二项分布的期望方差证明

二项分布的期望方差证明

$$\begin{gathered} P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k},k=0,1,2,..,n,q=1-p \\ EX=\sum _{k=0}^{n}k\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k} \\ =\sum _{k=1}^{n}k\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k} \\ =\sum _{k=1}^{n}k\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k} \\ =np\sum _{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}{p}^{k-1}{q}^{(n-1)-(k-1)} \\ =np\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)p^{k-1}{q}^{(n-1)-(k-1)} \\ =np[(\genfrac{}{}{0ex}{}{n-1}{0})p^0{q}^{n-1}+(\genfrac{}{}{0ex}{}{n-1}{1})p^1{q}^{n-2}+...+(\genfrac{}{}{0ex}{}{n-1}{n-1})p^{n-1}{q}^0] \\ =np \end{gathered}$$

因为：$DX=E{X}^2-(EX{)}^2$

且，
$$\begin{gathered} E{X}^2=\sum _{k=1}^n{k}^2\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k},k=0,1,2,..,n,q=1-p \\ =\sum _{k=1}^n[k(k-1)+k]\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k} \\ =\sum _{k=1}^{n}k(k-1)\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}+\sum _{k=1}^{n}k\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k} \\ 其中，\sum _{k=1}^{n}k\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}=EX=np \\ \sum _{k=1}^{n}k(k-1)\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k} \\ =\sum _{k=1}^{n}k(k-1)\frac{n!}{k!(n-k)!}{p}^2{p}^{k-2}{q}^{n-k} \\ =\sum _{k=2}^{n}k(k-1)\frac{n!}{k!(n-k)!}{p}^2{p}^{k-2}{q}^{n-k} \\ 注：特别注意这里k=1时项为0，所以可以从k=2开始计算。 \\ =\sum _{k=1}^n\frac{n(n-1)(n-2)!}{(k-2)!(n-k)!}{p}^2{p}^{k-2}{q}^{[(n-2)-(k-2)]} \\ =n(n-1)p^2\sum _{k=2}^n\frac{(n-2)!}{(k-2)!(n-k)!}{p}^{k-2}{q}^{[(n-2)-(k-2)]} \\ =n(n-1)p^2\sum _{k=2}^n\left(\genfrac{}{}{0ex}{}{n-2}{k-2}\right)p^{k-2}{q}^{[(n-2)-(k-2)]} \\ =n(n-1)p^2 \\ \to E{X}^2=n(n-1)p^2+np \\ \to DX=E{X}^2-(EX{)}^2=np-n{p}^2=np(1-p) \end{gathered}$$

核心思想是转化为更小规模的组合数，这里没法直接用幂级数的和函数求解思路。

END.