Revision as of 02:36, 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> In de Moivre's description of the game, we can modify the definition of player A's fortune in such a way that the game is still a martingale (and the calculations are simpler). We do this by assigning nominal values to the counters in the same wa...")
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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]

In de Moivre's description of the game, we can modify the definition of

player A's fortune in such a way that the game is still a martingale (and the calculations are simpler). We do this by assigning nominal values to the counters in the same way as de Moivre, but each player's current fortune is defined to be just the value of the counter which is being wagered on the next game. So, if player A has [math]a[/math] counters, then his current fortune is [math](q/p)^a[/math] (we stipulate this to be true even if [math]a = 0[/math]). Show that under this definition, player A's expected fortune after one play equals his fortune before the play, if [math]p \ne q[/math]. Then, as de Moivre does, write an equation which expresses the fact that player A's expected final fortune equals his initial fortune. Use this equation to find the probability of ruin of player A.