4

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.)

Mark Eichenlaub
  • 6,733
  • 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
  • 1
    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

3 Answers3

4

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

wsw
  • 336
  • 1
  • 10
1

Good question. Usually it can be made by Ito formula (difficult in some situations), but I derived integration order replacement theorems in multiple Ito stochastic integrals, see the books:

http://www.math.spbu.ru/diffjournal/pdf/kuznetsov_book2.pdf

http://www.math.spbu.ru/diffjournal/pdf/kuznetsov_book3.pdf

Dmitriy F. Kuznetsov
  • 21
0

More correctly. We can replace integration order in multiple stochastic Ito integrals even, when all integrals are stochastic with dW(t). In this case we must to use a special definition of stochastic integral (at the first time this definition was introduced by R.L. Stratonovich (I say here not about Stratonovich stochastic integral)) to avoid the situation, when intergrand is not Ft-measurable. I derived these formulas. Please check

pages 329-349 in the book (in English):

http://www.math.spbu.ru/diffjournal/pdf/kuznetsov_book2.pdf

or pages 145-165 in the book (in Russian):

http://www.math.spbu.ru/diffjournal/pdf/kuznetsov_book3.pdf

Femkemilene
  • 13
  • 4
Dmitriy F. Kuznetsov
  • 1