Suppose we have a simple random walk starting from $S_0=0$, and $S_n=X_1+\dots+X_n$ such that $$\mathbb{P}(X_i=1)=p \hspace{1em}\mathbb{P}(X_i=0)=r \hspace{1em} \mathbb{P}(X_i=-1)=q$$ for positive $p,q$ and $r$ with $p+q+r=1$. Let $Z_n = \left(\frac{q}{p}\right)^{S_n}$ and $\mathcal{F}_n=\sigma(X_1, \dots, X_n)$.
I want to show that $(Z_n)$ is a Martingale with respect to filtration $(\mathcal{F}_n)$. Below is my attempt, I would appreciate any hints here.
For the Martingale property, we have $$\mathbb{E}[Z_{n+1} |\mathcal{F_n}] = \mathbb{E}\left[\left(\frac{q}{p}\right)^{S_{n+1}} |\mathcal{F_n}\right] = \left(\frac{q}{p}\right)^{S_n+1}\cdot p + \left(\frac{q}{p}\right)^{S_n}\cdot r + \left(\frac{q}{p}\right)^{S_n-1}\cdot q$$
$$= (Z_n \cdot \frac{q}{p} \cdot p) + (Z_n \cdot r) + (Z_n \cdot \left(\frac{q}{p}\right)^{-1} \cdot q) = Z_n (q+r+p) = Z_n.$$
To show $Z_n$ is adapted, i.e. for each $n, Z_n$ is $\mathcal{F_n}$-measurable, note that $Z_n$ is a function of $S_n$, which in turn is a function of $X_1, \dots, X_n$ and our filtration is the $\sigma$-algebra generated by $X_1, \dots, X_n$, hence $(Z_n)$ is adapted.
For integrability, i.e. $\mathbb{E}(|Z_n|) < \infty$, I am finding this difficult. I have thought that perhaps $S_n$ is a martingale and in particular $\mathbb{E}(|S_n|) < \infty$ and we can use this fact, although I am not sure how.
Thank you!
Best Answer
Integrability comes from the fact that $|S_n| \le n$. In the case that $p \ge q$, we have $\mathbb{E}[|Z_n|] \le \left(\frac qp\right)^{-n} < \infty$ for all $n$. In the case that $q \ge p$, we similarly have $\mathbb{E}[|Z_n|] \le \left(\frac qp\right)^{n} < \infty$.