reverse Markov inequalityReverseMarkovInequality2008-02-13 02:45:502008-02-13 02:48:49Definition% this is the default PlanetMath preamble. as your knowledge
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Let $X$ be a random variable that satisfies $\Pr(X \leq a) = 1$ for some constant $a$.
Then, for $d < E[X]$,
\[
\Pr( X > d) \geq \frac{E[X] - d}{a - d}
\]
{\em Proof:}
Apply the Markov's inequality to the random variable $\tilde{X} = a-X$,
\[\Pr(X\leq d) = \Pr(\tilde{X}\geq a-d)
\leq \frac{E[\tilde{X}]}{a-d}
= \frac{a-E[X]}{a-d}.
\]
Hence
\[
\Pr(X> d) \geq 1 - \frac{a-E[X]}{a-d} = \frac{E[X]-d}{a-d}.
\]