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定义

\[P(X = k) = q^{k - 1}p, \quad k = 1,2,..., \ 0 < p < 1, \ q = 1 - p, \]

记为 \(X \sim G(p)\).

期望

\[EX = \frac{1}{p}. \]

证明

\[EX = \sum_{k = 1}^{\infty }kq^{k - 1}p = p\sum_{k = 1}^{\infty }kq^{k - 1} = p\sum_{k = 1}^{\infty }\frac{dq^{k}}{dq} = p \cdot \frac{d\displaystyle \sum_{k = 1}^{\infty }q^{k}}{dq} = p \cdot \frac{d \displaystyle \frac{q}{1 - q}}{dq} = \frac{p}{(1 - q)^2}, \]

所以,

\[EX = \frac{1}{p}. \]

方差

\[DX = \frac{1 - p}{p^2}. \]

posted @ 2020-11-23 01:16  模糊计算士  阅读(816)  评论(0编辑  收藏  举报