Assume you have a Brownian $B_t$ motion equipped with its natural filtration $\{\mathcal{F}_t\}$ on a probability space.

Assume that $f: \mathbb{R}_+\times\Omega\rightarrow \mathbb{R}$. Is a jointly measurable, function, which for each t is adapted to the filtration above. Also assume that $E[\int_0^Tf^2dt]<\infty$.

Then it is shown that the stochastic integral $\int_0^tf(s,\omega)dB_s$ is a martingale with respect to the filtration $\{\mathcal{F}_t\}$.

Let $t \in [0,T]$. I wonder what we can say about indepence between $\int_0^tf(s,\omega)dB_s$ and $\int_t^Tf(s,\omega)dB_s?$

What I think I am able to prove for myself, is that when $f$ is deterministic, these integrals are independent, because of the independent increments of Brownian motion? Do you agree with this?

What I struggle with is the part when f may not be deterministic. Here I am not able to prove independence or find a counterexample. I tried for instance integrating the brownian motion with itself, then the two integrals are $(B_t^2-t)/2$ and $(B_T^2-B_t^2-T+t)/2$, however I am not sure how to find out if these are independent or not. I tried doing something numerically and they seem to not be correlated, but that does not exclude independence?:

> t = 0.5
> T = 1
> n=10000000
> B_t =rnorm(n,0,sqrt(t))
> B_T = B_t+rnorm(n,0,sqrt(T-t))
> I_1 = (B_t^2-t)/2
> I_temp =(B_T^2-T)/2
> I_2 = I_temp-I_1
> cor(I_1,I_2)
[1] 0.0005113802

So do you know the answer to this?, are the integrals independent when the integrand is deterministic? And what happens in the general case? Are they independent here also?


If $f$ is deterministic, i.e. $f(s,\omega)=f(s)$, then $\int_0^t f(s) \, B_s$ and $\int_t^T f(s) \, dB_s$ are independent; this follows from the independence of the increments of Brownian motion.

In general, however, the two random variables are not independent; for instance, this answer shows that $X_t := \int_0^t B_s^2 \,d B_s$ does not have independent increments. There are other counterexamples, but most of them require rather lengthy calculations to disprove the independence of the increments.

Although $\int_0^t f(s) \, dB_s$ and $\int_t^T f(s) \, dB_s$ are, in general, not independent, they are uncorrelated for any (twice integrable progressively measurable) function $f=f(s,\omega)$. This follows from Itô's isometry which states that

$$\mathbb{E} \left[ \left( \int_0^T f(s) \, dB_s \right) \left( \int_0^T g(s) \, dB_s \right) \right] = \mathbb{E} \left( \int_0^T f(s) g(s) \, ds \right).$$

(Note that the right-hand side equals $0$ if $f$ and $g$ have disjoint support.)

  • $\begingroup$ @user119615 You are welcome. $\endgroup$ – saz Oct 6 '16 at 17:26

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