Q.4.25 - A First Course in Probability

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 

Step-by-Step Solution

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 1: 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 $X=\sum _{i=1}^N{X}_i$

Step2 : Calculation (Part-a)

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

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

          $=n·P\left(X_1=1\right)$

           $=1$

Step 3 : Final Answer(Part-a)

The answer is$EX=1$

Step 4 : 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 $X=\sum _{i=1}^N{X}_i$

Step 5 : 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)}$

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

Step 6 :Final Answer (Part-b)

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