TITLE: Itō Integral has expectation zero
QUESTION [28 upvotes]: I have a question about the following property, which I didn't know so far: 

Why does the Itō integral have zero expectation? Is this true for every integrator and integrand? Or is this restricted to special processes, i.e. is
  $$\mathbb{E}\left[\int f \, \mathrm{d}M\right]=0$$
  for all local Martingales $M$ and predictable $f$, such that the integral is well defined? 

Thank you for clarification.

REPLY [29 votes]: This statement is wrong in general.
It may fail even when the integrator $M_t$ is a Brownian motion. In fact,

Given a probability distribution $P$ on $\mathbb{R}$, it is possible
  to find an adapted $t$-measurable process $f(\omega,t)$, with
  $\mathbb{P}\left(\int_0^1 f^2(\omega,t)\,dt<\infty\right)=1$ such
  that the random variable $$\int_0^1 f(\omega,t) \, dB_t$$ has
  distribution $P$.

This statement is known as Dudley's representation theorem (see the original paper). Hence, the expectation of the stochastic integral may take any real value, be infinite or not exist at all. 

Another counterexample arises from the stochastic differential equation $$dX_t = X^2_t\, dB_t, \quad X_0=x, \quad \textrm{where } x>0.$$
It may be shown that the solution exists, is unique, is a strictly positive local martingale, but $\mathbb{E} X_t \to 0$ as $t\to \infty$.
See the details in George Lowther's blog, where this example is taken from.

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

$f(\omega,s)$ is adapted, measurable in s, and
$\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$.

In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

If the integrator $M_t$ is an arbitrary martingale, and the integrand $f$ is bounded, then the integral is a martingale, and the expectation of the integral is again zero (proof).

Finally, if the integrator $M_t$ is a local martingale, very little can be said about the expectation of the integral. If $f(\omega,t)$ is sufficiently nice, the integral $\int_0^t f(\omega,s) \, dM_s$ is a local martingale, but that does not guarantee that the expectation is zero, as the second counterexample above shows.