For the match problem (Example 5m in Chapter 2), find

(a) the expected number of matches.

(b) the variance of the number of matches

Define $X_i$as the indicator random variable that i th man has picked his hat,$i=1....,N$Because all hats are equally likely to be picked by i th man, we have that

$P\left(X_i=1\right)=\frac{1}{n}$

The number of matches (call it X) can be written as$X=\sum _{i=1}^N{X}_i$

$E\left(X\right)=\sum _{i}E\left(X_i\right)$

$=\sum _{i}P\left(X_i=1\right)$

$=n·P\left(X_1=1\right)$

$=1$

The answer is$EX=1$

Define $X_i$as the indicator random variable that i th man has picked his hat,$i=1....,N$ Because all hats are equally likely to be picked by i th man, we have that

$P\left(X_i=1\right)=\frac{1}{n}$

The number of matches (call it X) can be written as$X=\sum _{i=1}^N{X}_i$

$Var\left(X\right)=Var\left(\sum _i{X}_i\right)=\sum _{i}Var\left(X_i\right)+2\sum _{i<j}Cov\left(X_i,X_j\right.$

$=nVar\left(X_1\right)+2\left(\begin{array}{l}n\\ 2\end{array}\right)Cov\left(X_1,X_2\right)$

$Var\left(X_1\right)=\frac{1}{n}\left(1-\frac{1}{n}\right)=\frac{n-1}{n^2}$

$Cov\left(X_1,X_2\right)=E\left(X_1{X}_2\right)-E\left(X_1\right)E\left(X_2\right)$

$P\left(X_1=1,X_2=1\right)=\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}$

$Cov\left(X_1,X_2\right)=\frac{1}{n(n-1)}-\frac{1}{n^2}=\frac{1}{n^2(n-1)}$

$Var\left(X\right)=$$n·\frac{n-1}{n^2}+2·\frac{n(n-1)}{2}·\frac{1}{n^2(n-1)}$

$=\frac{n-1}{n}+\frac{1}{n}=1$

The answer is$Var\left(X\right)=1$