\beta_i$ and the likelihood of $Y_i$, but I'm having some difficulties. I think we have the following prior of $\beta_i$ and likelihood of $Y_i$: $$p(\beta_i\mid\tau_i^2) = \dfrac{1}{\sqrt{2\pi}}\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2}\bigg), \,\,p(Y_i\mid\beta_i\sigma^2) = \dfrac{1}{\sqrt{2\pi}}\exp\bigg(-\dfrac{(Y_i -\beta_i)^2}{2}\bigg)$$

If this is correct, the posterior probability density function looks like this: $$p(\beta_i\mid Y_i,\tau_i) = \dfrac{1}{\sqrt{2\pi}}\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2}\bigg)\dfrac{1}{\sqrt{2\pi}}\exp\bigg(-\dfrac{(Y_i -\beta_i)^2}{2}\bigg) \\= \dfrac{1}{2\pi}\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2} - \dfrac{Y_i^2 - 2Y_i\beta_i + \beta_i^2}{2}\bigg)$$ This is the point at which I'm stuck; I don't see how the distribution corresponding to this density would have an expected value of $(1 - u_i)Y_i$...

Question: How do I solve this exercise? Am I going in the right direction?

Edit: I was wondering whether I'm allowed to just consider $$\dfrac{1}{2\pi}\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2} - \dfrac{Y_i^2 - 2Y_i\beta_i + \beta_i^2}{2}\bigg)\propto\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2} + Y_i\beta_i - \dfrac{\beta_i^2}{2}\bigg)$$ I don't think I am; the expectation of a constant times a normal distribution is not the same as the expectation of the normal distribution itself right? I thought I could mention it anyway since I might be mistaken.

You definitely went in the right direction. If we start form your second equation we find that $$\dfrac{1}{2\pi}\exp\bigg(-\dfrac{\beta_i^2}{2\tau_i^2} - \dfrac{Y_i^2 - 2Y_i\beta_i + \beta_i^2}{2}\bigg) \propto \exp\bigg(-\frac{1}{2} \frac{\tau_i^2+1}{\tau_i^2}\left(\beta_i - \frac{\tau_i^2}{\tau_i^2+1}Y_i\right)^2 \bigg)$$ From there you can easily derive the distribution of $\beta \mid Y,\tau$, which is normal and thus $$\beta \mid Y,\tau \sim N\left(D(D+I)^{-1}Y, D(D+I)^{-1}\right)$$
• Since $\beta_i\mid Y_i, \tau_i$ is normally distributed with known mean, you can use this to find $E[\beta_i\mid Y_i, \tau_i]$ (the expression I gave is more general). Regarding your edit, you are allowed to "throw away" factors that do not depend on $\beta$ since they do not determine the distribution. You can always find the "correct" distribution in the end by normalizing. – Tim Dikland Apr 25 '18 at 15:43