Markov's and Chebyshev's Inequalities

Last updated Dec 4, 2021 Edit Source

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$X$ 是一个非负的随机变量. 对于任意正实数 $a$ , 有 $$ P(X \geq a) \leq \frac{E(X)}{a} $$

• 证明里我们假设$X$是离散随机变量, 对于连续随机变量只需要将求和变成求积分即可.

按照定义, $E(X)=\sum_{x} x P(X=x)$. We’ll split this sum into two pieces, depending on whether or not $x \geq a$. $$ \begin{aligned} E(X) &=\sum_{x \geq a} x P(X=x)+\sum_{x<a} x P(X=x) \\ & \geq \sum_{x \geq a} a P(X=x)+0 \quad(\text { since in the first sum we assume } x \geq a) \\ &=a \sum_{x \geq a} P(X=x) \\ &=a P(X \geq a) \end{aligned} $$

Let $X$ be any random variable with finite expected value and variance. Then for every positive real number a, $$ P(|X-E(X)| \geq a) \leq \frac{\operatorname{Var}(X)}{a^{2}} . $$

We can prove it using Markov’s Inequality:

Let $Y=(X-E(X))^{2}$. Recall the definition of the variance of $X$: $$\operatorname{Var}(X)=\mathrm{E}\left[(X-E(X))^{2}\right]$$

Then $Y$ is a non-negative valued random variable with expected value $E(Y)=\operatorname{Var}(X)$. By Markov’s inequality, $$ P\left(Y \geq a^{2}\right) \leq \frac{E(Y)}{a}=\frac{\operatorname{Var}(X)}{a^{2}} . $$ But notice that the event $Y \geq a^{2}$ is the same as $|X-E(X)| \geq a$, so we conclude that $$ P(|X-E(X)| \geq a) \leq \frac{\operatorname{Var}(X)}{a^{2}} $$