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Given any probability distribution $D$ that does not put probability $1$ on a single point, and let $X$ be a random variable drawn from this distribution. Is it true that there always exist positive constant $p,r$ such that $$\text{Pr}[X>(1+r)E[X]]>p?$$

Clearly, this fails for a distribution that puts probability $1$ on a single point. But it works for the uniform distribution, binomial, Poisson, exponential, and other distributions that I can think of, including distributions that put masses of probability on individual (but more than one) points. So I suspect that the statement is true. However, I am not sure how to prove it rigorously.

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    $\begingroup$ Go by contradiction. Suppose the above is false for all positive $p$ and $r$. Then $P(X>E(X))=0$ and hence $P(X<E(X))=0$ and hence $P(X=E(X))=1$. You can convert it to a constructive argument starting with $P(X>E(X))>0$. $\endgroup$ – A.S. Nov 7 '15 at 16:16
  • $\begingroup$ @A.S. You might want to write your solution as an answer. $\endgroup$ – Element118 Dec 29 '15 at 13:27

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