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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 $t > s$.

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  • $\begingroup$ Try to use $W_t=W_t-W_s+W_s$ and $W_t-W_s$ is independent of $\mathcal{F}_s$. $\endgroup$ Sep 14, 2021 at 14:46
  • $\begingroup$ 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 $\mu$. $\endgroup$
    – plm
    Jun 5, 2023 at 11:30
  • $\begingroup$ Also, i don't quite see what Surb proves in his answer. He leaves an expectation $\mathbb Ee^{๐œŽ(๐‘Š_๐‘กโˆ’๐‘Š_๐‘ )}$ which is not trivial to compute, and his formula has at least one mistake anyway. $\endgroup$
    – plm
    Jun 5, 2023 at 11:36

1 Answer 1

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Hint

Using the fact that $W_t-W_s$ is independent of $\mathcal F_s$ yields

$$\mathbb E[\alpha e^{\sigma W_t-\sigma ^2t/2}\mid \mathcal F_s]=\mathbb E[e^{\sigma (W_t-W_s)}]e^{\alpha W_s+\sigma ^2t/2}.$$

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