Calculating Expectation of Exponential Brownian motion

Question

I want to simplify and get the answer for following expression: $$E[e^{-r(t_2-t_1)}e^{\int_{t_1}^{t_2}(\mu_u-\frac{\sigma_u^2}{2})du+\sigma_udW_u}|F(t_1)]$$ Where $r$ is a constant and $W_u$ is standard Brownian motion.

Here what I tried. First I rewrote above expression as $$E[e^{\int_{t_1}^{t_2}(\mu_u-r-\frac{\sigma_u^2}{2})du+\sigma_udW_u}|F(t_1)]$$

Then I defined $$\hat{W_u}=\bigg(\frac{\mu_u-r}{\sigma_u}\bigg)u+W_u$$ which is distributed normally with mean $\bigg(\frac{\mu_u-r}{\sigma_u}\bigg)u$ and variance $\sigma_u u$. This allows me to write expectation expression as $$E[e^{\int_{t_1}^{t_2}(-\frac{\sigma_u^2}{2})du+\sigma_ud\hat{W}_u}|F(t_1)]$$ Now I am stuck, since $\mu_u, \sigma_u$ are not constant.

Answer 1

I suppose $F(t) = \sigma(W_s,s\leq t)$. I recall that $M_t = \int_0^t\sigma_s dW_s$ is a Wienner integral. Therefore:

• $M_t \sim \mathcal{N}\left(0, \int_0^t\sigma_s^2ds\right)$
• for $s\leq t$, $M_t - M_s $ is independant of $F_s$

We have then : \begin{align} E[e^{-r(t_2-t_1)}e^{\int_{t_1}^{t_2}(\mu_u-\frac{\sigma_u^2}{2})du+\sigma_udW_u}|F(t_1)] &= e^{-r(t_2-t_1)+\int_{t_1}^{t_2}(\mu_u-\frac{\sigma_u^2}{2})du}E[e^{\int_{t_1}^{t_2}\sigma_udW_u}|F(t_1)] \\ &= \exp\left\{-r(t_2-t_1)+\int_{t_1}^{t_2}(\mu_u-\frac{\sigma_u^2}{2})du + \frac12\int_{t_1}^{t_2}\sigma_u^2du\right\} \\ &= \exp\left\{-r(t_2-t_1)+\int_{t_1}^{t_2}\mu_udu\right\} \end{align}