概率期望小结论

对于一个概率 $p$，设它能提供的期望值为命中此概率的次数。那么保持这个概率直至命中此概率的期望值为 $\frac{1}{p}$

证明：

$$\begin{array}{rl}\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast p\ast i& =p\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast i\end{array}$$

先省略前面的 $p$， 看后面的部分。

$$\begin{array}{rl}\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast i& =1+2(1-p)+3(1-p{)}^2+4(1-p{)}^3+\cdots +\mathrm{\infty }(1-p{)}^{\mathrm{\infty }-1}\\ & =(1+(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1})+((1-p)+2(1-p{)}^2+\cdots +(\mathrm{\infty }-1)(1-p{)}^{\mathrm{\infty }-1})\end{array}$$

设 $A=1+(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1}$，

$B=(1-p)+2(1-p{)}^2+\cdots +(\mathrm{\infty }-1)(1-p{)}^{\mathrm{\infty }-1}$

$$\begin{array}{rl}A& =1+(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1}\\ (1-p)A& =(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1}+(1-p{)}^{\mathrm{\infty }}\\ (1-p)A-A& =(1-p{)}^{\mathrm{\infty }}-1\\ -pA& =(1-p{)}^{\mathrm{\infty }}-1\\ A& =\frac{(1-p{)}^{\mathrm{\infty }}-1}{-p}\end{array}$$

$\because 0\le p\le 1$

$\therefore (1-p{)}^{\mathrm{\infty }-1}\approx 0$

可得：$A=\frac{-1}{-p}=\frac{1}{p}$

再看 $B$ ：

$$\begin{array}{rl}B& =(1-p)+2(1-p{)}^2+\cdots +(\mathrm{\infty }-1)(1-p{)}^{\mathrm{\infty }-1}\\ & =((1-p)+(1-p{)}^2++\cdots +(1-p{)}^{\mathrm{\infty }-1})+((1-p{)}^2+(1-p{)}^3+\cdots +(\mathrm{\infty }-2)(1-p{)}^{\mathrm{\infty }-1})\end{array}$$

设 $C=(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1}$，$D=(1-p{)}^2+(1-p{)}^3+\cdots +(\mathrm{\infty }-2)(1-p{)}^{\mathrm{\infty }-1}$

可用等比数列求得：

$$C=\frac{(1-p)}{p}$$

而 $D$ 可以继续按上述方法分解。

整个式子分解得到：

$$\begin{array}{rl}\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast i& =\frac{1}{p}+\frac{(1-p)}{p}+\frac{(1-p{)}^2}{p}+\cdots +\frac{(1-p{)}^{\mathrm{\infty }-1}}{p}\\ & =\frac{1+(1-p)+(1-p{)}^2+(1-p{)}^3+\cdots +(1-p{)}^{\mathrm{\infty }-1}}{p}\end{array}$$

继续使用等比数列，设 $R=1+(1-p)+(1-p{)}^2+(1-p{)}^3+\cdots +(1-p{)}^{\mathrm{\infty }-1}$

$$\begin{array}{rl}R& =1+(1-p)+(1-p{)}^2+(1-p{)}^3+\cdots +(1-p{)}^{\mathrm{\infty }-1}\\ (1-p)R& =(1-p)+(1-p{)}^2+\cdots +(1-p{)}^{\mathrm{\infty }-1}+(1-p{)}^{\mathrm{\infty }}\\ (1-p)R-R& =(1-p{)}^{\mathrm{\infty }}-1\\ -pR& =(1-p{)}^{\mathrm{\infty }}-1\\ R& =\frac{(1-p{)}^{\mathrm{\infty }}-1}{-p}\end{array}$$

$\because 0\le p\le 1$

$\therefore (1-p{)}^{\mathrm{\infty }-1}\approx 0$

可得：

$$\begin{array}{rl}R& =\frac{-1}{-p}=\frac{1}{p}\\ \sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast i& =\frac{R}{p}\\ & =\frac{1}{p^2}\end{array}$$

将前面省略的 $p$ 加上：

$$\begin{array}{rl}\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast p\ast i& =p\sum _{i=1}^{\mathrm{\infty }}(1-p{)}^{i-1}\ast i\\ & =p\ast \frac{1}{p^2}\\ & =\frac{1}{p}\end{array}$$

得证。

将此结论运用于题目：

Q：求 $1$~$n$ 中随机选取一个整数，可以重复，整数 $x\in {\mathbb{Z}}^{+},1\le x\le n$ ，求 $x$ 被随机选取到的期望次数是多少？

就可列出式子：

$$E=\sum _{i=1}^{\mathrm{\infty }}(1-\frac{1}{n}{)}^{i-1}\ast \frac{1}{n}\ast i=n$$

期望次数就为 $n$。