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抛 n 次

What is the expected product of the number of heads and the number of tails?

专题
Probability / 概率
难度
L4

题目详情

概率题：抛 n 次：正面数×反面数的期望。

英文原题

A fair coin is tossed nn times. What is the expected product of the number of heads and the number of tails?

解析

设正面数为 HH，反面数为 T=n−HT=n-H，则

HT=H(n−H)=nH−H2.HT=H(n-H)=nH-H^2.

对公平硬币，H∼Bin(n,1/2)H\sim\mathrm{Bin}(n,1/2)，所以

E[H]=n2,qquadVar(H)=n4,qquadE[H2]=Var(H)+E[H]2=n4+n24.\mathbb{E}[H]=\frac{n}{2},\\qquad \mathrm{Var}(H)=\frac{n}{4},\\qquad \mathbb{E}[H^2]=\mathrm{Var}(H)+\mathbb{E}[H]^2=\frac{n}{4}+\frac{n^2}{4}.

因此

E[HT]=n⋅n2−(n4+n24)=n2−n4.\mathbb{E}[HT]=n\cdot\frac{n}{2}-\left(\frac{n}{4}+\frac{n^2}{4}\right)=\boxed{\frac{n^2-n}{4}}.

英文解析

HT=H(n−H)=nH−H2.HT=H(n-H)=nH-H^2. E[H]=n2,qquadVar(H)=n4,qquadE[H2]=Var(H)+E[H]2=n4+n24.\mathbb{E}[H]=\frac{n}{2},\\qquad \mathrm{Var}(H)=\frac{n}{4},\\qquad \mathbb{E}[H^2]=\mathrm{Var}(H)+\mathbb{E}[H]^2=\frac{n}{4}+\frac{n^2}{4}. E[HT]=n⋅n2−(n4+n24)=n2−n4.\mathbb{E}[HT]=n\cdot\frac{n}{2}-\left(\frac{n}{4}+\frac{n^2}{4}\right)=\boxed{\frac{n^2-n}{4}}.