As said in the title, for $(W_t)$ a Brownian motion and $\mathcal{F}_s $ the natural filtration of it. I'd like to compute $\mathbb{E}(e^{r(W_t-W_s)} \mid\mathcal{F}_s)$.
The only thing I can think of is to use the fact that $W_t-W_s$ is independent of $\mathcal{F}_s$, and thus $W_t-W_s\mid\mathcal{F}_s$ has the same distribution as $W_t-W_s$, that is $$W_t-W_s\mid\mathcal{F}_s\sim N(0,t-s).$$
But how could I apply this result to the above computation?
Thank you
So the question is: If $X\sim\operatorname N(0,\sigma^2)$ and $r$ is constant (i.e. not random) then what is $\operatorname E(\exp(rX))\text{?}$ In this case $\sigma^2=t-s$ and $X= W_t-W_s.$
\begin{align} \operatorname E(e^{rX}) & = \int_{-\infty}^\infty e^{rx} f_X(x) \,dx \\[8pt] & = \int_{-\infty}^\infty e^{rx} \frac 1 {\sqrt{2\pi}} e^{-x^2/(2\sigma^2)} \, \frac{dx} \sigma. \tag 1 \end{align} So work on the exponent: \begin{align} rx - \frac{x^2}{2\sigma^2} = {} & -\frac{x^2 -2\sigma^2rx}{2\sigma^2} \\[8pt] = {} & - \frac{x^2 - 2\sigma^2 rx + \sigma^4 r^2}{2\sigma^2} + \frac{\sigma^2 r^2} 2 \\[8pt] = {} & - \frac{(x - \sigma^2 r)^2}{2\sigma^2} + \frac{\sigma^2 r^2} 2 \end{align} Thus line $(1)$ above becomes $$ \int_{-\infty}^\infty \frac 1 {\sqrt{2\pi}} e^{-(x-\sigma^2 r)^2/(2\sigma^2)} \, \frac{dx}\sigma \cdot e^{\sigma^2 r^2/2} = 1\cdot e^{\sigma^2 r^2/2}. $$ And now remember that $\sigma^2 = t-s.$