Is there any way of simplifying the covariance of a function of a random variable and a random variable? For example, if we had random variables $X$ and $Y$ and we know that $corr(X,Y)=\rho$, how would you solve for $Cov(e^X,Y)$?
 A: The following is given,
$$
\int dxdy \ p(x, y)(x - \mu_x)(y - \mu_y) = \sigma_x\sigma_y\rho,
$$
where $\mu_x(\mu_y)$ and $\sigma_x(\mu_y)$ are the mean and standard deviation of $X(Y)$,
the requested covariance is given by
$$
\begin{aligned}
{\rm Cov}(e^X, Y) &= \int dxdy \  p(x, y)(e^x - \mu_{e^x})(y - \mu_y)\\
&=\int dxdy \  p(x, y)\left(1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots - \mu_{e^x}\right)(y - \mu_y)
\end{aligned}
$$
The mean $\mu_{e^x}$
$$
\begin{aligned}
\mu_{e^x} &= \int dxdy \  p(x, y) e^x\\
&= M_X(t=1)\\
&= 1 + \mu_x + \frac{1}{2!}m_{2,x} + \frac{1}{3!}m_{3,x} + \cdots
\end{aligned}
$$
where $M_X$(t) is the moment generating function (see the definition here), and $m_{n,x}$ is the $n$-th moment of $p_X(x) = \int dy \ p(x, y)$.
We can further simplify the expression as
$$
\begin{aligned}
{\rm Cov}(e^X, Y) &= \int dxdy \  p(x, y)(y - \mu_y)\left(1 + x + \frac{1}{2!}x^2 + \cdots - 1 - \mu_x - \frac{1}{2!}m_{2,x} - \cdots\right)\\
&= \sigma_x\sigma_y\rho + \sum_{n=2}^{\infty}\frac{1}{n!}\int dxdy \  p(x, y)(y - \mu_y)(x^n - m_{n,x}).
\end{aligned}
$$
Therefore we would need a lot more information to obtain the desired correlation.
