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?