马尔可夫不等式

2. 定义

${\begin{array}{c} P(\boldsymbol{X} \geq a) \leq \frac{E(\boldsymbol{X})}{a} \end{array}}$

证明

${\begin{array}{c} E(\boldsymbol{X}) = \int_{-\infty}^{\infty} xf(x) \mathrm{d}x \end{array}}$

${\begin{array}{rl} E(\boldsymbol{X}) & = \int_{-\infty}^{\infty} xf(x) \mathrm{d}x = \int_{0}^{\infty} xf(x) \mathrm{d}x \\ & = \int_{0}^{a} xf(x) \mathrm{d}x + \int_{a}^{\infty} xf(x) \mathrm{d}x \\ & \geq \int_{0}^{\infty} xf(x) \mathrm{d}x \\ & \geq \int_{a}^{\infty} af(x) \mathrm{d}x \\ & = a P(\boldsymbol{X} \geq a) \end{array}}$

${\begin{array}{c} P(\boldsymbol{X} \geq a) \leq \frac{E(\boldsymbol{X})}{a} \end{array}}$

3. 推论

3.1 切比雪夫不等式

${\begin{array}{c} P((\boldsymbol{X} - E(\boldsymbol{X}))^2 \geq a^2) \leq \frac{Var(\boldsymbol{X})}{a^2} \\ \Downarrow \\ P(|\boldsymbol{X} - E(\boldsymbol{X})| \geq a) \leq \frac{Var(\boldsymbol{X})}{a^2} \end{array}}$

3.2 分位函数

${\begin{array}{c} Q_{\boldsymbol{X}}(1-p) \leq \frac{E(\boldsymbol{X})}{p} \end{array}}$

证明

${\begin{array}{c} p \leq 1 - P(\boldsymbol{X} \lt Q_{\boldsymbol{X}}(1-p)) = P(\boldsymbol{X} \geq Q_{\boldsymbol{X}}(1-p)) \leq \frac{E(\boldsymbol{X})}{Q_{\boldsymbol{X}}(1-p)} \end{array}}$