# Conditional expectation of a geometric brownian motion

I'm reviewing stuff from the past and I'm very confused all of a sudden. Some verification would help about the following.

$$\mathbb{E}[e^{\sigma W(t)}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s) + W(s))}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s))}|{\cal F}_s]e^{W(s)}$$ $$=\mathbb{E}[e^{\sigma (W(t) - W(s))}]e^{W(s)} =\frac{1}{2}\sigma^2 (t-s) e^{W(s)}$$

Is this true?

We have

$$\sigma \left( W_t - W_s \right) \sim \mathcal{N} \left( 0, \sigma^2 (t - s) \right).$$

Let $X \sim \mathcal{N} \left( 0, \xi^2 \right)$, then

\begin{eqnarray} \mathbb{E} \left[ e^{X} \right] & = & \frac{1}{\sqrt{2 \pi} \xi} \int_\mathbb{R} \exp \left\{ x -\frac{x^2}{2 \xi^2} \right\} \mathrm{d}x\\ & = & \frac{1}{\sqrt{2 \pi} \xi} \int_\mathbb{R} \exp \left\{ -\frac{x^2 - 2 x \xi^2 \pm \xi^4}{2 \xi^2} \right\} \mathrm{d}x\\ & = & \frac{1}{\sqrt{2 \pi} \xi} e^{\xi^2 / 2} \int_\mathbb{R} \exp \left\{ -\frac{\left( x - \xi^2 \right)^2}{2 \xi^2} \right\} \mathrm{d}x\\ & = & e^{\xi^2 / 2}, \end{eqnarray}

where we recognize the integrand in the second last line as the density of of a $\mathcal{N} \left( \xi^2, \xi^2 \right)$ normal random variable which integrates to one. Thus

$$\mathbb{E} \left[ e^{\sigma \left( W_t - W_s \right)} \right] = \exp \left\{ \frac{1}{2} \sigma^2 (t - s) \right\}.$$

Your other steps are correct, i.e.

$$\mathbb{E} \left[ \left. e^{W_t} \right| \mathcal{F}_s \right] = \exp \left\{ W_s + \frac{1}{2} \sigma^2 (t - s) \right\}$$

Depending on how well you are familiar with stochastic calculus, another way of getting to this result is to recognise that: $$e^{\sigma W_t} = \mathcal{E}[\sigma W_t] e^{\frac{1}{2}\sigma^2 t}$$ where $$\mathcal{E}(X_t) = \exp\left(X_t - \frac{1}{2}\langle X \rangle_t\right)$$ denotes the Doléans-Dade exponential (stochastic exponential) of an Itô process $X_t$, which is well-known to verify the martingale property i.e. $$\Bbb{E}_0\left[\mathcal{E}(X_t)\right] = \exp(X_0)$$ (of course the proof relies on @LocalVolatility's answer)

Using this result along with the fact that $W_0=0$ by definition for a standard Brownian motion, one gets: $$\Bbb{E}_0 \left[ e^{\sigma W_t} \right] = \underbrace{\Bbb{E}_0 \left[ \mathcal{E}(\sigma W_t) \right]}_{=\exp(\sigma W_0)=1} \Bbb{E}_0 \left[ e^{\frac{1}{2}\sigma^2 t} \right] = e^{\frac{1}{2}\sigma^2 t}$$