exercise:85a2ae80e0: 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> In the gambler's ruin problem, assume that the gambler initial stake is 1 dollar, and assume that her probability of success on any one game is <math>p</math>. Let <math>T</math> be the number of games until 0 is reached (the gambler is ruined)....") |
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\newcommand{\secstoprocess}{\all} | \newcommand{\secstoprocess}{\all} | ||
\newcommand{\NA}{{\rm NA}} | \newcommand{\NA}{{\rm NA}} | ||
\newcommand{\mathds}{\mathbb}</math></div> In the gambler's ruin problem, assume that the gambler initial stake | \newcommand{\mathds}{\mathbb}</math></div> In the gambler's ruin problem, assume that the gambler initial stake is 1 dollar, and assume that her probability of success on any one game is <math>p</math>. Let <math>T</math> be the | ||
is 1 dollar, and assume that her probability of success on any one game is <math>p</math>. Let <math>T</math> be the | |||
number of games until 0 is reached (the gambler is ruined). Show that the generating | number of games until 0 is reached (the gambler is ruined). Show that the generating | ||
function for <math>T</math> is | function for <math>T</math> is | ||
| Line 17: | Line 16: | ||
h(1) = \left \{ \begin{array}{ll} | h(1) = \left \{ \begin{array}{ll} | ||
q/p, & \mbox{if $q \leq p$}, \\ | q/p, & \mbox{if $q \leq p$}, \\ | ||
1, & \mbox{if | 1, & \mbox{if $q \geq p,$} | ||
\end{array} | \end{array} | ||
\right. | \right. | ||
| Line 26: | Line 25: | ||
h'(1) = \left \{ \begin{array}{ll} | h'(1) = \left \{ \begin{array}{ll} | ||
1/(q - p), & \mbox{if $q > p$}, \\ | 1/(q - p), & \mbox{if $q > p$}, \\ | ||
\infty, & \mbox{if | \infty, & \mbox{if $q = p.$} | ||
\end{array} | \end{array} | ||
\right. | \right. | ||
Latest revision as of 01:54, 15 June 2024
[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]
In the gambler's ruin problem, assume that the gambler initial stake is 1 dollar, and assume that her probability of success on any one game is [math]p[/math]. Let [math]T[/math] be the
number of games until 0 is reached (the gambler is ruined). Show that the generating function for [math]T[/math] is
[[math]]
h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2pz}\ ,
[[/math]]
and that
[[math]]
h(1) = \left \{ \begin{array}{ll}
q/p, & \mbox{if $q \leq p$}, \\
1, & \mbox{if $q \geq p,$}
\end{array}
\right.
[[/math]]
and
[[math]]
h'(1) = \left \{ \begin{array}{ll}
1/(q - p), & \mbox{if $q \gt p$}, \\
\infty, & \mbox{if $q = p.$}
\end{array}
\right.
[[/math]]
Interpret your results in terms of the time [math]T[/math] to reach 0. (See also Example.)