Covariation of these processes

Question

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 $d(N+M)$?

Answer 1

(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:

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