Integral on interval $[-\infty,W_t]$, $W_t$ is Brownian motion

Question

Basicaly I have an expectation of an integral on the interval which contains Brownian motion and it look like this. $$ E\left[e^{W_t}\cdot\int_{-\infty}^{W_t} e^{-z^2}dz\right] $$

$W_t$ is Brownian motion, $t$ is fixed. My question is: Can I write this is any other form or, can calculate this?

Any help would be appreciated. Thanks!

Answer 1

Hint::

For the function

$$ f(x):=\int\limits^{x}_{-\infty}e^{-r^2}\mathrm d r\,, $$

using Leibniz's rule and Ito's lemma tells us

$$ \mathrm df(W_t) = e^{-W_t^2}\mathrm dW_t-W_t~e^{-W_t^2}\mathrm dt\,. $$

Therefore,

$$ \int\limits^{W_t}_{-\infty}e^{-r^2}\mathrm d r = \dfrac{\sqrt{\pi}}{2}{}{}+{}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u-\int\limits_{0}^{t}W_u~e^{-W_u^2}\mathrm du\,. $$

This puts the argument of the expectation in a form that is "more appealing". Can you proceed from here?

Edit: This approach leads to a successive evaluation of expectations by successive use of Ito's lemma. The result is a series representation of the desired expectation. So, for instance, expansions like $$ \mathbb{E}\left[e^{W_t}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] = \mathbb{E}\left[\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] + \mathbb{E}\left[\int\limits_{0}^{t}e^{W_u-W_u^2}\mathrm du\right] + \frac{1}{2}\mathbb{E}\left[\int\limits_{0}^{t}e^{W_u}\mathrm du\int\limits_{0}^{t}e^{-W^2_u}\mathrm dW_u\right] $$

follow from observing $$ e^{W_t}=1 + \int\limits_{0}^{t}e^{W_u}\mathrm dW_u + \frac{1}{2}\int\limits_{0}^{t}e^{W_u}\mathrm du\,. $$

This is akin to situations in ordinary calculus where integrals are solved by obtaining a Taylor expansion for the solution. However, this is more involved than it needs to be. Certainly, a more direct approach would be to use the properties of the standard normal CDF, $\Phi$, and $W_t\sim\mathcal{N}\left(0,t\right)$ to justify the following:

$$ \begin{eqnarray*} \mathbb{E}\left[e^{W_t}\int\limits^{W_t}_{-\infty}e^{-u^2}\mathrm du\right] = \mathbb{E}\left[\sqrt{\pi}e^{W_t}\Phi(\sqrt{2}W_t)\right] = \sqrt{\pi}\frac{1}{\sqrt{2\pi t}}\int\limits_{-\infty}^{\infty}\Phi(\sqrt{2}x)e^{-\frac{1}{2}\frac{x^2}{t}+x}\mathrm dx = \sqrt{\pi}\frac{e^{\frac{1}{2}t}}{\sqrt{2\pi(2t)}}\int\limits^{\infty}_{-\infty}\Phi(y)e^{-\frac{1}{4t}\left(y-\sqrt{2}t\right)^2}\mathrm dy = \sqrt{\pi}e^{\frac{1}{2}t}\Phi\left(\frac{\sqrt{2}t}{\sqrt{1+2t}}\right), \end{eqnarray*} $$

resulting in a final form for the solution originally suggested by @Did.