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

A Poisson distribution with parameter $\lambda \ge 0$ hasFor $\lambda =0$ it is concentrated at zero. For $\lambda >0$, shifting the index in the defining series gives $\mathbb{E}[X]=\lambda$ and $\mathbb{E}[X(X-1)]={\lambda }^2$, so $\mathrm{Var}X={\lambda }^2+\lambda -{\lambda }^2=\lambda$.

Independence and convolution giveby the binomial theorem. Thus

By the addition rule for independent Poisson random variables, $X_n$ has the same distribution as a sum of $n$ independent variables $Z_j\sim \mathrm{Poisson}(\lambda )$. Each has expectation and variance $\lambda$. The central limit theorem therefore givesFor $\lambda =0$, every variable is zero, and the conclusion is the degenerate normal law with zero variance.

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 $1-\Phi (1)\approx 0.159$.