# Covariation of these processes

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 $$d(N+M)$$?

• This is likely non-trivial, I believe there are a few different ways to construct correlated Poisson processes. Jun 28 at 10:50
• But something to note: $d(N+M)^2\neq d(N+M)$, so your last equality is wrong. Jun 28 at 13:52
• @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 $(d(N+M))^2 = d(N+M)$? Jun 28 at 14:34
• 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/… Jun 28 at 15:35

## 1 Answer

(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:

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

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