TITLE: Covariance $X$ and $e^{X}$
QUESTION [5 upvotes]: How can I prove that covariance of random variable $X$ and $e^{X}$ is non-negative regardless distribution of $X$. I assume it is true.

REPLY [5 votes]: Suppose that $X$ and $Y$ are two independent and identically distributed random variables. Since $e^x$ is an increasing function, $X-Y$ and $e^{X}-e^Y$ are co-sign. So the following holds:
$$
\mathbb E[(X-Y)(e^X-e^Y)]\geq 0.
$$
Using the assumption that $X$ and $Y$ are identically distributed, we have $\mathbb E(e^X)=\mathbb E(e^Y)$ and $\mathbb E(X)=\mathbb E(Y)$. Hence:
$$
\mathbb E[(X-Y)(e^X-e^Y)]=2\mathbb E(Xe^X)-2\mathbb E(X)\mathbb E(e^X)\geq 0.
$$
The last one is two times the covariance of $X$ and $e^X$.

The statement can be generalized for any increasing function $f$. The reverse holds for any decreasing function.