exercise:12d2388018: Difference between revisions
From Stochiki
(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...") |
No edit summary |
||
| Line 1: | Line 1: | ||
Suppose that <math>n</math> people have their hats returned at random. | |||
Let | 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>X_i = 1</math> if the <math>i</math>th person gets his or her own hat back and 0 otherwise. Let | ||
| Line 11: | Line 5: | ||
\sum_{i = 1}^n X_i</math>. Then <math>S_n</math> is the total number of people who get their own hats | \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 | back. Show that | ||
<ul><li> <math>E(X_i^2) = 1/n</math>. | <ul style="list-style-type:lower-alpha"><li> <math>E(X_i^2) = 1/n</math>. | ||
</li> | </li> | ||
<li> <math>E(X_i \cdot X_j) = 1/n(n - 1)</math> for <math>i \ne j</math>. | <li> <math>E(X_i \cdot X_j) = 1/n(n - 1)</math> for <math>i \ne j</math>. | ||
Latest revision as of 21:06, 14 June 2024
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].