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The questions are simple:

Does the process $ X(t) = \int_0^t B(s)ds$ have independent increments?

What about $X(t) = \int_{t-r}^{t}B(s)ds$?

Here $B$ denotes the standard Brownian motion. Thanks a lot!

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2 Answers 2

Where $B(s)$ is a Brownian motion? No, the increments are certainly not independent for either of those, because the increment from $t$ to $t+\epsilon$ and the increment from $t+\epsilon$ to $t+2\epsilon$ are both close to $B(t) \epsilon$.

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Yes $B$ is a Brownian motion, sorry I forgot to say it. :) mm.. it's counterintuitive that $\int_s^t B(u)du$ is not independent of $\int_0^s B(u)du$ hehe thanks! –  Daniel Oct 25 '12 at 19:28
What is true is that (for $0 < s < t$) $\int_0^s B(u)\ du$ and $\int_s^t B(u)\ du$ are conditionally independent given $B(s)$. –  Robert Israel Oct 25 '12 at 20:51

Well I am not sure to fully understand the heuristic argument of Robert Israel so I give my answer hoping that someone can spot my mistakes if any.

So first let's remark that thanks to the representation of $X_t$ as the sum of two gaussian processes that are jointly gaussian.

Indeed, using integration by part formula (or Itô's lemma if you want to) we have :

$X_t=t.B_t+\int_0^tr.dB_r$ (the fact that a Wiener integral is a gaussian process is considered known the jointy gaussian of the process $tB_t$ and $\int_0^tr.dB_r$ is also not difficult to derive but I can elaborate on this if asked for).

So now, are increments of this gaussian process independents ? To get to the conclusion it suffices to examine the covariation of the increments. I get for $0<s<t<u<v$ the following boring calculation :


Here all the terms are null (this can be deduced for terms from Itô's isometry) except the fifth which is equal to $(t-s).(v-u).min(u,s)=(t-s).(v-u).s$

Whereas from the fact that $Y$ is centered :

As both expressions are not equal this is enough to prove that independence of increments does not occur.

The proof is rather inelegant though,

Best regards

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