Markov's inequality

• If $X$ is random variable that takes only non-negative values then for any value $a>0$, then $P(X\ge a)\le \frac{E[X]}{a}$
• Proof:
  • Let $I$ be indicator variable for $X\ge a$
  • $P(X\ge a)=E[I]$
  • As $X\ge a$ when $I=1$ and $X\ge 0$ when $I=0\to I\le \frac{X}{a}\to E[I]\le E[\frac{X}{a}]$
  • $P(X\ge a)\le \frac{E[X]}{a}$

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