Consider a functional on Wiener space $F\in L^2(\Omega, \Bbb R)$. Then by Ito's representation theorem we have that $$F=E[F]+\int_0^T\phi(s)dB(s)$$ for some nice $\phi$. Question, when is $\phi$ deterministic? I know that if $\phi$ is deterministic then $F$ must be Gaussian as the Ito integral of a deterministic function is Gaussian.
However is the converse true? If $F\sim \mathcal N(\mu,\sigma^2)$ can we find a deterministic $\phi$ so that $$F=\mu+\int_0^T \phi(s) dB(s)$$
What about just letting $\phi(s)\equiv \sigma/\sqrt{T}$?
Example: Take $$ \phi(s) = \cases{ 1,& $0\le s\le T/2$;\cr \xi,& $T/2<s\le T$.\cr} $$ where $\xi$ is the sign of $B_{T/2}$. This $\phi$ is non-deterministic. But, using the facts that $\xi$ is (i) $\mathcal F_{T/2}$ measurable and therefore (ii) independent of the post-$T/2$ increments of the Brownian motion, one can check that $F=\int_0^T\phi(s)\,dB_s=B_{T/2}+\xi\cdot(B_T-B_{T/2})$ has the normal distribution with mean $0$ and variance $T$.