# evaluating conditional expectation of exponential of brownian motions

For $s \leq t \leq T$, I want to evaluate $E\left[\exp\left(iz(W_t-W_s)+izb(s-t) + bW_T-b^2T/2 \right)|\mathcal{F_s}\right]$, where $W$ is Brownian motion on $\mathcal{F}$, and $b \in \mathbb{R}$. Clearly, $W_t -W_s$ is independent of $\mathcal{F}_s$, and $E[W_T|\mathcal{F}_s] = W_s$.

I would like to be able to write \begin{aligned} &E\left[\exp\left(iz(W_t-W_s)+izb(s-t) + bW_T-b^2T/2 \right)|\mathcal{F_s}\right]\\ &=E[\exp\left(iz(W_t-W_s)+izb(s-t) \right) | \mathcal{F}_s]\times E[\exp\left(bW_T-b^2T/2 \right) | \mathcal{F}_s]\\ & = E[\exp\left(iz(W_t-W_s)+izb(s-t) \right)] \times E[\exp(bW_T-b^2T/2)]\\ &? = \exp\left(-\frac{1}{2}z^2(t-s)^2\right) \times E[\exp(bW_T-b^2T/2)] \end{aligned}

Which would be as far as I need to go for my purposes. My first question is if I am able to split the conditional expectation into the product of two conditional expectations. Am I allowed to do this because $W_t-W_s$ is independent of $\mathcal{F}_s$ and $W_T$ isn't?

Also, are my steps after the splitting up of the expectation correct?

Thanks very much.

• $W_T=W_T-W_s+W_s$. – Gordon May 24 '18 at 20:35