Past exam of the options I course of the University of Oxford / 2013 / M (Probability)

A Poisson process has a Poisson distribution for the count in an interval of length $t$, with mean equal to rate times length. Fifteen minutes is $1/4$ hour, so

The first arrival time $T_1$, measured in hours after opening, has exponential distribution of rate twelve. ThereforeEquivalently, there must be no arrivals in the first quarter hour and at least one in the next; independent increments give $e^{-3}(1-e^{-3})$.

Let $I_2=T_2-T_1$ and $I_3=T_3-T_2$ be the interarrival times adjacent to the second arrival. They are independent exponential random variables of rate twelve per hour. The second customer meets the first if $I_2<1/12$ hour, and meets a later customer if $I_3<1/12$ hour. If the third arrives too late, every subsequent customer does also. Thus the only way to meet nobody is for both gaps to be at least five minutes:Hence $\boxed{\mathbb{P}(\text{meeting})=1-e^{-2}}$. Exact equality of arrival and departure times has probability zero. The calculation includes both customers already present and customers arriving during the visit; it is a two-sided interarrival test for overlap.

Independent marking with probability $5/6$ is Poisson thinning and gives a Poisson process of chip-buying arrivals, of rate $12(5/6)=10$ per hour. Its first arrival time therefore satisfiesFor example, its survival probability is $\mathbb{E}[(1/6{)}^{N(t)}]=\exp (12t(1/6-1))=e^{-10t}$, which also derives the thinning conclusion directly.

For $0\le z\le 1$, independence and the uniform distribution giveSince $M_n\le 1$, for $0<\epsilon <1$,For $\epsilon \ge 1$ the probability is zero. This is exactly convergence in probability of $M_n$ to one.

Put $Y_n=n(1-M_n)\ge 0$. For fixed $x\ge 0$ and sufficiently large $n>x$,For $x<0$ its distribution function is zero. The limiting distribution function is therefore $F(x)=0$ for $x<0$, $F(x)=1-e^{-x}$ for $x\ge 0$, the exponential distribution of rate one. Since this distribution function is continuous everywhere,

For the uniform distribution on $[0,1]$, the mean is $1/2$ and the variance is $1/12$. The central limit theorem for independent identically distributed variables with finite nonzero variance givesChoose $\boxed{\alpha =1/2,\beta =1/2}$. ThenThe standard normal distribution function is continuous and strictly increasing, so its value at the finite positive number $\sqrt{12}$ is strictly between $1/2$ and one, as required.