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?
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?
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.