Is the sum of two compensated poisson processes always a martingale? Let $M^{1}_t=N^{1}_t-t\lambda^{1}$ and $M^{2}_t=N^{2}_t-t\lambda^{2}$ be two compensated poisson processes, where $\lambda^{1}$ and $\lambda^{2}$ are the constant intensities of $N^{1}_t$ and $N^{2}_t$. Hence $M^{1}_t$ and $M^{2}_t$ are martingales with respect to their own filtration $\sigma(M^{1}_t)$ and $\sigma(M^{2}_t)$.  Furthermore $N^{1}_t$ and $N^{2}_t$ are not independent! What can we say about the process $Z_t=M^{1}_t+M^{2}_t$. In particular, is $Z_t$ a martingale with respect to its own filtration $\sigma(Z_t)$? Is $Z_t$ a martingale with respect to the joint filtration $\sigma(M^{1}_t)\vee\sigma(M^{2}_t)$?
 A: In general, the sum of two compensated poisson processes is not a martingale.
Example Let $(N_t^1)_{t \geq 0}$ a Poisson process with intensity $\lambda^1 =1$. If we set $N_t^2 := N_{t/2}^1$, then $(N_t^2)_{t \geq 0}$ is a Poisson process with intensity $\lambda^2 = 1/2$. For the joint filtration $\mathcal{F}_t := \sigma(N_s^1,N_s^2; s \leq t)$ it holds that $\mathcal{F}_t := \sigma(N_s^1; s \leq t)$. If we choose $s:=t/2$, then
$$\mathbb{E}(Z_t \mid \mathcal{F}_s) = \mathbb{E}(M_t^1 \mid \mathcal{F}_s) + \mathbb{E}(M_t^2 \mid \mathcal{F}_s) = M_s^1 + M_{t}^2 = Z_s + (M_t^2-M_s^2)$$
as $M_t^2 = N_{t/2}^1 - t/2$ is $\mathcal{F}_{s}$-measurable. This shows that $(Z_t,\mathcal{F}_t)_{t \geq 0}$ is not a martingale.
In fact, $(Z_t)_{t \geq 0}$ is not a martingale with respect to any (admissible) filtration $(\mathcal{G}_t)_{t \geq 0}$. Indeed, if $(Z_t,\mathcal{G}_t)_{t \geq 0}$ was a martingale, then
$$\mathbb{E}((Z_t-Z_s) \cdot Z_s) = \mathbb{E}\bigg[ \mathbb{E} \big( (Z_t-Z_s) \cdot Z_s  \mid \mathcal{G}_s\big) \bigg] = \mathbb{E} \big[ (\mathbb{E}(Z_t \mid \mathcal{G}_s)-Z_s) \cdot Z_s \big] =0$$
for any $s \leq t$. This means that
$$\mathbb{E}((Z_t-Z_s) \cdot Z_s)=0, \qquad s \leq t$$
is a necessary condition for $(Z_t)_{t \geq 0}$ being a martingale. Using that covariance is known - as $(N_t^1)_{t \geq0}$ is a Poisson process-, it is not difficult to calculate this expectation - and to see that is does not equal $0$ for $s=t/2$.
