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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?

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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 $s<t$, \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.

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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*}

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