exercise:95907a1e66: 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 text, it was shown that if <math>q < p</math>, there is a positive probability that a gambler, starting with a stake of 0 dollars, will never return to the origin. Thus, we will now assume that <math>q \ge p</math>. Using Exercise exe...")
 
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\newcommand{\mathds}{\mathbb}</math></div> In the text, it was shown that if <math>q  <  p</math>, there is a positive probability
\newcommand{\mathds}{\mathbb}</math></div> In the text, it was shown that if <math>q  <  p</math>, there is a positive probability that a gambler, starting with a stake of 0 dollars, will never return to the origin.  Thus, we will now assume that <math>q \ge p</math>.  Using [[exercise:E6fe303a8b |Exercise]], show that if a gambler starts with a stake of 0 dollars, then the expected number of times her stake equals <math>M</math> before returning to 0 equals <math>(p/q)^M</math>, if <math>q  >  p</math> and 1, if <math>q = p</math>.  (We quote from Feller:  “The truly amazing implications of this result appear best in the language of fair games.  A perfect coin is tossed until the first equalization of the accumulated numbers of heads and tails.  The gambler receives one penny for every time that the accumulated number of heads exceeds the accumulated number of tails by <math>m</math>. <i>The ‘fair entrance fee’ equals 1 independent of <math>m</math></i>.”)
that a gambler, starting with a stake of 0 dollars, will never return to the origin.  Thus, we
will now assume that <math>q \ge p</math>.  Using Exercise [[exercise:E6fe303a8b |Exercise]], show that if a gambler starts with a
stake of 0 dollars, then the expected number of times her stake equals <math>M</math> before returning to
0 equals <math>(p/q)^M</math>, if <math>q  >  p</math> and 1, if <math>q = p</math>.  (We quote from Feller:  “The truly amazing
implications of this result appear best in the language of fair games.  A perfect coin is tossed
until the first equalization of the accumulated numbers of heads and tails.  The gambler receives
one penny for every time that the accumulated number of heads exceeds the accumulated number of
tails by <math>m</math>. ''The `fair entrance fee' equals 1 independent of <math>m</math>.''”)

Latest revision as of 01:02, 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 text, it was shown that if [math]q \lt p[/math], there is a positive probability that a gambler, starting with a stake of 0 dollars, will never return to the origin. Thus, we will now assume that [math]q \ge p[/math]. Using Exercise, show that if a gambler starts with a stake of 0 dollars, then the expected number of times her stake equals [math]M[/math] before returning to 0 equals [math](p/q)^M[/math], if [math]q \gt p[/math] and 1, if [math]q = p[/math]. (We quote from Feller: “The truly amazing implications of this result appear best in the language of fair games. A perfect coin is tossed until the first equalization of the accumulated numbers of heads and tails. The gambler receives one penny for every time that the accumulated number of heads exceeds the accumulated number of tails by [math]m[/math]. The ‘fair entrance fee’ equals 1 independent of [math]m[/math].”)