Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / M / 1 / d

A rate-two Poisson process starts at $N_0=0$, has independent increments and stationary increments, and has $N_{t+h}-N_t\sim \mathrm{Poisson}(2h)$ for $h\ge 0$ when time is measured in minutes. Equivalently it is a counting process with independent stationary increments satisfying, uniformly in the start time,as $h\downarrow 0$. Thus an average rate alone is insufficient; the independence and increment assumptions are part of the model.

The one-minute increment has Poisson distribution with parameter two, independently of where that interval occurs. Hence

The full-day count has Poisson distribution with mean and variance $2\cdot 450=900$. Its normal approximation is $N(900,900)$, with standard deviation $30$. With the usual continuity correction,Without the half-unit correction the central-limit approximation gives $1-\Phi (1)\approx 0.159$.