Revision as of 03:33, 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> 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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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]

Show that [math]w_x[/math] of Exercise Exercise satisfies

the following conditions:

  • [math]w_x = pw_{x + 1} + qw_{x - 1}[/math] for [math]x = 1[/math], 2, \ldots,\ [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.