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up vote 4 down vote accepted

The same approach works. Consider the random walk $(X_n)$ whose steps are $+1$ with probability $p$ and $-1$ with probability $1-p$. Then $\mathrm E(\mathrm e^{tX_{n+1}}\mid (X_k)_{k\leqslant n})=u(t)\mathrm e^{tX_{n}}$ with $u(t)=p\mathrm e^t+(1-p)\mathrm e^{-t}$ hence $(M_n(t))_n$ is a martingale with $M_n(t)=u(t)^{-n}\mathrm e^{tX_{n}}$ for every $n$. Expanding $M_n(t)$ as a power series in $t$ then yields a sequence of martingales, just like in the other case. The first ones are $$ 1,\quad X_n-(2p-1)n,\quad X_n^2-2(2p-1)nX_n+(2p-1)^2n^2-4p(1-p)n. $$

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Thanks a lot! I'm looking for one that helps me find the expectation of $T^2$. Sadly none of these seem to work. Any idea? – wircho Oct 27 '11 at 23:36
Any idea? Yes, but this is another question. If what you want are the moments of $T$, the method above is not the right one, already to compute $E_0(T)$. – Did Oct 28 '11 at 0:08
Hmmm. Thanks for pointing out that martingales are not the right method for the moments. Could you direct me towards the right idea? Thanks a lot! – wircho Oct 28 '11 at 0:28
See your other question. – Did Oct 28 '11 at 1:08

I don't know what you are trying to calculate, but if it is something related to stopping times of an asymmetric random walk then consider the following.

Let $ X_n = \sum_{i=1}^n \xi_i $, where ($\xi_i, i >0$) are i.i.d with distribution $P(\xi_i =1)=p$ and $P(\xi_i= -1) = q, \quad \forall i >0$

You can easily verify that $$ Y_n = (\frac{q}{p})^{X_n}$$ is a martingale.

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