exercise:870fedb4e2: 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> Assume in the gambler's ruin problem that <math>p = q = 1/2</math>. <ul><li> Using Equation, together with the facts that <math>q_0 = 1</math> and <math>q_M = 0</math>, show that for <math>0 \le z \le M</math>, <m...") |
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\newcommand{\mathds}{\mathbb}</math></div> Assume in the gambler's ruin problem that <math>p = q = 1/2</math>. | \newcommand{\mathds}{\mathbb}</math></div> Assume in the gambler's ruin problem that <math>p = q = 1/2</math>. | ||
<ul><li> Using [[guide:30289c2ffb#eq 12.2.2 |Equation]], together with the facts that <math>q_0 = 1</math> and <math>q_M = 0</math>, | <ul style="list-style-type:lower-alpha"><li> Using [[guide:30289c2ffb#eq 12.2.2 |Equation]], together with the facts that <math>q_0 = 1</math> and <math>q_M = 0</math>, | ||
show that for <math>0 \le z \le M</math>, | show that for <math>0 \le z \le M</math>, | ||
Latest revision as of 01:05, 15 June 2024
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Assume in the gambler's ruin problem that [math]p = q = 1/2[/math].
- Using Equation, together with the facts that [math]q_0 = 1[/math] and [math]q_M = 0[/math],
show that for [math]0 \le z \le M[/math],
[[math]] q_z = {{M - z}\over M}\ . [[/math]]
- In Equation, let [math]p \rightarrow 1/2[/math] (and since [math]q = 1 - p[/math], [math]q
\rightarrow 1/2[/math] as well). Show that in the limit,
[[math]] q_z = {{M - z}\over M}\ . [[/math]]Hint: Replace [math]q[/math] by [math]1-p[/math], and use L'Hopital's rule.