Past exam of the options I course of the University of Oxford / 2011 / M (Probability)

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

Condition on the intermediate state. Starting at $+1$, a return to $+1$ after two transitions occurs through $-1$ with probability $q(1-p)$ or through $+1$ with probability $(1-q{)}^2$. These alternatives are disjoint, so

A stationary distribution is a probability row vector $\pi$ satisfying $\pi P=\pi$. Its coordinates obey ${\pi }_{-}(1-p)={\pi }_{+}q$ and ${\pi }_{-}+{\pi }_{+}=1$. ThusBecause both transitions between distinct states have positive probability, the finite Markov chain is irreducible; its positive self-transition probabilities also make it aperiodic.

The finite irreducible aperiodic Markov chain converges to its stationary distribution, regardless of its initial law. Therefore $\mathbb{E}[X_j]\to {\pi }_{+}-{\pi }_{-}$. Taking the arithmetic means of this convergent sequence givesThis uses linearity of expectation and convergence of Cesàro means; it does not require the variables $X_j$ to be independent.

With probability $2/3$, the chain starts at $+1$ and $T=1$. Conditional on starting at $-1$, it takes a geometric number $G$ of transitions to first reach $+1$, with success probability $1-p$, so $T=1+G$. The probability generating function is thereforeThe parameter $q$ is absent because the process stops at its first visit to $+1$. At $s=1$ the expression equals one, confirming eventual hitting with probability one.