Answer

Let \(X\) denote the number of people who pick their own hat. We have been asked \(E[X]\). Let \(X_{i}\) be a binary random variable denoting whether the \(i^{th}\) person picked their own hat, i.e.,

\[ \begin{alignat}{2} X_{i} &= \begin{cases} 1 &\mbox{if $i^{th}$ person picks their own hat}\newline 0 &\mbox{otherwise} \end{cases} \newline P(X_{i} = 1) &= \frac{1}{n} \newline E[X_{i}] &= 1 \times \frac{1}{n} + 0 \times (1 - \frac{1}{n}) = \frac{1}{n}\newline \end{alignat} \]

Consequently

\[ \begin{aligned} E[X] = E[\sum_{i=1}^{n} X_{i}] = \sum_{n=1}^{n}E[X_{i}] = 1 \end{aligned} \]

It is interesting to see the variance of X. Note that the formula for variance is \(E[X^{2}] - E[X]^{2}\). Thus,

\[ \begin{aligned} X^{2} = (\sum_{i=1}^{n} X_{i})^{2} = \sum_{i=1}^{n} X_{i}^{2} + \sum_{i=1}^{n} \sum_{j=1, j\neq i}^{n} X_{i}X_{j} \newline E[X^{2}] = \sum_{i=1}^{n}E[X_{i}^{2}] + \sum_{i=1}^{n} \sum_{j=1, j\neq i}^{n} E[X_{i}X_{j}] \end{aligned} \]

Note that \(X_{i}\) and \(X_{j}\) are not independent since after the first person has picked the hat, only \(n-1\) hats remain

\[ \begin{aligned} {3} X_{i}X_{j} &= \begin{cases} 1 &\mbox{if $X_{i} = X_{j} = 1$}\newline 0 &\mbox{otherwise} \end{cases} \newline P(X_{i}X_{j} = 1) &= P(X_{i} = 1) P(X_{j} = 1 \vert X_{i} = 1) &&= \frac{1}{n} \times \frac{1}{n-1}\newline E[X_{i}X_{j}] &= 1 \times (\frac{1}{n} \times \frac{1}{n-1}) + 0 \times (1 - \frac{1}{n} \times \frac{1}{n-1}) &&= \frac{1}{n(n-1)}\newline E[X_{i}^2] &= 1^{2} \frac{1}{n} + 0^{2} (1-\frac{1}{n}) &&= \frac{1}{n} \end{aligned} \]

Putting these values in the original equation for variance

\[ \begin{aligned} {2} E[X_{2}] &= n \frac{1}{n} + \frac{1}{n} \frac{1}{n-1} (\frac{n(n-1)}{2} \times 2) &&= 2\newline Var(X) &= 2 - 1^{2} &&= 1 \end{aligned} \]