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The answer is probably gonna be very straightforward but I'm missing it. I want to prove the following claim:

Suppose we play a game where we win with probability $p$ and lose with probability $q = 1 - p$, where $p \neq \frac{1}{2}$. Denote by $X_i$ the outcome of game $i$ ($X_i = 1$ if we win and $X_i = -1$ if we lose). Prove that $$M_n = (\frac{p}{q})^{S_n}$$ is a martingale, where $S_n = \sum_{i=1}^{n}X_i$ and $M_0 = 1$.

My attempt:

We want to show that $E(M_n|\mathcal{F}_{n-1}) = M_{n-1}$. So:

$$E(M_n|\mathcal{F}_{n-1}) = \prod_{i=1}^{n-1}(\frac{p}{q})^{X_i} E((\frac{p}{q})^{X_n})$$

I omitted the $\mathcal{F}_{n-1}$ since it has no influence on the outcome of the $n$-th game.We are done if we can show that $E((\frac{p}{q})^{X_n}) = 1$.

Now, $$E((\frac{p}{q})^{X_n}) = (\frac{p}{q})^{1}P(X_n = 1) + (\frac{p}{q})^{-1}P(X_n = -1) = \frac{p^2}{q} + \frac{q^2}{p} = \frac{p^3 + q^3}{pq}$$

At this point I'm having this feeling I am making things more difficult than it is. Can anyone give an easier way of showing this?

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$p + q = 1$ is implied? – Eelvex Mar 11 '11 at 18:00
Yes, I'll add that to my post. – Stijn Mar 11 '11 at 18:01
There is a typo; the martingale should be $M_n=(q/p)^{S_n}$. That is, use $q$ over $p$, not $p$ over $q$. – Byron Schmuland Mar 11 '11 at 18:02
@Eelvex $M_n=(q/p)^{S_n}$ is also a martingale if $p+q<1$ and $P(X=0)=1-(p+q)$. – Byron Schmuland Mar 11 '11 at 18:09
up vote 1 down vote accepted

This is wrong. $$p^3+q^3 = (p+q) \times (p^2-p q+q^2)$$ Since $p+q=1$, $$p^3+q^3 = p^2-p q+q^2 = (p-q)^2 + p q = (1-2 q)^2 + p q $$ And what you would expect to be equal to 1 is not unless $p=q=\frac{1}{2}$ $$\frac{p^3+q^3}{p q} = 1 + \frac{(1-2 q)^2}{p q}$$

Edit to clarify a point: If this is taken from an exercise book, there is a mistake in the claim: see Byron Schmuland's comment for the correction, and Eelvex's comment for the name of the martingale (after correcting the claim).

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This is de Moivre's martingale and it sure is a martingale for $p \neq .5$. – Eelvex Mar 11 '11 at 19:38
See Byron Schmuland's comment for the correct de Moivre's martingale. This is not what is stated in the question, and I show this is wrong using Stijn's early thoughts. – Wok Mar 12 '11 at 13:16
off course you're right. I didn't notice the $p\leftrightarrow q$ switching. (Please edit to remove the downvote) – Eelvex Mar 12 '11 at 14:25

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