reverse Markov inequality

Let $X$ be a random variable that satisfies $\mathrm{Pr}(X\le a)=1$ for some constant $a$. Then, for ,

$$\mathrm{Pr}(X>d)\ge \frac{E[X]-d}{a-d}$$

Proof: Apply the Markov’s inequality to the random variable $\tilde{X}=a-X$,

$$\mathrm{Pr}(X\le d)=\mathrm{Pr}(\tilde{X}\ge a-d)\le \frac{E[\tilde{X}]}{a-d}=\frac{a-E[X]}{a-d}.$$

Hence

$$\mathrm{Pr}(X>d)\ge 1-\frac{a-E[X]}{a-d}=\frac{E[X]-d}{a-d}.$$

Title	reverse Markov inequality
Canonical name	ReverseMarkovInequality
Date of creation	2013-03-22 17:48:08
Last modified on	2013-03-22 17:48:08
Owner	kshum (5987)
Last modified by	kshum (5987)
Numerical id	6
Author	kshum (5987)
Entry type	Definition
Classification	msc 60A99
Related topic	MarkovsInequality