6
$\begingroup$

Let $\{N_t|0<t\leqslant T \}$ and $\{M_t|0<t\leqslant T \}$ 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 $$dN_t dM_t = 0.$$

Can anyone please help me with the proof of this equation?

$\endgroup$
9
$\begingroup$

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.)

$\endgroup$
6
  • 1
    $\begingroup$ 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)? $\endgroup$
    – Alex
    Jun 26 at 18:30
  • 3
    $\begingroup$ @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. $\endgroup$
    – ir7
    Jun 26 at 18:48
  • 3
    $\begingroup$ @Alex This resource has the table, Table 20.1 (but not the cross-term proofs, I think). $\endgroup$
    – ir7
    Jun 26 at 19:01
  • $\begingroup$ cool, thank you very much for the table. That's really interesting. Thank you! $\endgroup$
    – Alex
    Jun 26 at 19:15
  • 1
    $\begingroup$ @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. $\endgroup$
    – ir7
    Jun 27 at 19:29

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.