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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.

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

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

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

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.

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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
The expectation of a local martingale is not necessarily constant, so this is not always true. – Georgy Ivanov Jun 10 '14 at 16:40

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.

[EDIT:] But, as pointed out by others, this intuitive answer does not necessarily hold: the Ito integral might turn out to be only a local martingale, not a martingale. However, being a local martingale means it has mean zero "locally": see the definition.

As for a more 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. If this is a martingale, then $\mathbb{E}[N_t]=0$ for all $t>0$. Since in general it's only a local martingale, instead there is a sequence of increasing stopping times $\tau_k$ tending to $\infty$ such that $\mathbb{E}[N_{\min(t,\tau_k)}] = 0$ for each $k$.

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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
The fact that $N_t$ is a continuous local martingale does not imply that its expectation is zero. – Georgy Ivanov Jun 10 '14 at 16:41
oops, yep. fixed. – petrelharp Jun 13 '14 at 3:41

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