Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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 $B(t) \epsilon$.

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.