# solve for $\theta \neq 0$ for exponential martingale $e^{\theta S_n}$

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 $$\theta.$$