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 $d(N+M)$?

  • $\begingroup$ This is likely non-trivial, I believe there are a few different ways to construct correlated Poisson processes. $\endgroup$ Jun 28 at 10:50
  • 1
    $\begingroup$ But something to note: $d(N+M)^2\neq d(N+M)$, so your last equality is wrong. $\endgroup$ Jun 28 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 $(d(N+M))^2 = d(N+M)$? $\endgroup$ Jun 28 at 14:34
  • 1
    $\begingroup$ 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 at 15:35

(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:

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

  • $\begingroup$ 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 at 16:29
  • $\begingroup$ Yes. The method above using only standard Poisson processes has that shortcoming. $\endgroup$
    – ir7
    Jun 28 at 16:48
  • $\begingroup$ Okay I see, thanks. $\endgroup$ Jun 28 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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