# Geometric Brownian motion is a martingale

Why is the geometric Brownian motion, given by

$$\alpha \exp \left( \sigma W_t - \frac{\sigma^2}{2} t \right)$$

a martingale?

I just have problems to show the point: $$\mathbb{E} [X_t \mid \mathcal{F}_s] = X_s \ \ \ \ \mathbb{P}$$-a.s for all $$t > s$$.

• Try to use $W_t=W_t-W_s+W_s$ and $W_t-W_s$ is independent of $\mathcal{F}_s$. Sep 14, 2021 at 14:46
• Note that this is a very restricted instance of a brownian motion, with drift $\mu=0$, this does yield $\mathbb E[X_t,F_s]=X_s\mathbb E\exp(\sigma W_{t-s})\cdot\exp(-\frac{\sigma^2}{2}(t-s))=X_s$ because $\mathbb Ee^{\alpha W_t}=e^{\frac{1}{2}\alpha^2 W_t}$ and $\mathbb EX_t=X_0 e^{\mu t}$ using Îto's lemma to compute this expectation. In general GBM is a strict submartingale or supermartigale, thus only a semimartingale and not a martingale, as you can see from its expectation for general $\mu$.
– plm
Jun 5, 2023 at 11:30
• Also, i don't quite see what Surb proves in his answer. He leaves an expectation $\mathbb Ee^{๐(๐_๐กโ๐_๐ )}$ which is not trivial to compute, and his formula has at least one mistake anyway.
– plm
Jun 5, 2023 at 11:36

Using the fact that $$W_t-W_s$$ is independent of $$\mathcal F_s$$ yields
$$\mathbb E[\alpha e^{\sigma W_t-\sigma ^2t/2}\mid \mathcal F_s]=\mathbb E[e^{\sigma (W_t-W_s)}]e^{\alpha W_s+\sigma ^2t/2}.$$