Revision as of 02:33, 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> Consider a random walker who moves on the integers 0, 1, \ldots, <math>N</math>, moving one step to the right with probability <math>p</math> and one step to the left with probability <math>q = 1 - p</math>. If the walker ever reaches 0 or <math...")
BBy Bot
Jun 09'24
Exercise
[math]
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Consider a random walker who moves on the integers 0, 1,
\ldots, [math]N[/math], moving one step to the right with probability [math]p[/math] and one step to the left with probability [math]q = 1 - p[/math]. If the walker ever reaches 0 or [math]N[/math] he stays there. (This is the Gambler's Ruin problem of Exercise Exercise.) If [math]p = q[/math] show that the function
[[math]]
f(i) = i
[[/math]]
is a harmonic function (see Exercise Exercise), and if [math]p \ne q[/math] then
[[math]]
f(i) = \biggl(\frac {q}{p}\biggr)^i
[[/math]]
is a harmonic function. Use this and the result of Exercise Exercise to show that the probability [math]b_{iN}[/math] of being absorbed in state [math]N[/math] starting in state [math]i[/math] is
[[math]]
b_{iN} = \left \{ \matrix{
\frac iN, &\mbox{if}\,\, p = q, \cr
\frac{({q \over p})^i - 1}{({q \over p})^{N} - 1}, &
\mbox{if}\,\,p \ne q.\cr}\right.
[[/math]]
For an alternative derivation of these results see Exercise Exercise.