Show that the stochastic process $$Z_n := (−1)^n \cos(πY_n)$$ is a martingale with respect to the filtration $\{\mathcal{F_n}\}_{n\geq0}$,$\mathcal{F_n}\ =\sigma(X_1,\dots X_n)$ where $Y_n=\sum_{i=1}^n X_i$ is a symmetric random walk.
So I need to show $E(Z_n|\mathcal{F}_{n-1})=Z_{n-1}$. A symmetric walk is a martingale so $E(Y_n|\mathcal{F}_{n-1})=Y_{n-1}$.
I have tried to use properties of expectation, the addition formula for cosine and a telescope sum but I haven't gotten anywhere. Can someone point in the right direction?
Best Answer
Assuming that $X_i$ are $\pm k$, where $k$ is an odd integer, then:
$$E(Z_n|F_{n-1})= E[(-1)^n\cos(\pi Y_{n-1}+\pi X_n)|F_{n-1}).$$
Now use law of angle addition to get $\cos(Y_{n-1}+X_n)=\cos(\pi Y_{n-1})(-1)-0 = -\cos(\pi Y_{n-1}).$
So $E(Z_n|F_{n-1}) = E[Z_{n-1}|F_{n-1}]=Z_{n-1}.$