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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?

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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.)

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

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