In our discussion of the bivariate normal, there is an expression for $E(Y|X = x)$.
a. By reversing the roles of X and Y give a similar formula for $E(X|Y = y)$.
b. Both $E(Y|X = x)$ and $E(X|Y = y) $are linear functions. Show that the product of the two slopes is $\rho^2$.
My textbook defines the definition of the expected value of $Y$ as:
$E(Y|X=x)=\mu_2+\rho\sigma_2\frac{x-\mu_1}{\sigma_1}. $ For part a) I am not sure if this expected value flips or if the sigma's and the mu's swap. I am also not sure how to prove that the products of the slopes is $\rho^2$. Any suggestions?
Just to make things clear, we have $$\begin{pmatrix} X \\ Y \end{pmatrix} \sim N\left(\begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix},\begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2\end{pmatrix}\right),$$ where $\rho = \frac{\sigma_{12}}{\sigma_1\sigma_2}$.
Given $E[Y\,|\,X=x]$ to find $E[X\,|\,Y=y]$ we just swap over the covariances, means and the variable appropriately - this hopefully is intuitive. That is, $E[X\,|\,Y=y] = \mu_1 + \rho\frac{\sigma_1}{\sigma_2}(y-\mu_2).$
Now that we have computed $E[X\,|\,Y=y]$ it should now be simple to finish the other part. All you need to do is find the slopes (i.e. the gradient) of the lines and multiply them. Their product should be $\rho^2$.