for a random walk, $S_0$ = 0, $S_n = x_1 + ... x_n$, where $X_i$ iid increments. Assume $Ex_i < 0$ and $X_i$ has discrete distribution with finite number of support points, with at least one of the x's being positive.
Show there is a real number $\theta \neq 0$ such that $exp(\theta S_n$) is a Martingale.
I know to show it is a Martingale, it needs $E(S_{n+1} |F_n) = S_n$.
$E(S_{n+1} |F_n) = e^{\theta S_n} E(e^{\theta x_{n+1}}|F_n) = e^{\theta S_n} E(e^{\theta x_{n+1}})$. To make this a Martingale, I know that I need to find $\theta$ such that $E(e^{\theta x_{n+1}}) = 1$, expanding the expectation formula, I have $\sum_{i = 1}^k P(X = x_i) e^{\theta x_i} = 1$. But I get stuck in terms of how to solve for $\theta$.
You don't need to solve for $\theta$ in order to prove a solution exists. Observe that if $\theta$ is very large, then the term with a positive $x_i$ will become very large as well, and since any terms with negative $x_i$ will just go to zero, the MGF will get large. But at the origin, the derivative with respect to $\theta$ is negative (since this is just the expectation value), so the MGF will be less than $1$ for small $\theta>0.$ Thus, since the MGF is continuous, it will be equal to one again at some value of $\theta.$