martingale - 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?

Answer

We have

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

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

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

Your other steps are correct, i.e.

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