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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 ?

3 Answers 3

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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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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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    @user7406: I suggest posting your question at http://mathoverflow.net/2011-03-08
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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