04 - Markov's Inequality

Markov's Inequality

Let $X$ be random variable taking non-negative values so $\mathbb{P}(X\ge 0)=1$ then
For any $z>0$

$$\mathbb{P}(X\ge z)\le \frac{\mathbb{E}[x]}{z}$$

Proof

Consider random variable

$$X_z=z{1}_{\{X\ge z\}}$$

Hence

1. $X_z$ takes value $0$ whenever $X$ in $[0,z)$
2. $X_z$ takes value $z$ whenever $X$ in $z,\infty )$

Thus

$$X\ge X_z$$

Hence

$$\mathbb{E}[X]\ge \mathbb{E}[X_z]=z\mathbb{E}[1_{\{X\ge z\}}]=z\mathbb{P}(X\ge z)$$

So result follows