# Problem on the exponential martingale with the Brownian motion

Let $S$ be a stochastic process defined by $S_t$ = exp$(\int_{0}^{t}$ $z_s$ $dW_s$ $- \frac{1}{2}\int_{0}^{t}$$\vert z_s \vert^2$ $ds$), where $W$ is a standard brownian motion and $z$ is deterministic with $\int_{0}^{T} \vert z_s \vert^2 ds$ $\lt$ $\infty$

I want to show that $S$ is a martingale:

I think the integrability follows from Hölder's inequality using $\mathbb{E}$[exp($\nu$)] = exp($\frac{\sigma^2}{2}$), if $\nu$ ~ $\mathcal{N}$($0$, $\sigma^2$).

Furthermore, I think adaptability is trivially given since the brownian motion is also a martingale.

However, I am unable to show the decisive martingale property $\mathbb{E}$[$S_{t+1}$|$\mathcal{F_t}$] = $S_t$. Is the independence of the increments $\int_{0}^{t} z_t dW_t - \int_{0}^{s} z_s dW_s$ needed?

I am grateful for any tip and piece of advice

• Do you know Ito's lemma? Jul 20, 2018 at 15:48
• No I unfortunately do not - is there a way to solve the problem without using the lemma? Jul 20, 2018 at 15:56

Hint: Because $z$ is deterministic, the stochastic integral $\int_t^{t+u} z_s\,dW_s$ is independent of $\mathcal F_t$, for each $t>0, u>0$. Now write $\int_0^{t+u} z_s\,dW_s$ as $\int_0^{t} z_s\,dW_s+\int_t^{t+u} z_s\,dW_s$.