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

Is there an easy way to construct, on the same filtered probability space, a Brownian motion $W$ and a Poisson process $N$, such that $W$ and $N$ are not independent ?

share|cite|improve this question

In case (W,N) has independent increments, then W and N are also independent, since a Brownian Motion has no common jumps with the Poisson process. This of course doesn't say, there is no probability space, where (W,N) has dependent increments, but it gives you a hint, how it might be constructed.

See e.g. Kallenberg - Foundations of Modern Probability 13.6 for a proof

share|cite|improve this answer

Consider the following. Let $Y$ be a Brownian motion, and $N$ an independent Poisson process with rate $\lambda$. Let $t$ satisfy $e^{-\lambda t} = 1/2$. Define $W$ as follows. If $N_t = 0$, then $W = {\rm sgn}(Y_1) Y$; if $N_t > 0$, then $W = -{\rm sgn}(Y_1) Y$.

Anyone is welcome to give opinion on this suggestion.

share|cite|improve this answer
Thanks for the suggestion. I might be missing something but I don't think this works. Let ${\cal F}_t$ be the filtration generated by $N$ and $W$. First I assume that $Y_1$ should be $Y_t$ in your formula (if not, for instance for $t<1$, $sgn(W_1)$ is ${\cal F}_t$-measurable). So you have $W_t>0$ iff $N_t=0$. But if $N$ is a ${\cal F}_t$-Poisson, for s<t, $\mathbb{P}[N_t=O|{\cal F}_s]=e^{\lambda(t-s)} 1_{\left\{N_s=0\right\}}$. And if $W$ is a BM for the same filtration, $\mathbb{P}[W_t>O|{\cal F}_s]=\Phi(- W_s/\sqrt{t-s} )$. Clearly these can't be equal a.s. ? – pgassiat Mar 7 '11 at 23:24
@user7406: $Y_1$ should be $Y_1$... However, I have ignored the "filtered" part. It's just an example where $W$ and $N$ are not independent. I can give further details if you wish. – Shai Covo Mar 7 '11 at 23:40
I understand your answer. Indeed your construction is nice if you just want nonindependent $W$ and $N$, but I was more interested in the case where the filtration is the same for both, unfortunately I have no idea how to proceed. – pgassiat Mar 8 '11 at 9:06
@user7406: I suggest posting your question at – Shai Covo Mar 8 '11 at 9:47
Thanks. I have asked it there. – pgassiat Mar 8 '11 at 10:49

Hi I can suggest the following, start with independent $W$ and $N$ processes, you can build up a "dependent" couple $(W,N)$ by specifying a family of copulas for the finite dimensional law of the increments of the processes, then use Kolmogoroff extension theorem to finish the job.

Take a look here Chapter 5.

Hope this helps

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.