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Let $W$ be a random variable such that $\mathbb{P}(W > 0) = 1$ and $\mathbb{E}(W) = 1$. Is there an interpretation or motivation for the condition $$\mathbb{E}(W \log (W) ) < c$$ where $c \in (0,\infty)$ is a positive constant? Perhaps if we additionally assume that $W$ is absolutely continuous (with respect to the Lebesgue measure)?

If $W$ is a log-normal variable for example, i.e. $W = \exp(\sigma X - \sigma^2 / 2)$ where $X$ is a standard normal random variable, the condition $$\mathbb{E}(W \log (W) ) < c$$ is equivalent to $\sigma^2 < 2*c$. This is somehow a condition that 'most of the mass is concentrated around one'. Can this motivation/interpretation be generalized?

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Assuming that $\mathrm E(W)=1$, Markov inequality yields $\mathrm P(W\geqslant w)\leqslant1/w$ for every $w\gt1$. Assuming further that $\mathrm E(W\log W)=c$ is finite, one gets $\mathrm P(W\geqslant w)\leqslant c/(w\log w)$ for every $w\gt1$, hence a faster convergence to zero of the tail of the distribution of $W$.

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up vote 0 down vote accepted

The best I came up with is the following:

The convex map $\Phi( s ) = \mathbb{E}(W^s)$ for $s \approx 1$ describes how the mass of $W$ is distributed around one. Now, $\Phi'(s) = \mathbb{E}( \log(W) W^s )$ (possibly under certain regularity conditions) and hence $$\mathbb{E}( \log(W) W ) < c$$ is a condition on the derivative of $\Phi$ at $s = 1$.

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