Question

3) 100 people are all wearing name tags, and no two people share the same name. They decide to randomly permute their name tags. We will compute the expected number of people who get their name tag back, as well as the variance a) Let X be the indicator random variable for the it person getting their original name tag back. Find EX and E equal to j separately) consider the cases i and i not b) Let X be the number of people who get their original name tag. Write X in terms of the random variables X c) Find E[X] using linearity of expectation. d) Find E[X2] using linearity of expectation, and compute Var(x)

Answer 1

We know different people can permute their name tags in ways.

Let the event $A_i$ be the event that $i\left ( 1\leqslant i\leqslant n \right )$ th man picks his own name tag He can get his hat in $\left (n-1 \right )!$ ways.

Then $P\left ( A_i \right )=\frac{\left (n-1 \right )!}{n!}=\frac{1}{n},1\leqslant i\leqslant n$

a)Since $X_i$ is the indicator random variable of the event $A_i$ then

$P\left (X_i =1\right )=P\left ( A_i \right )=\frac{1}{n}\\ P\left (X_i =0\right )=1-\frac{1}{n};1\leqslant i\leqslant n$

So

$E\left ( X_i \right )=1\times P\left (X_i =1\right )+0\times P\left (X_i =0\right )\\ E\left ( X_i \right )=\left (\frac{1}{n}\right )+0\left (1-\frac{1}{n}\right )\\ E\left ( X_i \right )=\frac{1}{n}\\ {\color{Blue} E\left ( X_i \right )=\frac{1}{100};1\leqslant i\leqslant 100}$

Now probability that both and person get their own tags is

$P\left (A_i\cap A_j \right ) =\frac{\left (n-2 \right )!}{n!}\\ P\left (A_i\cap A_j \right ) =\frac{1}{n\left ( n-1 \right )}\\ P\left (A_i\cap A_j \right ) =\frac{1}{9900} ;1\leqslant i,j\leqslant 100,i\neq j$

So $P\left (X_i=1\cap X_j=1 \right ) =\frac{1}{9900} ;1\leqslant i,j\leqslant 100,i\neq j$

Now the expectation,

$E\left ( X_iX_j \right )=1\times P\left (X_i=1\cap X_j=1 \right ) +0\times P\left (\overline{X_i=1\cap X_j=1 }\right ) \\ {\color{Blue} E\left ( X_iX_j \right ) =\frac{1}{9900} ;1\leqslant i,j\leqslant 100,i\neq j}\\ {\color{Blue} E\left ( X_iX_j \right ) =\frac{1}{100} ;1\leqslant i,j\leqslant 100,i=j}$

Hence,

$E\left ( X_i^2\right ) =\frac{1}{100} ;1\leqslant i\leqslant 100$

b) The number of people who gets their own tag in terms of the indicator random variables $X_i$ is

${\color{Blue} X=\sum_{i=1}^{100}X_i}$

c) Using linearity of expectation,

$E\left ( X \right )=\sum_{i=1}^{100}E\left (X_i \right )\\ E\left ( X \right )=\sum_{i=1}^{100}\frac{1}{100}\\ {\color{Blue} E\left ( X \right )=1}$

d) Now,

$X^2=\left (\sum_{i=1}^{100}X_i \right )^2\\ X^2=\sum_{i=1}^{100}X_i ^2+2\sum_{i=1}^{100}\sum_{j=1,j<i}^{100}X_iX_j$

Using linearity of expectation,

$E\left (X^2 \right )=\sum_{i=1}^{100}E\left (X_i ^2 \right )+2\sum_{i=1}^{100}\sum_{j=1,j<k}^{100}E\left (X_iX_j \right )\\ E\left (X^2 \right )=\frac{100}{100}+2\binom{100}{2}\frac{1}{9900}\\ E\left (X^2 \right )=\frac{100}{100}+\frac{9900}{9900}\\ {\color{Blue} E\left (X^2 \right )=2}$

$Var\left ( X \right )=E\left (X^2 \right )-E\left ( X \right )^2\\ Var\left ( X \right )=2-1^2\\ {\color{Blue} Var\left (X \right )=1}$

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