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2 players A and B start with x & y dollars respectively, and they bet against each other 1 dollar each time by tossing a fair coin.

I let $X_n = x + \sum_{i=1}^{n}\xi_i$ where $\xi_i$ are i.i.d. with $P(\xi_i=\pm 1)=\frac{1}{2}$. Let $\tau_0 = \inf{\{n:X_n = 0}\}$, $\tau_{x+y} = \inf{\{n:X_n = x+y}\}$, and $\tau=\tau_0\wedge \tau_{x+y}$, which are all stopping times w.r.t. the martingale $(X_n)$.

I know that $E[\tau]=xy$, and $E[X_\tau]=x$.

Am i able to find $E[\tau^2]$ by using any moment generating function in this case? Any suggestion is welcome! Thanks!

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Why do you specify the method? The variance is simpler than the moment generating function. –  Douglas Zare Mar 20 '13 at 6:27
    
In case you know moment-generating function of $\tau$, you can simply use the fact that $$ \mathsf E[\tau^2] = m''(0) $$ where $m(t) = \mathsf E[\mathrm e^{\tau t}]$ is MGT of $\tau$ –  Ilya Mar 20 '13 at 7:34
    
@Ilya: Of course. My point was that it seems easier to compute the variance without first finding the moment generating function. So, why specify an awkward method? Was this assigned as an exercise in a class which just covered moment generating functions? –  Douglas Zare Mar 20 '13 at 16:45
    
@Douglas: completely agree, but since OP asked how to find a moment using MGT, I thought it's worth reminding how to do this. –  Ilya Mar 28 '13 at 16:16
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2 Answers

I understand that in your solution you use an approach by considering martingales $X_n$ and ${X^2}_n - n$ and applying optimal stopping theorem.

To calculate the mgf, you need to use a discrete analog of an exponential martingale. As in here

http://www.google.com/url?sa=t&rct=j&q=discrete%20time%20martingales%20and%20concentration%20inequalities%20-%20springer&source=web&cd=1&cad=rja&ved=0CDEQFjAA&url=http%3A%2F%2Fwww.springer.com%2Fcda%2Fcontent%2Fdocument%2Fcda_downloaddocument%2F9781441996336-c1.pdf%3FSGWID%3D0-0-45-1137941-p174105368&ei=_OlJUcayCIaHrAfqlID4Bg&usg=AFQjCNEPlTfoyxEoaOQ_EVuCnrzhxLaQIA&bvm=bv.44011176,d.bmk

page 476.

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You can continue to play the same "game" that lead to the probabilities for hitting $0$ and $x+y$ and the expectations for $\tau$ and $X_\tau$. Just construct martingales to which you can apply the optional stopping theorem. Note that $X_n$, $X_n^2-n$, $X_n^3-3nX_n$, $X_n^4-6nX_n^2+n(3n+2)$, are all martingales. Applying the optional stopping theorem to the first two yields

$P(X_\tau=0)=\frac{y}{x+y}$, $P(X_\tau=x+y)=\frac{x}{x+y}$, $E(\tau)=xy$, as you already know.

With the next one you can show that

$E(\tau X_\tau)=\frac{1}{3}xy(2x+y)$.

Conditioning on $X_\tau=0$ and $X_\tau=x+y$ gives

$E(\tau X_\tau)=\frac{y}{x+y}E(\tau X_\tau|X_\tau=0) + \frac{x}{x+y}E(\tau X_\tau|X_\tau=x+y)$,

which tells you that

$E(\tau|X_\tau=x+y) =\frac{1}{3}y(2x+y)$, $E(\tau|X_\tau=0)=\frac{1}{3}x(x+2y)$,

i.e. now you know the expectations of the hitting times conditional on player A winning or losing.

Applying the optional stopping theorem to the fourth martingale will give you the second moment of the hitting time...

You can construct martingales that are higher order polynomials in $X_n$ and $n$, or, since after a while this gets a little bit tedious, you might want to put them together and use mgf's, to describe the law of $\tau$ (conditionally on hitting 0 and $x+y$). Just observe that

$E(\exp(\lambda \xi_i))=\cosh(\lambda)$,

so

$M_n(\lambda) = \frac{\exp(\lambda X_n)}{\cosh(\lambda)^n}$

is a martingale.

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