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Asking for $X \ge 0$ under any probability measure $P$, does $\int X \, dP = 1 \implies \int |X\log(X)|\,dP < \infty?$

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    $\begingroup$ The reason the space between $X$ and $\log$ was too small, so that you saw $X\text{log}(X)$ instead of $X\log(X),$ is that you typed X\text{log}(X) instead of X\log(X). I corrected that. $\endgroup$ – Michael Hardy Mar 8 at 19:25
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No. Consider

$$ P\left(X=\frac6{\pi^2}\cdot\frac{2^n}{n^2}\right)=2^{-n} $$

for all positive integers $n$. Then $E(X)=1$, but

\begin{eqnarray} E(|X\log X|) &=& \sum_{n=1}^\infty2^{-n}\cdot\frac6{\pi^2}\cdot\frac{2^n}{n^2}\log\left(\frac6{\pi^2}\cdot\frac{2^n}{n^2}\right) \\ &=& \sum_{n=1}^\infty\frac6{\pi^2}\cdot\frac1{n^2}\left(\log\frac6{\pi^2}+n\log2-2\log n\right) \\ &=& \infty\;, \end{eqnarray}

as the first and third term yield convergent series whereas the second one yields a multiple of the divergent harmonic series.

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