Let $W_t,W_s$ be the Brownian Motion and $\mathcal{F}_s$ the natural filtration. Why is the following true

$$E(e^{-\alpha(W_t-W_s)}\mid\mathcal{F}_s)=e^{\frac{1}{2}\alpha^2 (t-s)}$$

I understand that the increment is independent so $$E(e^{-\alpha(W_t-W_s)}\mid\mathcal{F}_s)=E(e^{-\alpha(W_t-W_s)})$$

(I'm trying to prove that $e^{\alpha W_t -\frac{1}{2}\alpha^2 t}$ is a martingale)


1 Answer 1


Recall that if $X$ is normally distributed with mean zero and variance $\sigma^2$, then the moment generating function of $X$ is $$ \mathbb{E}[e^{uX}]=e^{\frac{u^2\sigma^2}{2}}$$

Now apply this to $X=W_t-W_s$ and set $u=-\alpha$.


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