I'm reading through some notes for stochastic calculus and the author says the following:
Since $\xi_{k-1}\in \mathcal{F}_{t_{k-1}}\subseteq \mathcal{F}_s$ and $B_t-B_s$ is independent of $\mathcal{F}_s$, we find that $$\mathbb{E}[\exp(\xi_{k-1}(B_t-B_s))|\mathcal{F}_s]=\exp \left(\frac{1}{2}\xi_{k-1}^2(t-s)\right)$$
Note: $B$ is a Brownian motion, and $\xi_k$ is a bounded $\{\mathcal{F}_k\}$-measurable random variable for each $k\ge1$.
I am unsure how to make the jump to the right hand side. I get that we should still have $\xi_{k-1}$ by the fact that it is $\{\mathcal{F}_s\}$-measurable, and also that $B_t-B_s$ is independent of $\mathcal{F}_s$.
My rough attempt is that since $\exp(\xi_{k-1}(B_t-B_s))$ is of the form $f(X, Y)$, we indeed have that $\mathbb{E}[f(X,Y)|\mathcal{F_s}] = \mathbb{E}[f(X,Y)] = \mathbb{E}[\exp(\xi_{k-1}(B_t-B_s))]$.
Then we also have that,
$\begin{align} \mathbb{E}[\exp(\xi_{k-1}(B_t-B_s))] &= \mathbb{E}[\mathbb{E}[\exp(\xi_{k-1}(B_t-B_s))|\xi_{k-1}]]\\ &= \mathbb{E}[\exp(\xi_{k-1}^2(t-s))]\\ \end{align}$
Where we use the fact that the moment generating function of a normal variable is $\exp(\mu t + \frac{1}{2}\sigma^2 t^2)$, and that $\mathbb{V}(B_t-B_s) = t-s$.
What I am confused about is how the expected value vanishes, just leaving the inside of it. Intuitively I know that the mistake is on my end since $\xi_{k-1}$ should come outside of the expected value at some point.
Since $\xi_{k-1}\in \mathcal{F}_{t_{k-1}}\subseteq \mathcal{F}_s$ we can treat $\xi_{k-1}$ as a constant throughout. Just compute $\mathbb{E}[\exp(u(B_t-B_s))|\mathcal{F}_s]$ for fixed $u$ and replace $u$ by $\xi_{k-1}$ at the end. All you need the MGF of normal distribution to prove this identity. Conditional expectation goes away because $B_t-B_s$ is independent of $\mathcal F_s$.