# Sign of Correlation between $X$ and $\log X$

Suppose $$\text{supp}(X)\subseteq \mathbb{R}_{\geq 1}.$$ Can we say $$\text{Cov}(X,\log X)\geq 0?$$

On one hand, we can say by monotonicity of log and Jensen's inequality that $$X\geq E[X]\implies \log X\geq \log E[X]\geq E[\log X].\quad (1)$$

Now if it also holds that $$\log X\geq E[\log X]\implies X\geq E[X]\quad (2)$$

then $$\text{sign}(X-E[X])=\text{sign}(\log X-E[\log X])$$ and we are done, but I don't think $$(2)$$ necessarily holds.

• Here is a suggestion. Try to prove/disprove it when $X$ is discrete. If it is true for in the discrete case, then you can try to generalize by using discrete approximations. In other words, find $(X_n)_{n\geq 0}$, a sequence of random variables, which are monotonically increasing, and converge to $X$, then apply the monotone convergence theorem. Mar 16, 2023 at 5:30

Your sign requirement does not necessarily hold, but it's still possible to prove the result using an alternative method. Since $$x \log x$$ is convex and $$\log x$$ is concave (over the stipulated range), Jensen's inequality gives:

\begin{align} \mathbb{E}(X) \log(\mathbb{E}(X)) &\leqslant \mathbb{E}(X\log X), \\[6pt] \log(\mathbb{E}(X)) &\geqslant \mathbb{E}(\log X). \\[6pt] \end{align}

Applying each of these inequalities (in order) we get:

\begin{align} \mathbb{Cov}(X,\log X) &= \mathbb{E}(X \log X) - \mathbb{E}(X) \mathbb{E}(\log X) \\[6pt] &\geqslant \mathbb{E}(X) \log(\mathbb{E}(X)) - \mathbb{E}(X) \mathbb{E}(\log X) \\[6pt] &\geqslant \mathbb{E}(X) \mathbb{E}(\log X) - \mathbb{E}(X) \mathbb{E}(\log X) \\[6pt] &= 0. \\[6pt] \end{align}

• very nice...thanks ben! Mar 16, 2023 at 6:15
• Ben, I have added a helpful MathSE link. Hope you don't mind. Mar 16, 2023 at 6:25
• @User1865345: Don't mind at all.
– Ben
Mar 16, 2023 at 6:27