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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 ?

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3 Answers

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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.

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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 mathoverflow.net – Shai Covo Mar 8 '11 at 9:47
Thanks. I have asked it there. – pgassiat Mar 8 '11 at 10:49
up vote 1 down vote

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

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up vote 0 down vote

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

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