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I'm trying to see what conditions are necessary such that the following holds for a general random variable $X$: $$\mathrm{Pr}[X\ge E[X]] \ge \frac{1}{2}.$$ This seems to be intuitively true if the range of $X$ is somewhat "uniform". Is there anything more specific known for this condition to hold?

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up vote 4 down vote accepted

One would ask that the distribution of $X$ is symmetric with respect to its mean, that is, that ($X$ is integrable and that) there exists $m$ such that $m-X$ and $X$ are equidistributed.

Then the only possible value of $m$ (when $X$ is integrable) is $m=2\mathbb E(X)$ and the inequality $\mathbb P(X\geqslant \mathbb E(X))\geqslant\frac12$ follows from the remark that $\mathbb P(X\geqslant \mathbb E(X))=\mathbb P(A)+\mathbb P(B)$ where $A=[X\gt E(X)]$, $B=[X=E(X)]$, $C=[X\lt E(X)]$, and from the identities $\mathbb P(A)+\mathbb P(B)+\mathbb P(C)=1$ (always true) and $\mathbb P(A)=\mathbb P(C)$ (due to the symmetry).

In the absence of symmetry, such a bound cannot hold in general. For a simple counterexample, consider $X$ such that $\mathbb P(X=-1)=x/(1+x)$ and $\mathbb P(X=x)=1/(1+x)$ for some $x\gt0$. Then $\mathbb E(X)=0$ and $\mathbb P(X\geqslant0)=1/(1+x)$ can take every value in $(0,1)$.

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