Prove equation $E(h(X)\,e^{-Y})=E(e^{-Y})\, E(h(X-\operatorname{Cov}(X,Y)))$ Prove:
$$
E\bigl(h(X)\, e^{-Y}\bigr) 
= E\bigl(e^{-Y}\bigr) \, E\bigl(h(X-\operatorname{Cov}(X,Y))\bigr)
$$
when $X$, $Y$ are both normally distributed, $h(X)$ is a function of $X$.
I think it can be proved by definition and maybe conditional distribution. But I am stuck in the calcultation of conditional expected value of $h(X)$ given $Y$ and cannot get result I want to see. Maybe this idea is wrong. Can you solve this problem or give a reachable idea? Thank you very much!
 A: In this answer, we will assume that $X$ and $Y$ are jointly normal (i.e., their joint distribution is multivariate normal). To begin with, define $\tilde{Y}$ by
$$\tilde{Y} = Y - \alpha X, \qquad \alpha = \frac{\mathbf{Cov}(X,Y)}{\mathbf{Var}(X)}.$$
Then $\mathbf{Cov}(X, \tilde{Y}) = 0$, and so, $\tilde{Y}$ and $X$ are independent. (The above construction of $\tilde{Y}$ is essentially the Gram–Schmidt orthogonalization.) Then
\begin{align*}
\frac{\mathbf{E}[h(X)e^{-Y}]}{\mathbf{E}[e^{-Y}]}
= \frac{\mathbf{E}[h(X)e^{-(\tilde{Y} + \alpha X)}]}{\mathbf{E}[e^{-(\tilde{Y} + \alpha X)}]}
= \frac{\mathbf{E}[h(X)e^{-\alpha X}]}{\mathbf{E}[e^{-\alpha X}]}.
\end{align*}
Notice that the last expression corresponds to the exponential tilting. In particular, if we write $X \sim \mathcal{N}(\mu, \sigma^2)$, then it is the same as $\mathbf{E}[h(X')]$ with $X' \sim \mathcal{N}(\mu - \alpha\sigma^2, \sigma^2)$, or equivalently,
$$ = \mathbf{E}[h(X - \alpha \sigma^2)] = \mathbf{E}[h(X - \mathbf{Cov}(X,Y))]. $$
(Note that $\alpha\sigma^2 = \mathbf{Cov}(X, Y)$.) This result can also be verified by a direct computation. Using the MGF of the normal distribution, we get $\mathbf{E}[e^{-\alpha X}] = e^{-\alpha \mu + \sigma^2\alpha^2/2}$. So,
\begin{align*}
\frac{\mathbf{E}[h(X)e^{-\alpha X}]}{\mathbf{E}[e^{-\alpha X}]}
&= e^{\alpha \mu - \sigma^2\alpha^2/2} \int_{\mathbb{R}} h(x) e^{-\alpha x} \frac{e^{-(x-\mu)^2/2\sigma^2}}{\sqrt{2\pi}\sigma} \, \mathrm{d}x \\
&= \int_{\mathbb{R}} h(x) \frac{e^{-(x-\mu+\alpha\sigma^2)^2/2\sigma^2}}{\sqrt{2\pi}\sigma} \, \mathrm{d}x \\
&= \int_{\mathbb{R}} h(x - \alpha\sigma^2) \frac{e^{-(x-\mu)^2/2\sigma^2}}{\sqrt{2\pi}\sigma} \, \mathrm{d}x \\
&= \mathbf{E}[h(x - \alpha\sigma^2)].
\end{align*}
