I was trying to calculate the expectation of $T^2$ using some martingale and got that I needed the expectation of $TS_T$. Any idea?
Expectation of $TS_T$ where $T$ is the absorption time at $\{a,-a\}$ of a simple symmetric random walk $\{S_n\}$
1 Answers
If $(S_n)$ is a simple symmetric random walk on the integers, then $S_n^3-3nS_n$ is a martingale. Applying the optional stopping theorem for martingales, we get $$ \mathbb{E}_x(S_T^3-3TS_T)=\mathbb{E}_x(S_0^3-0)=x^3.$$ Here $-a\leq x\leq a$ is the starting point of the random walk and $T$ is the hitting time of the boundary $\{-a,a\}$. Standard results tell us that $$\mathbb{P}_x(S_T=a)={a+x\over 2a}\quad\mbox{ and }\quad\mathbb{P}_x(S_T=-a)={a-x\over 2a}$$ so that $\mathbb{E}_x(S_T^3)=a^2 x$. It follows that $\mathbb{E}_x(T\ S_T)=(a^2x-x^3)/3$.
To find $\mathbb{E}_x(T^2)$, you could use the martingale $S_n^4-6nS_n^2+2n+3n^2$.
Update: The non-symmetric case.
Here is an infinite family of exponential martingales. For any real $\theta$ you can check that $\exp(\theta S_n)/[m_X(\theta)]^n$ is a martingale. Here, $m_X$ is the moment generating function of the increment $X$. For a non-symmetric random walk this will be $m_X(t)=q \exp(-\theta)+p \exp(\theta)$.
Starting with the exponential martingales, you can generate other martingales by differentiating with respect to $\theta$, then setting $\theta=0$. The first two martingales you get this way are $S_n-n\mu$ and $(S_n-n\mu)^2-n\sigma^2$ where $\mu,\sigma^2$ are the mean and variance of the increment $X$.
For more on this topic, take a look at Example M.1.3 (page 3) of Professor Glynn's notes on martingales.
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3[This MO answer](http://mathoverflow.net/questions/55092/martingales-in-both-discrete-and-continuous-setting/55101#55101) may help understand why these martingales @Byron is using, and many others, exist and how they may be computed. – 2011-10-26
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0@Didier That is a very impressive MO answer. I hadn't noticed it before, so thanks for the reference. – 2011-10-26
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0Wow thanks a lot for your answer Byron, and for the awesome link Didier! – 2011-10-26
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0I just realized the original question from the source does not specify it must be symmetric. Where can I find a list of martingales for the asymmetric case? – 2011-10-27