Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to compute the expectation and variance of the following stochastic process: $$ Z_t = \exp \left( \frac{1}{2} \int_0^t W_s \, dW_s \right) $$ where $W_t$ is a standard Brownian motion. I have tried the following to compute the expectation:

Let $Y_t = \int_0^t W_s \, dW_s$, then $dY_t = W_t \ dW_t$, and applying the Ito formula to $Z_t = \exp\left( \frac{1}{2} Y_t \right) $ I get $$ dZ_t = \frac{1}{2} Z_t W_t \, dW_t + \frac{1}{8} Z_t W_t^2 \, dt $$ Writing this in integral form and taking expectations, the $dW_t$ integral vanishes and I find that $$ \mathbb{E}(Z_t) = 1 + \frac{1}{8} \int_0^t \mathbb{E}(Z_s W_s^2) \, ds $$

So it seems like in order to compute $ \mathbb{E}(Z_t) $, I need to compute $ \mathbb{E}(Z_t W_t^2)$. That involves expressing $ Z_t W_t^2$ as an Ito process, which in turn involves expressing $Z_t W_t^n$ as an Ito process for some higher power of $n$, and this process does not seem to terminate. Is there some trick I can use to simplify the computation of this expectation?

I am running into the same issues when computing $\mathbb{E}(Z_t^2) $ to find the variance as well, so suggestions for that computation would also be very welcome.

share|improve this question
    
Does this help: $\int_0^t W_s d W_s = \frac{1}{2} (W_t^2 - t)$? –  Fabian Jan 24 '13 at 0:30
    
That makes the problem really easy; I feel really stupid now. How did you think of coming up with that? –  Jonas Jan 24 '13 at 0:32
    
See my hint.... –  Fabian Jan 24 '13 at 6:26
add comment

1 Answer 1

up vote 3 down vote accepted

Hints:

  • Using Ito's rule you can show that $$\int_0^t W_s d W_s = \frac{1}{2} (W_t^2 - t).$$ To find this releation define $X_t = W_t^2$. Then we have (using Ito) $$d W_t^2 = 2 W_t dW_t + dt .$$ Integrating this SDE, we obtain $$W_t^2= 2 \int_0^t W_s dW_s + t .$$

  • With that the problem reduces to evaluating moments of $$\exp((W_t^2 -t)/4)$$ where $W_t$ is simply a Gaussian variable and $t$ a constant...

Edit: For completeness, I present the results of the expectation value and variance below. Note that the probability $p(x)$ distribution of $X_t= W_t^2$ assumes the form $p(x) =\frac1{\sqrt{2\pi t}} e^{-x^2/2t}$. So we have to evaluate the expectation value $$\left\langle \exp\left(\int_0^t W_s d W_s \right)\right\rangle = e^{-t/4}\int dx\,p(x) e^{x^2/4} = \frac{e^{-t/4}}{\sqrt{1-t/2}} $$ and the second moment $$\left\langle \exp\left(\int_0^t W_s d W_s \right)^2\right\rangle = e^{-t/2}\int dx\,p(x) e^{x^2/2} = \frac{e^{-t/2}}{\sqrt{1-t}} $$ thus the variance is $$\left\langle\left\langle \exp\left(\int_0^t W_s d W_s \right)^2\right\rangle\right\rangle = e^{-t/2} \left( \frac{1}{\sqrt{1-t}} - \frac{1}{1-t/2}\right).$$

share|improve this answer
    
...Whose $k$th moment is finite only if $t\lt1/(2k)$. (+1) –  Did Jan 24 '13 at 6:46
    
@Did: thanks for the appendum. I did not realize this fact... –  Fabian Jan 24 '13 at 7:29
    
Sorry: ...only if $t\lt2/k$. –  Did Jan 24 '13 at 8:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.