Consider a positive random variable $X$ and call $E[X]$ its expectation. For any positive $a \in \mathbb{R}$, an upper bound for the probability of $P(X>a)$ is provided by the Markov's Inequality, $$ P(X>a) \leq \frac{E[X]}{a}, $$ Is there an analogous lower bound that is based only on the knowledge of the expectation?
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6$\begingroup$ After a Google en.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality $\endgroup$– JP McCarthyOct 16, 2014 at 15:56
1 Answer
Consider a distribution with $$P(X=m\delta)=1-\delta$$ $$P\left(X=m\tfrac{1-\delta+\delta^2}{\delta}\right)=\delta$$ then $E[X]=m$ but $P(X\gt a )$ can be made arbitrarily small by choosing $\delta \lt \frac{a}{m}$ and if necessarily smaller.
So to get something useful, you would need some constraint on $X$ such as a given maximum or a given variance.