Solution: Covariation processes of martingales

Let \(\lambda_1(t)\) and \(\lambda_2(t)\) denote the intensity processes for \(N_1\) and \(N_2\), respectively, and \(\Lambda_1(t) = \int_0^t \lambda_1 (u)du\) and \(\Lambda_2(t) = \int_0^t \lambda_2 (u)du\) the corresponding cumulative intensity processes. The Doob-Meyer decompositions of \(N_1\) and \(N_2\) are thus \(N_1(t) = M_1(t) + \Lambda_1 (t)\) and \(N_2(t) = M_2(t) + \Lambda_2 (t)\), respectively, where \(M_1\) and \(M_2\) are mean-zero martingales. Since \(N_1\) and \(N_2\) do not jump simultaneously, \(N = N_1+N_2\) is also a counting process, and \(M = M_1 + M_2\) is also a mean-zero martingale.

Since \(\lambda_1 (t)\) and \(\lambda_2 (t)\) are both predictable processes, so is \(\lambda (t)=\lambda_1 (t)+\lambda_2 (t)\) . Further we have \[\Lambda (t) = \int_0^t \lambda (u)du = \int_0^t (\lambda_1 (u) + \lambda_2 (u))du = \int_0^t \lambda_1 (u)du+ \int_0^t \lambda_2 (u)du = \Lambda_1(t)+\Lambda_2(t).\] We get that \[N(t) = N_1(t) + N_2(t) = M_1(t) + \Lambda_1 (t) + M_2(t) + \Lambda_2 (t) = M_1(t) + M_2(t) + \Lambda_1 (t) + \Lambda_2 (t) = M(t) + \Lambda(t),\] which thus is the unique Doob-Meyer decomposition of \(N\).

Next, \((2.42)\) yields \[[M_1+M_2]=[M] = N = N_1+N_2, \quad [M_1] = N_1, \quad [M_2] = N_2,\] while \((2.43)\) yields \[\langle M_1+M_2\rangle=\langle M\rangle = \Lambda = \Lambda_1+\Lambda_2, \quad \langle M_1\rangle = \Lambda_1, \quad \langle M_2\rangle = \Lambda_2.\] Finally we obtain from \((2.28)\) and the results above that \begin{align*}\langle M_1+M_2\rangle &= \langle M_1 \rangle + \langle M_2 \rangle + 2\langle M_1,M_2 \rangle \\ &= \Lambda_1+\Lambda_2 + 2\langle M_1,M_2 \rangle \\ &= \langle M_1+M_2\rangle + 2\langle M_1,M_2 \rangle,\end{align*} and so \[\underline{\underline{\langle M_1,M_2 \rangle (t) = 0.}}\]

Likewise, from the corresponding result to \((2.28)\) for optional variation processes and the results above we get \begin{align*}[M_1+M_2] &= [M_1] + [ M_2 ] + 2[ M_1,M_2 ] \\ &= N_1+N_2 + 2[ M_1,M_2 ] \\ &= [ M_1+M_2] + 2[ M_1,M_2 ],\end{align*} and so \[\underline{\underline{[ M_1,M_2 ] (t) = 0.}}\]