# Ito multiplication

Let $$\{N_t|0 and $$\{M_t|0 be two Poisson processes with intensities $$\lambda_n, \lambda_m>0$$, respectively.

Based on the implicit results of Corollaries 1 and 2 of this article and Theorem 1 of this article, I think we should be able to write $$dN_t dM_t = 0.$$

If $$M$$ and $$N$$ are independent (your references appear to make this assumption), then $$M+N$$ is also a Poisson process. So, using the polarization identity:

$$dMdN = 2^{-1}\left[(d(M+N))^2 - (dM)^2 - (dN)^2\right]$$

$$= 2^{-1}\left[d(M+N) - dM - dN \right] = 0$$

(A proof of $$(dX)^2 = dX$$ for a Poisson process $$X$$ is available here.)

• Great answer (+1)! It made me think: For a Brownian motion, we have $dWdt=0$, $dW^2=dt$ and $dW^3=0$. Are there analogues for Poisson processes? Your link suggests $dN^2=dN$ and $dN^3=dN$, etc.? What about cross-terms like $dNdt$ or $dWdN$ (if $W$ and $N$ are independent)?
– Alex
Jun 26 at 18:30
• @Alex Yes, $(dN)^3 = (dN)^2\cdot (dN) = dN\cdot dN = dN$. And, yes, based on the so-called "Ito multiplication table for Brownian motion and jumps" (aka "and Poisson process"), the two cross-terms you mentioned are 0. Make the cross-terms a SE Quant question if you are looking for some sort of proofs or proof references.
– ir7
Jun 26 at 18:48
• @Alex This resource has the table, Table 20.1 (but not the cross-term proofs, I think).
– ir7
Jun 26 at 19:01
• cool, thank you very much for the table. That's really interesting. Thank you!
– Alex
Jun 26 at 19:15
• @VultraUiolet Make it a SE Quant question, basically, how does one create correlated Poisson processes (common shock model comes to mind; just stating different constant intensities does not imply process dependence) and then how does one compute their quadratic covariation.
– ir7
Jun 27 at 19:29