What is the Malliavin derivative of $\int_0^T f(B(t))dB(t)$?

119 Views Asked by At

What is the Malliavin derivative of $F=\int_0^T f(B(t))dB(t)$?

I know if $f$ is deterministic, then the Malliavin derivative of $F$ is just $f(t)$. But what if $f$ depends on the path, can we say anything then?

1

There are 1 best solutions below

0
On BEST ANSWER

Assume that $f$ is good enough so that everything I'll write is well defined, in that case we have that

$$D_t \int_0^T f(B(s)dB(s)=f(B(t))+\int_t^T f'(B(s))dB(s).$$

This is a particular case of proposition 1.3.8. of Nualart's book.