Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 5

With independent Bernoulli trials of success probability $p$, $N$ counts the failures before the first success. There are $n$ failures followed by one success with probability $(1-p{)}^{n}p$, giving the stated geometric distribution on $\{0,1,2,\dots \}$. It counts failures, rather than the total number of trials including the success.

Writing $q=1-p$, the probability generating function is a geometric series:The displayed restriction ensures absolute convergence.

Differentiate the probability generating function inside its radius of convergence:Since these are $\mathbb{E}[N]$ and $\mathbb{E}[N(N-1)]$, respectively,

Take independent Bernoulli random variables $X_i$ of success probability $r$, independently of $N$. On the event $N=n$, the random sum ${\sum }_{i=1}^N{X}_i$ has the binomial distribution with parameters $(n,r)$, including the empty sum zero at $n=0$. It therefore has exactly the specified joint distribution with $N$.

Thus one may realize $\boxed{Z=\sum _{i=1}^N{X}_i}$ on a suitable probability space. This is a distributional construction; it does not require a pre-existing $Z$ to be independent of $N$.

For an independent nonnegative integer-valued count $N$ and identically distributed independent summands, the random sum probability generating function is $G_Z(s)=G_N(G_{X_1}(s))$. Conditioning on $N=n$ proves this, since the conditional probability generating function is $G_{X_1}(s{)}^n$.

Here $G_{X_1}(s)=1-r+rs$, soThis identity holds throughout $|s|\le 1$, and its power-series extension has radius $[p+(1-p)r]/[(1-p)r]$.

Put $p_{\ast }=p/[p+(1-p)r]$. ThenUniqueness of the probability generating function therefore givesThis is Bernoulli thinning of a geometric count, using the convention of counting failures before success.

In each trial the presence indicators of $A$ and $B$ are independent Bernoulli random variables, with probabilities $p$ and $r$. Independence across trials makes their two entire indicator sequences independent. Let $N$ be the number of trials before the first $A$; then $N\sim {\mathrm{Geom}}_0(p)$. Among these $N$ preceding trials, the $B$ indicators are independent with probability $r$.

Thus the count is the random sum from part i, andAn $AB$ outcome on the stopping trial contributes nothing, because the question counts only outcomes strictly before the first $A$.