Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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 $W$.

share|improve this answer
add comment
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 $s = 1$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.