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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Stochastic Processes for Physicists by Jacobs says that we can exchange the order of a multiple Ito stochastic integral, giving the example:

enter image description here

I don't see how this works either for a regular integral or a stochastic integral.

For a regular integral, suppose I let $f=1$ and $W(s) = s^2$ so $dW(s) = 2s ds$. I'm getting that (3.88) evaluates to $T^3/3$ while (3.89) evaluates to $2T^3/3$. Am I missing something?

For a stochastic Ito integral, if $f=1$ and $dW(s)$ is Gaussian, then (3.89) simplifies to

$$I = \int_0^T t dW(t)$$

which gives a Gaussian with zero mean and variance $T^3/3$.

I don't completely understand the meaning of (3.88) in this context. Does it also imply that $I$ is Gaussian with a variance $T^3/3$?

What are the missing steps implied in "discretizing the integral"? (I tried writing this out, but didn't find something that was obviously the same sum.)

share|cite|improve this question
How can $dW_s = 2s \, ds$? By definition $dW_s \equiv W_{s+ds} - W_s$. – wsw Oct 10 '13 at 15:39
$(s+ds)^2 - s^2 = 2s ds + ds^2 = 2s ds$ – Mark Eichenlaub Oct 10 '13 at 15:44
Mark: $dW_s \sim \mathcal{N}(0, ds)$ is a random variable, while $ds$ is a deterministic quantity. For example, $\text{Var}(ds) = 0$. – wsw Oct 10 '13 at 15:57
Mark: I think I know what the problem is. You simply cannot say $W_s = s^2$ since $W_s$ is Brownian motion. – wsw Oct 10 '13 at 17:59
In the portion where I set W=s^2, I am not talking about stochastic integrals, but instead just a regular integral, so your criticisms do not make sense. – Mark Eichenlaub Oct 10 '13 at 22:02
up vote 3 down vote accepted

Mark: I got it -- there's a typo in the book. Note that the integration region is triangular. In other words, $$ \int_0^T \left( \int_0^t g_{s,t} \, dy_s \right) dt = \int_0^T \left( \int_s^T g_{s,t} \, dt \right) dy_s \, . $$

Let's use your example of $y = s^2$, where $dy = 2s \, ds$.

$$ \int_0^T \int_0^t 2s \, ds \, dt = \int_0^T t^2 \, dt = \frac{T^3}{3}. $$

For the other way, $$ \int_0^T \int_s^T dt \, 2s \, ds = \int_0^T (T-s) 2s \, ds = T^3 - \frac{2}{3}T^3 = \frac{T^3}{3}. $$

share|cite|improve this answer
I edited your answer because your LaTeX code wasn't right. I hope you don't mind. – Cameron Williams Oct 11 '13 at 0:38

Your Answer


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.