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Let $N_t \sim \text{Poisson}(\lambda t)$ and $M_t \sim \text{Poisson}(\theta \lambda t)$.

We know that if $N$ and $M$ were independent, $dNdM = 0$ using polarization identity. We also know that $(dN)^2 = dN$; but now that these two processes are correlated, how can we calculate $dNdM$ ?

I though about polarization identity and putting it in differential notations and given that $N+M$ is also a Poisson process, we can write \begin{align*} dNdM &= \frac{1}{2}\left[ \left(d(N+M)\right)^2 - (dN)^2 - (dM)^2 \right] \\ &= \frac{1}{2}\left[ d(N+M) - dN - dM \right] \end{align*} But how can we calculate $d(N+M)$?

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  • $\begingroup$ This is likely non-trivial, I believe there are a few different ways to construct correlated Poisson processes. $\endgroup$ Jun 28, 2021 at 10:50
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    $\begingroup$ But something to note: $d(N+M)^2\neq d(N+M)$, so your last equality is wrong. $\endgroup$ Jun 28, 2021 at 13:52
  • $\begingroup$ @DaneelOlivaw Could you please clarify? If both $N$ and $M$ are Poisson processes, can't we say that $N+M$ is also a Poisson process, regardless of the correlation, hence $(d(N+M))^2 = d(N+M)$? $\endgroup$
    – user57062
    Jun 28, 2021 at 14:34
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    $\begingroup$ Ok I see, I am not sure then, your comment makes sense. But you should have $d(N+M)=dN+dM$, so you would end up with $dNdM=0$ as in your previous question, which does not seem correct. I've found this thesis, might be relevant to you: citeseerx.ist.psu.edu/viewdoc/… $\endgroup$ Jun 28, 2021 at 15:35

1 Answer 1

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(Special case only.)

One special way to create correlated Poisson processes is using a common 'shock' model idea.

For $X$, $Y$, and $Z$ independent Poisson processes, let's define:

$$ M = X+ Z, \; \; N = Y+Z.$$

We note that $M$ and $N$ are Poisson processes, but that $M+N$ is not ($2Z$ is not a Poisson process).

We also note that the Pearson correlation between $M_t$ and $N_t$ is not time-dependent and it is always positive (since intensities are positive):

$$ \rho(M_t, N_t) = \frac{\lambda_Z}{\sqrt{(\lambda_X+\lambda_Z)((\lambda_Y+\lambda_Z)}} $$

Formally we also get:

$$ dMdN = (dX +dZ)(dY+dZ) = dXdY+dXdZ +dYdZ + (dZ)^2 = dZ. $$

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  • $\begingroup$ I know some constructions of correlated Poisson processes put a constraint on the possible values of correlation, is it the case with this one? $\endgroup$ Jun 28, 2021 at 16:29
  • $\begingroup$ Yes. The method above using only standard Poisson processes has that shortcoming. $\endgroup$
    – ir7
    Jun 28, 2021 at 16:48
  • $\begingroup$ Okay I see, thanks. $\endgroup$ Jun 28, 2021 at 17:45

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