$\newcommand{\e}{\operatorname E}$"Obviously" if $g$ is a weakly increasing function and $X$ and $g(X)$ are both random variables with finite variance, then the covariance (and hence the correlation) between $X$ and $g(X)$ is non-negative. But there is the question of how to prove this "obvious" proposition. My question is whether something simpler than what I wrote should be used instead of what I wrote.
\begin{align}
\operatorname{cov}(X,g(X)) = {} & \e\Big( \big(X-\e(X)\big)\big(g(X) – \e(g(X))\big) \Big) \\[6pt]
= {} & \e\Big( \big(X-\e(X)\big) \big( (g(X) – g(\e(X))) + (g(\e(X)) – \e(g(X)))\big) \Big) \\[6pt]
= {} & \e\Big(\big(X-\e(X)\big)\big( g(X) – g(\e(X))\big)\Big) \\
& + \e\Big( \big( X-\e(X)\big)\big( \underbrace{g(\e(X)) – \e(g(X)} \big) \Big)
\end{align}
Now an essential point is that the expression over the $\underbrace{\text{underbrace}}$ is constant, i.e. not random. Therefore it can be pulled out of the outermost expectation operator just after the "plus" sign. Then we're left with
$$
\e\big( X-\e(X)\big)
$$
and that is zero. In the first term we now have the expected value of
$$
\big( X-\e(X)\big)\big( g(X) – g(\e(X))\big)
$$
and that random variable is nonnegative with probability $1.$
This leaves me feeling that I may have missed a simpler proof, just because the proposition stated just after the scare-quoted "Obviously" above seems obvious.
Did I miss a simpler proof?
Best Answer
Let $X_i$ be independent random variables with the same distribution as $X.$ Because $g$ is weakly increasing if and only if $(g(x_2)-g(x_1))(x_2-x_1)\ge 0$ for all real numbers $x_i,$
$$\operatorname{Cov}(X,g(X)) = \frac{1}{2} E[(X_2-X_1)(g(X_2)-g(X_1))] \ge \frac{1}{2}E[0] = 0,$$
QED. (See How would you explain covariance to someone who understands only the mean? for an explanation of this formula for covariance.)
An alternative is to exploit the basic invariance property of covariance with respect to changes of location. Observe that $g$ is weakly increasing if and only if $x\to g(x)+a$ is weakly increasing for any constant $a.$ Thus, without any loss of generality you may assume $x g(x)\ge 0$ for all $x$ by choosing $a = -g(0).$ Consequently, initially shifting $X$ if necessary to make $E[X]=0,$
$$\operatorname{Cov}(X,g(X)) = E[X g(X)]\ge E[0] = 0,$$
QED.
Illustrating both approaches is this visual proof without words.