For a simple symmetric random walk $S_n$, the process $f(S_n)$ is a martingale if and only if $$D^2 f = f(x+1) - 2f(x) + f(x-1) = 0.$$ Where this condition comes from I have no idea. I have never seen this anywhere and cannot find it anywhere. That being said, I need to find a condition for some function $f(n, S_n)$ using this fact and then prove that $Y_n = \rho ^ {-n} e^{S_n}$ is a martingale where $$\rho = \frac{e + e^{-1}}{2}.$$
Now, I have been looking at this problem for days and have no idea how to do it. I think we can write $f(n,S_n)$ as $$f(n, S_n) = g(n)f(S_n).$$ Then we have that $g(n) D^2 f = 0$, so $g(n)$ can be anything at all. Obviously this is not true.
I truly have no idea. Any help is appreciated.
The $D^2f = 0$ condition can be rewritten as $f(x) = \frac 12 (f(x+1)+f(x-1))$. We compute \begin{align*} \mathbb{E}[f(S_{n+1})|S_n] = \mathbb{E}[f(S_n + X_n) | S_n] = \frac 12 f(S_n + 1) + \frac 12 f(S_n-1), \end{align*} so we can see that the martingale condition $\mathbb{E}[f(S_{n+1})|S_n] = f(S_n)$ is equivalent to the $D^2f=0$ condition.
Do you see how to apply a similar method to find a condition for $f(n,S_n)$ to be a martingale?