二项分布方差推导

$\begin{aligned} \operatorname{var}(m) & =\mathrm{E}\left[m^2\right]-\mathrm{E}^2[m] \\ & =\sum_{m=0}^N \frac{m^2 N !}{(N-m) ! m !} \mu^m(1-\mu)^{N-m}-N^2 \mu^2 \\ & =\sum_{m=1}^N \frac{m^2 \mu N !}{(N-m) ! m !} \mu^{m-1}(1-\mu)^{N-m}-N^2 \mu^2 \\ & =\sum_{m=1}^N \frac{m \mu N !}{(N-m) !(m-1) !} \mu^{m-1}(1-\mu)^{N-m}-N^2 \mu^2\end{aligned}$

令 $u=m-1, T=N-1$ 代入上面

• $\begin{aligned} \operatorname{var}(m) & =\sum_{u=0}^T \frac{(u+1) \mu(T+1) !}{(T-u) ! u !} \mu^u(1-\mu)^{T-u}-N^2 \mu^2 \\ & =(T+1) \mu \sum_{u=0}^T u \frac{T !}{(T-u) ! u !} \mu^u(1-\mu)^{T-u}+(T+1) \mu \sum_{u=0}^T \frac{T !}{(T-u) ! u !} \mu^u(1-\mu)^{T-u}-N^2 \mu^2 \\ & =(T+1) T \mu^2+(T+1) \mu-N^2 \mu^2 \\ & =N(N-1) \mu^2+N \mu-N^2 \mu^2 \\ & =N \mu-N \mu^2 \\ & =N \mu(1-\mu)\end{aligned}$