Question About Proving $\exp(aW(t)-\frac{a^{2t}}{2})$ is Martingale I'm trying to prove that $Z(t)=\exp(aW(t)-\frac{a^{2t}}{2})$ is a martingale.
($W(t)$ is a Wiener process, $a$ is a constant).
My attempt:
$$E[Z(t+s)] = E\left[\exp\left[aW(t+s)-\frac{a^{2(t+s)}}{2}\right]\right]$$ I was told that we can write $\exp[aW(t+s)]$ as $\exp[aW(t)+aW(s)]$, could any expert explain why it is the case?
 A: Let $(W_t)$ be a standard Brownian motion and $a>0$. We define $X_t=\exp\left(aW_t-\frac{1}{2}a^2t\right)$. Then, the process $(X_t)$ is adapted and integrable which are the first two conditions of being a martingale. Finally, for any $s<t$, 
\begin{align*}
\mathbb{E}\left[ X_t\mid\mathcal{F}_s\right] &= \mathbb{E}\left[ \exp\left(aW_t-\frac{1}{2}a^2t\right)\mid\mathcal{F}_s\right] \\
&= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ \exp\left(aW_t-\frac{1}{2}a^2s\right)\mid\mathcal{F}_s\right] \\
&= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ e^{a(W_t-W_s)}\exp\left(aW_s-\frac{1}{2}a^2s\right)\mid\mathcal{F}_s\right] \\
&= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ e^{a(W_t-W_s)}\right] \exp\left(aW_s-\frac{1}{2}a^2s\right) \\
&= \exp\left(aW_s-\frac{1}{2}a^2s\right) \\
&= X_s.
\end{align*}
Note that the increment $W_t-W_s\sim N(0,t-s)$ is independent of $\mathcal{F}_s$ and hence the conditioning on $\mathcal{F}_s$ can be dropped. Furthermore, $\mathbb{E}\left[ e^{a(W_t-W_s)}\right]=e^{\frac{1}{2}a^2(t-s)}$. On the other hand, $W_s$ is $\mathcal{F}_s$ measurable (i.e. known at time $s$ and thus may be taken out of the conditional expectation.
