I want to calculate $$\mathbb{E} \left( \int_0^T \exp(s + B_s) dB_s \right)^2,$$ where $(B_t)_{t \geq 0}$ is a Brownian motion. It makes sense that we wish to apply Ito isometry. For this, we need the process $s \mapsto \exp(s+B_s)$ to be in $H^2$. It is clearly adapted, being a continuous function of the brownian motion, but I don't see how $$\mathbb{E} \left( \int_0^t \left| \exp(s+B_s) \right|^2 ds \right) < \infty.$$

Suppose for the moment that I was able to show that $F: s \mapsto \exp(s+B_s)$ was in $H^2$, I am still having trouble applying the Ito isometry. Since \begin{eqnarray*} \mathbb{E} \left( \int_0^T \exp(s+B_s) dB_s \right)^2 &=& \mathbb{E} \left( \int_0^T \exp(2s + 2B_s) ds \right) \\ &=& \int_0^T \mathbb{E}(\exp(2s+2B_s)) ds. \end{eqnarray*} It would make sense to then calculate $\int_{-\infty}^{\infty} \exp(2s+2B_s)p(x,s) ds dx,$ where $p(s,x)$ is pdf of the Gaussian, but we don't know the mean or variance of $\exp(2s+2B_s)$.


You seem a bit confused. In order to compute $\mathbb{E}\exp(2s+2B_s)$, it suffices to notice that $B_s$ has normal distribution $\mathcal{N}(0, s)$ and hence

$$ \mathbb{E}\exp(2s+2B_s) = \int_{\mathbb{R}} e^{2s+2x} \cdot \frac{1}{\sqrt{2\pi s}} e^{-\frac{x^2}{2s}} \, dx = e^{4s}.$$

So it follows that

$$ \int_{0}^{T} \mathbb{E}\exp(2s+2B_s) \, ds = \int_{0}^{T} e^{4s} \, ds = \frac{e^{4T}-1}{4} < \infty. $$

This also justifies that the process is in $H^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy