Question

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

(a) the expected number of matches.

(b) the variance of the number of matches

Short answer

In the given information the answers of part (a) is $E\left(X\right)=1$

part(b) is$Var\left(X\right)=1$

Step-by-step solution

Given Information (Part-a)

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$X=\sum _{i=1}^N{X}_i$

Calculation (Part-a)

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

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

$=n·P\left(X_1=1\right)$

$=1$

Final Answer(Part-a)

The answer is$EX=1$

Given Information (Part-b)

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$X=\sum _{i=1}^N{X}_i$

Calculation (Part-b)

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

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

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

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

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

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

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

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

:Final Answer (Part-b)

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