On the conditional expectation of Ito integral

Question

In general, one can prove that \begin{equation} \mathbb{E}[f(X_t)|X_t=x]=f(x)\tag{1} \end{equation} where $X_t$ is the Ito process: $\mathrm{d}X_t=a(t,X_t)\mathrm{d}t+b(t,X_t)\mathrm{d}W_t$. Suppose now I have to evaluate $$ \mathbb{E}\left[\int_{0}^Tc(t,X_t)\mathrm{d}W_t\bigg|X_t=x\right] $$

Is it $\displaystyle \int_0^T c(t,x)\mathrm{d}W_t$ or $\displaystyle\mathbb{E}\left[\int_0^Tc(t,x)\mathrm{d}W_t\right]=0$? The correct answer should be the second but is this coherent with $(1)$?

Answer 1

The statement you want to prove is valid if the integral is a true martingale.

Indeed it is an elementary property of the stochastic integral to have null expectation if it is a martingale.

A suffcient condition for the integral $\int_{0}^{t}c(\omega, s) dM_s$ to be a martingale on $[0,T]$ is that:

1. $c(w,s)$ is adapted and measurable in $s$
2. $\mathbb{E}[\int_{0}^{T}c^2(\omega, s)ds] < \infty$

If, in particular, $M_t$ is a martingale and $c$ is bounded, then it can be proved that the integral is also a martingale.

If in general $M_t$ is only a local martingale we can't say much about that expectation.