exercise:D9107332f5: Difference between revisions
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(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> The following related discrete problem also gives a good clue for the answer to Exercise Exercise. Randomly select with replacement <math>t_1</math>, <math>t_2</math>, \dots, <math>t_r</math> from the set <math>(1/n, 2/n,...") |
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The following related discrete problem also gives a good clue for the answer to Exercise [[exercise:2d6dc4d7a2 |Exercise]]. Randomly select with replacement <math>t_1</math>, <math>t_2</math>,..., <math>t_r</math> from the set <math>(1/n, 2/n, \dots, n/n)</math>. Let <math>X</math> be the smallest value of | |||
clue for the answer to Exercise [[exercise:2d6dc4d7a2 |Exercise]]. Randomly select with replacement | |||
<math>t_1</math>, <math>t_2</math>, | |||
<math>r</math> satisfying | <math>r</math> satisfying | ||
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</math> | </math> | ||
Then <math>E(X) = (1 + 1/n)^n</math>. To prove this, we can just as well choose | Then <math>E(X) = (1 + 1/n)^n</math>. To prove this, we can just as well choose | ||
<math>t_1</math>, <math>t_2</math>, | <math>t_1</math>, <math>t_2</math>, ..., <math>t_r</math> randomly with replacement from the set <math>(1, 2, | ||
\dots, n)</math> and let <math>X</math> be the smallest value of <math>r</math> for which | \dots, n)</math> and let <math>X</math> be the smallest value of <math>r</math> for which | ||
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t_1 + t_2 +\cdots+ t_r > n\ . | t_1 + t_2 +\cdots+ t_r > n\ . | ||
</math> | </math> | ||
<ul><li> Use | <ul style="list-style-type:lower-alpha"><li> Use [[exercise:173c44e9bd|Exercise]] to show that | ||
<math display="block"> | <math display="block"> | ||
Latest revision as of 17:17, 14 June 2024
The following related discrete problem also gives a good clue for the answer to Exercise Exercise. Randomly select with replacement [math]t_1[/math], [math]t_2[/math],..., [math]t_r[/math] from the set [math](1/n, 2/n, \dots, n/n)[/math]. Let [math]X[/math] be the smallest value of [math]r[/math] satisfying
[[math]]
t_1 + t_2 +\cdots+ t_r \gt 1\ .
[[/math]]
Then [math]E(X) = (1 + 1/n)^n[/math]. To prove this, we can just as well choose [math]t_1[/math], [math]t_2[/math], ..., [math]t_r[/math] randomly with replacement from the set [math](1, 2, \dots, n)[/math] and let [math]X[/math] be the smallest value of [math]r[/math] for which
[[math]]
t_1 + t_2 +\cdots+ t_r \gt n\ .
[[/math]]
- Use Exercise to show that
[[math]] P(X \geq j + 1) = {n \choose j}{\Bigl(\frac {1}{n}\Bigr)^j}\ . [[/math]]
- Show that
[[math]] E(X) = \sum_{j = 0}^n P(X \geq j + 1)\ . [[/math]]
- From these two facts, find an expression for [math]E(X)[/math]. This proof is due to Harris Schultz.[Notes 1]
Notes