Martingale $c^{W_t}$ where $W$ is Brownian motion

Question

I have the process $c^{W_{t}}$ where $c$ is a constant and $W$ is Brownian motion. I would like to check if $\mathbb E[c^{W_{t+1}}|F_t]=c^{W_t}$. Dividing the right site yield $\mathbb E[c^{W_{t+1}-W_{t}}|F_t]=c^{0}=1$. Is that correct?

Answer 1

Yes, if $c$ is such that $\mathbb{E}[c^{W_{t+1}} \mid F_t] = c^{W_t}$ then it will also be true that $E[c^{W_{t+1}-W_t} \mid F_t] = 1$. This is simply the well-known property of conditional expectation that if $Y$ is measurable with respect to $\mathcal{G}$, then $Y E[X \mid \mathcal{G}] = E[XY \mid \mathcal{G}]$.

If I am not mistaken, you will find that the only positive $c$ for which this statement holds is $c=1$.