Solution: Expected number of events in a sample

Under the null hypothesis we have that \(Z_1(t) = N_1(t)−E_1(t)\) is a mean zero martingale. This follows from \((3.53)\), where it is shown that \(Z_1(t)\) can be expressed as a difference of two stochastic integrals (whose properties are discussed in section 2.2.2). Because of this, we have \[E[Z_1(t)] = E[N_1(t)−E_1(t)] = 0\] for all \(t \in [0,t_0]\), and in particular \[ E[N_1(t_0)−E_1(t_0)] = 0\] so that \[\underline{\underline{E[N_1(t_0)] = E[E_1(t_0)].}}\]