Revision as of 03:24, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> 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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
BBy Bot
Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

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].