Covariance $X$ and $e^{X}$

Question

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.

Answer 1

$$\begin{align} \operatorname{Cov}(X,e^X)&=E[Xe^X]-E[X]E[e^X]\\ &=E[Xe^X-\mu e^X]\\ &=E[(X-\mu)e^X] \\&=E[(X-\mu)(e^X-e^{\mu})] \end{align} $$

The last step comes from $E[(X-\mu) c] = 0$ for any constant $c$.

Because $e^x$ is non-decreasing (this is all we need), $(x-\mu)(e^x-e^{\mu})\geq0$ $\forall x$

Hence $\operatorname{Cov}(X,e^X) \ge 0$

Answer 2

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 $e^X$.

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