 Compensated Poisson integral

For a Poisson random measure $N$ with sigma-finite intensity measure $\mu$, a real deterministic simple integrand $f={\sum }_i{a}_i{1}_{A_i}$ on disjoint sets of finite intensity has compensated Poisson integralIts expected value is zero. Independence of disjoint counts and the Poisson variance formula give $\mathbb{E}[I(f{)}^2]=\int f^{2}d\mu$.

This isometry extends the integral by completion to deterministic $f\in L^2(\mu )$. If $f$ is not also integrable, writing two separate finite integrals $\int fdN-\int fd\mu$ can be invalid; the compensated integral is defined through the joint centered limit. For a constant-rate Poisson process, taking $f=1_{[0,t]}$ gives $N_t-\lambda t$ with variance $\lambda t$.^[ref]