引理 5.1.1 (切比雪夫不等式)

引理 5.1.1 (切比雪夫 (Чебышев) 不等式)

设随机变量 $X$ 的方差存在,则对任 $\epsilon >0$ 有

$$P\left(\left|X-E\left[X\right]\right|\ge \epsilon \right)\le \frac{\mathrm{Var}\left[X\right]}{{\epsilon }^2}.\text{\hspace{1em}(5.1.4)}$$

证明

只就 $X$ 为连续型随机变量的情形证明.

$$\begin{array}{rl}& P\left(\left|X-E\left[X\right]\right|\ge \epsilon \right)\\ & ={\int }_{|x-E\left[X\right]|\ge \epsilon }{f}_X\left(x\right)\mathrm{d}x\\ & \le {\int }_{|x-E\left[X\right]|\ge \epsilon }\frac{{\left(x-E\left[X\right]\right)}^2}{{\epsilon }^2}{f}_X\left(x\right)\mathrm{d}x\\ & \le {\int }_{-\infty }^{\infty }\frac{{\left(x-E\left[X\right]\right)}^2}{{\epsilon }^2}{f}_X\left(x\right)\mathrm{d}x\\ & =\frac{1}{{\epsilon }^2}{\int }_{-\infty }^{\infty }{\left(x-E\left[X\right]\right)}^2{f}_X\left(x\right)\mathrm{d}x\\ & =\frac{\mathrm{Var}\left[X\right]}{{\epsilon }^2}.\end{array}$$