8.
某同学在课外阅读时了解到概率统计中的切比雪夫不等式,该不等式可以使人们在随机变量X的期望E(X)和方差D(X)存在但其分布未知的情况下,对事件"|X-E(X)|\geqε”的概率作出上限估计,其中ε为任意正实数切比雪夫不等式的形式为:P(|X-E(X)|\geqε) \leq f(D(X), ε) ,其中f(D(X), ε)是关于D(X)和ε的表达式. 由于记忆模糊,该同学只能确定f(D(X),ε)的具体形式是下列四个选项中的某一种.请你根据所学相关知识,确定该形式是
A. D(X)\cdot ε^2
B. \dfrac{1}{D(X)\cdot ε^2}
C. \dfrac{ε^2}{D(X)}
D. \dfrac{D(X)}{ε^2}
解答：当X为离散型随机变量时：
P\left( {\left| {X - E\left( X \right)} \right| \ge \varepsilon } \right) = \sum\limits_{\left| {X - E\left( X \right)} \right| \ge \varepsilon } {{p_i}} \le \sum\limits_{\left| {X - E\left( X \right)} \right| \ge \varepsilon } {\frac{{{{\left( {X - E\left( X \right)} \right)}^2}}}{{{\varepsilon ^2}}}{p_i}} \le \sum\limits_{i = 1}^n {\frac{{{{\left( {X - E\left( X \right)} \right)}^2}}}{{{\varepsilon ^2}}}{p_i} = \frac{{D\left( X \right)}}{{{\varepsilon ^2}}}}
当X为连续型随机变量时，同理：
P\left( {\left| {X - E\left( X \right)} \right| \ge \varepsilon } \right) = \int\limits_{\left| {X - E\left( X \right)} \right| \ge \varepsilon } {{p_i}} \le \int\limits_{\left| {X - E\left( X \right)} \right| \ge \varepsilon } {\frac{{{{\left( {X - E\left( X \right)} \right)}^2}}}{{{\varepsilon ^2}}}{p_i}} \le \int\limits_{ - \infty }^{ + \infty } {\frac{{{{\left( {X - E\left( X \right)} \right)}^2}}}{{{\varepsilon ^2}}}{p_i} = \frac{{D(X)}}{{{\varepsilon ^2}}}}
显然答案选D
