Let (X,Y) be a bivariate distribution with $E[X] = E[Y] =0$, and let $f$ be a function with $|f(x)| + |f'(x)| \leq Ae^{\alpha|x|}, x \in \mathbb{R}$. Use conditional expectations and properties of normal distributions to show $E[Xf(Y)] = Cov(x,Y)E[f'(Y)]$.
Honestly, I havent gotten very far as this is totally mysterious to me. I've tried to use conditional expectations to do something: $$E[Xf(Y)] = E[E[Xf(Y)|X] = E[XE[f(Y)|X]]$$ or $$E[Xf(Y)] = E[E[Xf(Y)|Y] = E[f(Y)E[X|Y]] = E[f(Y) (\mu_x+ \frac{\rho\sigma_x}{\sigma_y}(Y-\mu_y)]$$ Using the formula for $\rho$, I can get this to: $$(\mu_x - \frac{Cov(X,Y)}{\sigma_Y^2}\mu_y)E[f(Y)] + \frac{cov(X,Y)}{\sigma_y^2}E[f(Y)Y]$$ But I don't think this is the right approach, I'm just blindly throwing formulas at it. How can I get $f'$ to come into it? If anyone could give me some pointers it would be fantastic! Thanks
This generalises to $X$ and $Y$ with non-zero mean. Suppose $X \sim N(\mu_X, \sigma^2_X)$, $Y \sim N(\mu_Y, \sigma_Y^2)$ and $\text{cov}(X, Y) = \rho$. First prove the following: $$ \mathbb E[f(X)(X - \mu_X)] = \sigma_X^2 \mathbb E[f'(X)] $$ Then prove that $$ \mathbb E[Y \mid X] = \mu_y + \frac{\rho}{\sigma_X^2}[X - \mu_X] $$ Then we have that \begin{align*} \text{cov}(f(X), Y) &= \mathbb E[(f(X) - \mathbb E[f(X)])(Y - \mu_Y)] \\ &= \mathbb E(\mathbb E[(f(X) - \mathbb E[f(X)])(Y - \mu_Y) \mid X]) \\ \end{align*} Now from this point use the first two formulas to show $\text{cov}(f(X), Y) = \mathbb E[f'(X)] \text{cov}(X, Y)$. This is Stein's lemma if you wish to research more.