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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.

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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
up vote 3 down vote accepted


  • 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 $$\left\langle\left\langle \exp\left(\int_0^t W_s d W_s \right)^2\right\rangle\right\rangle = e^{-t/2} \left( \frac{1}{\sqrt{1-t}} - \frac{1}{1-t/2}\right).$$

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...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

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