exercise:087f70d94a: 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> Show that <math>w_x</math> of Exercise Exercise satisfies the following conditions: <ul><li> <math>w_x = pw_{x + 1} + qw_{x - 1}</math> for <math>x = 1</math>, 2, \ldots,\ <math>T - 1</math>. </li> <li> <math>w_0 = 0</math...") |
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\newcommand{\mathds}{\mathbb}</math></div> Show that <math>w_x</math> of | \newcommand{\mathds}{\mathbb}</math></div> Show that <math>w_x</math> of [[exercise:8ae7bbfa06 |Exercise]] satisfies the following conditions: | ||
the | <ul style="list-style-type:lower-alpha"><li> <math>w_x = pw_{x + 1} + qw_{x - 1}</math> for <math>x = 1</math>, 2, ..., <math>T - 1</math>. | ||
following conditions: | |||
<ul><li> <math>w_x = pw_{x + 1} + qw_{x - 1}</math> for <math>x = 1</math>, 2, | |||
</li> | </li> | ||
<li> <math>w_0 = 0</math>. | <li> <math>w_0 = 0</math>. | ||
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</li> | </li> | ||
</ul> | </ul> | ||
Show that these conditions determine <math>w_x</math>. Show that, if <math>p = q = | Show that these conditions determine <math>w_x</math>. Show that, if <math>p = q =1/2</math>, then | ||
1/2</math>, then | |||
<math display="block"> | <math display="block"> | ||
Latest revision as of 23:04, 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]
Show that [math]w_x[/math] of Exercise satisfies the following conditions:
- [math]w_x = pw_{x + 1} + qw_{x - 1}[/math] for [math]x = 1[/math], 2, ..., [math]T - 1[/math].
- [math]w_0 = 0[/math].
- [math]w_T = 1[/math].
Show that these conditions determine [math]w_x[/math]. Show that, if [math]p = q =1/2[/math], then
[[math]]
w_x = \frac xT
[[/math]]
satisfies (a), (b), and (c) and hence is the solution. If [math]p \ne q[/math], show that
[[math]]
w_x = \frac{(q/p)^x - 1}{(q/p)^T - 1}
[[/math]]
satisfies these conditions and hence gives the probability of the gambler winning.