Conditional expectation of exponential Brownian motion

Question

Let $f\in L^2[0,T]$. Show that conditional expectation $$ \mathbb{E}\bigg[ e^{\int_{0}^T f(s)dB_s}\bigg| \mathscr{F}_t\bigg] =\exp \left(\int_0^t f(s)\ dB_s + \frac{1}{2} \int_0^T |f(s)|^2\ ds \right) $$

Where $(\mathscr{F}_t)_{t\in [0,T]}$ denotes the filtration generated by $(B_t)_{t\in [0,T]}$

I tried to follow the answer given in Expectation of Exponential of Stochastic Integral but couldn't get the answer.

Answer 1

Assuming that $B$ is a Brownian motion and $f \in L_2[0,T]$.

First method:

We know that $\forall t \in [0,T]$, $\int_0^t f(s)dB_s$ is a Wiener integral. Therefore, we can use the fact that $T\geq s \geq t$, $\int_t^s f(u)dB_u$ is independent of $\mathcal{F}_t$ (1) and follows a centered Gaussian distribution with variance $\int_s^t |f(u)|^2du$ (2). Thus, \begin{align} E\left[\exp\left(\int_0^Tf(s)dB_s\right)|\mathcal{F}_t\right] &=E\left[\exp\left(\int_t^Tf(s)dB_s\right)|\mathcal{F}_t\right]\exp\left(\int_0^tf(s)dB_s\right) \\ &=E\left[\exp\left(\int_t^Tf(s)dB_s\right)\right]\exp\left(\int_0^tf(s)dB_s\right) \\ &=\exp\left(\int_0^tf(s)dB_s+\frac12\int_t^T|f(s)|^2ds\right) \end{align} In the last equality, I used the following known result : $N\sim \mathcal{N}(\mu, \sigma^2), E[\exp(cN)] = \exp(c\mu +\frac12c^2\sigma^2)$.

Remarks:

• It seems that there is an issue in the writing of the problem. It should be $\int_\color{red}t^\color{red}T |f(s)|^2ds$.
• You can show with ease the property (1) and (2) by first assuming that $f$ is a step function and then use a density argument.

Second method:

Denoting $Y_t = \int_0^t f(s)dB_s - \frac12 \int_0^t |f(s)|^2ds$, we can apply Ito lemma to the function $\phi(x) = \exp(x)$ using $Y_t$. We have that \begin{align} d\exp(Y_t) &= \exp(Y_t)(dY_t + \frac12 d\langle Y_t \rangle ) \\ &= \exp(Y_t)(f(t)dB_t - \frac12f(t)^2dt + \frac12f(t)^2dt) \\ &=\exp(Y_t)f(t)dB_t \\ \end{align} We know that the RHS is a squared martingale. Therefore, the process $\lbrace{\exp(Y_t)\rbrace}_{0\leq t \leq T}$ is a martingale. For more details, look at this wiki page.