几何分布 (Geometric Distribution)

本文内容参考了此处.

若随机变量 $X$服从参数为 $p$的几何分布, 则有
$\Pr(X=k) = (1 - p)^kp$
其中 $k=0, 1, \cdots$

1. $E(X) = \frac{1-p}{p}$
   证明:
   $E(X) = \sum_{k=0}^{+\infty}k\Pr(X=k) = \sum_{k=0}^{+\infty}k(1-p)^kp$
   $=p(1-p)\sum_{k=0}^{+\infty}k(1-p)^{k-1}$
   令 $q=1-p$, 则有
   $\sum_{k=0}^{+\infty}k(1-p)^{k-1}$
   $=\sum_{k=1}^{+\infty}kq^{k-1}$
   $=[\sum_{k=1}^{+\infty}q^k]'$
   $=[\frac{q}{1-q}]'$
   $=\frac{1}{p^2}$
   $\therefore E(X)=p(1-p)\cdot\frac{1}{p^2} = \frac{1 - p}{p}$

2. $Var(X) = \frac{1-p}{p^2}$
   证明:
   令 $q = 1-p$, 则有
   $E(X^2) = \sum_{k=0}^{+\infty}k^2pq^k$
   $=pq\sum_{k=1}^{+\infty}k^2q^{k-1}$
   $=pq(\sum_{k=1}^{+\infty}k(k+1)q^{k-1} - \sum_{k=1}^{+\infty}kq^{k-1})$
   $=pq((\sum_{k=1}^{+\infty}q^{k+1})'' - \frac{1}{p^2})$
   $=pq((\frac{q^2}{1-q})'' - \frac{1}{p^2})$
   $=pq(\frac{2}{p^3} - \frac{1}{p^2})$
   $=\frac{2q - pq}{p^2}$
   $\therefore Var(X) = E(X^2) - \mu^2$
   $=\frac{2q -pq}{p^2} - \frac{q^2}{p^2}$
   $=\frac{q}{p^2}=\frac{1-p}{p^2}$