Let $N=\{N_t:t\in\mathbb R_+\}$ be a homogeneous Poisson process with intensity $\alpha$ and $M_t=N_t-\alpha t$ the compensated process. I'd like to show that $N$ is not a natural process, i.e. that the equality $$\mathbb E\left[\int_{(0,t]} M_s\,\mathsf dN_s \right] = \mathbb E\left[\int_{(0,t]} M_{s-}\,\mathsf dN_s \right]\tag1 $$ does not hold for each bounded cádlág martingale $M$, with respect to the completion of the canonical filtration of $N$. Intuitively, I know that the quadratic variation is $[M]_t=N_t$ but the predictable quadratic variation $\langle M\rangle_t = \alpha t$; indeed, the Poisson process is not predictable, and therefore is not natural (as it is increasing).
To prove this from the definition, consider the exponential martingale $$X_t = \exp\left((-\lambda N_t + \alpha t(1-e^{-\lambda})\right), $$ where $\lambda>0$ is fixed. For a given $t>0$, the process $X_{t\wedge T}$ where $T>t$ is then a bounded cádlág martingale. But how would I actually compute, e.g.
$$\mathbb E\left[ \int_{(0,t]}\exp\left((-\lambda N_t + \alpha t(1-e^{-\lambda})\right)\,\mathsf dN_s \right] \ \mathrm{?}$$
My thought is to use the fact that, conditioned on $N_t = k$, the jumps $$\Delta_i = \inf\{t>\Delta_{i-1}:N_s> N_{s-}\},\ 1\leqslant i\leqslant k $$ (with $\Delta_0\equiv0$) are uniformly distributed on $[0,t]$. So my questions are:
- Is this approach correct?
- Is there another approach that would generalize to non-Poisson counting processes?