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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!

Kyoto
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    $$\sqrt{\pi\mathrm e^t},\Phi(\sqrt{2/(1+2t)})$$ – Did Feb 09 '15 at 16:08
  • @Did, tahnk you! But I'm kind of new to stochastics. How did you get this result? – Kyoto Feb 09 '15 at 20:00
  • ?? I do not get it: you "accepted" an answer hence everything is clear to you, no? – Did Feb 09 '15 at 23:53
  • I accepted answer, yes. But I do not understand how did you get your result. :( – Kyoto Feb 10 '15 at 06:20
  • As I understand things, you accepted this answer and it ends by "you proceed from here" hence you "proceeded from here" and got a result. What is this result already? – Did Feb 10 '15 at 08:25
  • Result that you have written in the first comment. I have supposed that is the answer to my question. But I don't know how you got it. – Kyoto Feb 10 '15 at 08:52
  • (Last comment from me.) The question is to know the result YOU arrived at. – Did Feb 10 '15 at 09:48
  • @Did, Hi. I wanted to let you know that your final solution form is referenced in my latest response to the OP, as it results rather nicely from a more direct approach (than using successive Ito approximations) to solving the problem. Please let me know if the acknowledgment is lacking in any way and I will amend it. Thank you. – ki3i Feb 10 '15 at 11:59
  • @ki3i Well done (and a factor $t$ was missing from the formula in my comment, apparently). All this leaves fully open the question of which result the OP "arrived at", since they refuse to say. – Did Feb 10 '15 at 12:13
  • @Did, Thank you. Concerning the OP, it might be that he/she is new to the site and does not know that accepting an answer means that the answer has given you what you need to solve your problem? So they accepted the answer before fully answering the question themselves. I can only speculate. However, I hope they find this useful. – ki3i Feb 10 '15 at 12:41

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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.

ki3i
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  • Thanks for this. But can I proceed from this in the way that I "lose" the expectation? – Kyoto Feb 09 '15 at 19:58
  • @Kyoto, Proceeding from here will result in Ito integrals whose expectations will need to be evaluated. For instance, $$\mathbb{E}\left[e^{W_t}\frac{\sqrt{\pi}}{2}\right] = e^{\frac{1}{2}t}\frac{\sqrt{\pi}}{2},.$$ Another term is $$\mathbb{E}\left[\int\limits_{0}^{t}e^{W_u-W_u^2}\mathrm du\right] = \frac{t}{\sqrt{1+2t}}e^{\frac{1}{2}\frac{t}{1+2t}},.$$ Admittedly, this gives a series solution via the evaluation of multiple integrals. A more direct approach would be to use the standard normal CDF properties and $W_t\sim\mathcal{N}(0,t)$; doing so gives Did's compact result. – ki3i Feb 10 '15 at 05:27
  • Can this expectation be evaluated $$E\left[e^{W_t}\int_0^te^{−W^2_u}dWu\right]?$$ I don't undestand how can I lose $dW_u$. :( – Kyoto Feb 10 '15 at 06:35
  • @Kyoto, please see edits. – ki3i Feb 10 '15 at 11:22
  • My last question. I don't understand how do you get the last equality.

    $$ \frac{1}{\sqrt{2\pi(2t)}}\int_{-\infty}^\infty\Phi(y)e^{-\frac{1}{2(2t)}\left(y-\sqrt{2}t\right)^2}dy=\Phi\left(\frac{\sqrt{2}t}{\sqrt{1+2t}}\right) $$

     – Kyoto Feb 10 '15 at 14:11
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    @Kyoto, Suppose $X\sim\mathcal{N}(0,1)$ and $Y\sim\mathcal{N}(\sqrt{2}t,2t)$ are independent random variables. Then, the equality follows by computing $P(X<Y)$ in two different ways. Can you see it from the property used here: http://math.stackexchange.com/questions/1102336/find-the-distribution-of-sum-and-product-of-standard-normal-random-variables/1102864#1102864 ? Note: $X-Y$ is normally distributed. – ki3i Feb 10 '15 at 15:18