Suppose that [math]n[/math] people have their hats returned at random. Let [math]X_i = 1[/math] if the [math]i[/math]th person gets his or her own hat back and 0 otherwise. Let [math]S_n = \sum_{i = 1}^n X_i[/math]. Then [math]S_n[/math] is the total number of people who get their own hats back. Show that

• [math]E(X_i^2) = 1/n[/math].
• [math]E(X_i \cdot X_j) = 1/n(n - 1)[/math] for [math]i \ne j[/math].
• [math]E(S_n^2) = 2[/math] (using (a) and (b)).
• [math]V(S_n) = 1[/math].