Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 9F (Probability)

A geometric distribution with parameter $p$ on $\{1,2,\dots \}$ has probability mass functionIt counts the trial on which the first success occurs in independent trials with success probability $p$. Its probability generating function is $ps/[1-(1-p)s]$. Differentiating at $s=1$ givesThis support convention is important: a count of failures before the first success would start at zero.

This is the waiting time to observe both Bernoulli outcomes. Write $q=1-p$, and let $S$ record the sex of the first captured animal. That first capture is always needed. If $S$ is male, every subsequent trial independently produces the required female with probability $q$, so $N-1$ has a geometric distribution with parameter $q$. If $S$ is female, $N-1$ has a geometric distribution with parameter $p$. By the law of total expectation,Since $p+q=1$, this simplifies to $\boxed{EN=1/(pq)-1}$, orThe expectation includes both the first capture and the final capture completing the pair.

Use the same first-sex conditioning as in part (b). The conditional variances are $p/q^2$ after a male first capture and $q/p^2$ after a female first capture. ThusThe conditional expected values are $1+1/q$ and $1+1/p$, with probabilities $p,q$. A variable taking two values $a,b$ with probabilities $p,q$ has variance $pq(a-b{)}^2$, soThe law of total variance now givesAt $p=q=1/2$ this gives variance $2$, consistent with $N$ being one plus a geometric waiting time of parameter $1/2$.