# Expectation of the integral of e to the power a brownian motion with respect to the brownian motion

While following a proof on the uniqueness and existance of a solution to a SDE I encountered the following statement $$\mathbb{E}[\int_0^t e^{\alpha B_S}dB_s] = 0.$$ But since the exponential function is a strictly positive function the integral of this function should be greater than zero and thus the expectation as well? Is this statement true and how would I go about proving this?

• $B_s$ and $dB_s$ are independent. Thus the expectation of $e^{αB_s}dB_s$ at time $s$ is $e^{αB_s}$ times the expectation of $dB_s$, where the latter is zero. Commented Oct 27, 2017 at 9:17
• Sorry but do you remember how a stochastic integral $$\int_0^tX_sdB_s$$ is defined, already? Because if you do, then your sentence "since the exponential function is a strictly positive function the integral of this function should be greater than zero" is most odd.
– Did
Commented Oct 28, 2017 at 20:51

I like Gono's argument a lot. Here is a different one.

You know that if $h_s$ is adapted and $$E[ \int_0^t h_s^2 ds ] < \infty$$ then $M_t = \int_0^t h_s dW_s$ is a martingale.

Clearly $e^{aB_S}$ is adapted. By Tonelli $$E[ \int_0^t e^{(2a) B_s} ds ] = \int_0^t E[ e^{(2a)B_s} ] ds = \int_0^t e^{ 2 a^2 s} ds = \frac{ e^{2 a^2 t}-1}{2 a^2}<\infty$$

So since martingale $$E[ \int_0^t e^{ a B_s} dW_s] = E[ \int_0^0 e^{ a B_s} dW_s] = 0$$

• Should you be integrating with respect to a Brownian motion in the last display? Commented Nov 1, 2017 at 23:17
• @Theoretical Economist yes, thanks!
– htd
Commented Nov 2, 2017 at 0:06

To have a more "direct" way to show this you could use the well-known Itô formula for a suitable function $h$ $$h(B_t) = h(B_0) + \int_0^t h'(B_s) \, {\rm d} B_s + \frac{1}{2} \int_0^t h''(B_s) \, {\rm d}s$$

For $a=0$ the statement is clear, so we claim that $a\not= 0$.

Taking $h'(B_t) = e^{aB_t}$ we get $$\int_0^t e^{aB_s} \, {\rm d} B_s = \frac{1}{a}e^{aB_t} - \frac{1}{a}e^{aB_0} - \frac{1}{2} \int_0^t ae^{aB_s} \, {\rm d}s$$

Using expectation on both sides gives us the wanted result \begin{align*}E\left[\int_0^t e^{aB_s} \, {\rm d} B_s\right] &= \frac{1}{a}E\left[ e^{aB_t} \right] - \frac{1}{a}\cdot 1 - \frac{1}{2} E\left[ \int_0^t ae^{aB_s} \, {\rm d}s\right] \\ &= \frac{1}{a}\left(e^{\frac{a^2t}{2}} - 1\right) - \frac{a}{2}\int_0^t E\left[ e^{aB_s}\right] \, {\rm d}s \\ &= \frac{1}{a}\left(e^{\frac{a^2t}{2}} - 1\right) - \frac{a}{2}\int_0^t e^\frac{a^2s}{2} \, {\rm d}s \\ &= \frac{1}{a}\left(e^{\frac{a^2t}{2}} - 1\right) - \frac{1}{a}\left(e^{\frac{a^2t}{2}} - 1\right) = 0\end{align*}

where we can interchange expectation and integration in the second step by Fubini's theorem.