MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
6

4

One can read about Walsh's construction of martingale integral in the paper (pp.16-23) www.math.utah.edu/~davar/ps-pdf-files/SPDEBookDK.pdf

For $U,V\in \mathcal{B}(\mathbb{R}\times \mathbb{R}^+), \phi:U\to V$ is non-singular linear operator

I hope there was a formula like this:

$$ E\left|\iint_{\phi(U)}\xi(y,s)M(dy,ds)\right|^2=E\left|\iint_U \xi(\phi(x,t))|\det \phi|M(dx,dt)\right|^2 $$

or it's maybe more simple when $M$ is the 2-dimensional white noise (or Brownian sheet)

flag

1 Answer

1

In the case of white noise, if $\xi$ is deterministic, then \begin{align*} E\bigg|\iint_{\phi(U)}\xi(y,s)W(dy,ds)\bigg|^2 &= E\bigg|\iint\xi(y,s)1_{\phi(U)}(y,s)W(dy,ds)\bigg|^2\\ &= \iint|\xi(y,s)1_{\phi(U)}(y,s)|^2\,dy\,ds\\ &= \iint_{\phi(U)}|\xi(y,s)|^2\,dy\,ds\\ &= \iint_U |\xi(\phi(x,t))|^2\,|\det\phi|\,dx\,dt\\ &= E\bigg|\iint_U \xi(\phi(x,t))\,|\det\phi|^{1/2}\,W(dx,dt)\bigg|^2. \end{align*}

link|flag

Your Answer

Get an OpenID
or

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