# Proof that $\exp(aW(t)-0.5a^2t)$ is a martingale

I'm trying to prove that $$Z(t)=\exp(aW(t)-0.5a^2t)$$ is a martingale where $$W(t)$$ is a Wiener process and $$a$$ is a constant. Here is my attempt:

$$E[Z(t+s)] = E\left[\exp\left(aW(t+s)-0.5a^2(t+s)\right)\right].$$

I was told that we can write $$\exp(aW(t+s))$$ as $$\exp(aW(t)+aW(s))$$, could anyone explain why it is the case?

Let $$(W_t)$$ be a standard Brownian motion and $$a>0$$. We define $$X_t=e^{aW_t-\frac{1}{2}a^2t}$$. Then, the process $$(X_t)$$ is adapted and integrable which are the first two conditions of being a martingale. Finally, for any $$s, \begin{align*} \mathbb{E}\left[ X_t\mid\mathcal{F}_s\right] &= \mathbb{E}\left[ e^{aW_t-\frac{1}{2}a^2t}\mid\mathcal{F}_s\right] \\ &= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ e^{aW_t-\frac{1}{2}a^2s}\mid\mathcal{F}_s\right] \\ &= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ e^{a(W_t-W_s)}e^{aW_s-\frac{1}{2}a^2s}\mid\mathcal{F}_s\right] \\ &= e^{-\frac{1}{2}a^2(t-s)}\mathbb{E}\left[ e^{a(W_t-W_s)}\right] e^{aW_s-\frac{1}{2}a^2s} \\ &= e^{aW_s-\frac{1}{2}a^2s}\\ &= X_s. \end{align*} Note that the increment $$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$$ and thus may be taken out of the conditional expectation.
Set $$Z_t := f(W_t,t)$$ where $$f(x,t) = e^{ax -\frac{1}{2}a^2t}.$$ Then applying Ito's lemma we get $$dZ_t = aZ_tdW_t.$$ This means that $$Z_t$$ is a local martingale. For $$Z_t$$ to be a martingale you have to prove that $$E \int_0^t|a e^{aW_t -\frac{1}{2}a^2u}|^2 du <\infty.$$ For that you can use Fubini's theorem. \begin{align*} E \int_0^t|a e^{a W_u - \frac{1}{2}a^2u}|^2 du &= E \int_0^t a^2 e^{2aW_u -a^2u}|^2 du \\ &\leq a^2 E \int_0^t e^{2 a W_u}du \tag*{(since e^{2 a W_u -a^2u} \leq e^{2 a W_u })} \\ &= a^2 \int_0^t E e^{2aW_u}du \tag*{(Fubini's T)} \\ &= a^2 \int_0^t e^{2a^2 u}du \\ &< \infty. \end{align*}