# Itô Integral has expectation zero

I have a question about the following property, which I didn't know so far:

Why does the Itô 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.

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Ito Integral is a martingale. –  user60610 Mar 16 '13 at 10:58

Consider the related stochastic process given by the SDE $dX_t = f(t,X_t,M_t) dM_t$ and note that since $M_t$ is a local martingale, $dM_t$ will only contribute to the volatility of the process, not to the drift. Therefore, $dX_t$ has no drift and hence

$\mathbb{E} \left[\int_{t_0}^{t_1} dX_t\right] = \mathbb{E} \left[X(t_1) - X(t_0)\right] = 0$.

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What exactly do you mean by "$dM_t$ will only contribute to the vlatility of the process, not the drift"? –  user20869 Nov 8 '12 at 15:59
@hulik If you consider any SDE, e.g. $X_t = \mu(t,X_t) dt + \sigma(t,X_t) dB_t$ for some Brownian motion $B_t$, $\mu$ is called drift and $\sigma$ is called volatility. Since $M_t$ is a local martingale, $fdM$ has no drift, but drift is the only thing contributing to the expected value of the difference... –  gt6989b Nov 8 '12 at 16:04
thank you for your quick reply. But I think then the problem/question is just reformulated. I would then ask, why does the volatility not contribute to the expected value, or expected value of the difference? Thanks for your patience –  user20869 Nov 8 '12 at 16:12
Intuitive answer: for an Ito integral with respect to Brownian motion (and nice enough $f$), $\mathbb{E}\left[\int_0^t f(B_s) dB_s\right] = 0$ because each little $dB$ has mean zero - in fact, has distribution that is symmetric about zero (and, independent of where $B$ is!). You can think of the integral, just like a normal integral, as a weighted sum of lots of little $dB$'s; and the fact that you multiply them by a factor doesn't change the fact that their mean is zero. The fact that's being used here is exactly the martingale property.
As for the most general statement: in Kallenberg (15.12) I find that if $M$ is a continuous local martingale with (finite) quadratic variation process $[M]$, and $V$ is a progressive process (implies predictable) with $\mathbb{E}[\int_0^t V^2_s d[M]_s] < \infty$ for all $t>0$, then $N_t = \int_0^t V_s dM_s$ is a continuous local martingale, which implies $\mathbb{E}[N_t]=0$ for all $t>0$.
It is not clear to me why that would be so unless $f$ is constant. To be concrete, imagine that the taxman takes a cut of your gambling earnings but doesn't refund your losses. Then the integral would not be a martingale, would it? –  Pait Dec 30 '13 at 14:59
@pait : First, the taxman example doesn't work, since that's an integral against $dt$, not $dB_t$. Second, if you are convinced about constant functions, are you convinced about piecewise constant functions? If so, then, take limits of such to get it for arbitrary $f$ -- this is one way the theorem can be proved. –  petrelharp Dec 31 '13 at 15:58