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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.

Golden_Ratio
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    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. – Nicolas Bourbaki Mar 16 '23 at 05:30

1 Answers1

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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:

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

User1865345
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Ben
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