Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / b

Since $Y\ge 1$, its finite mean satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative, andThus $(q_k)$ is a probability mass function. It defines the size-biased distribution of $Y$: an outcome of size $k$ receives $k$ times its original relative weight.

By the definition of a probability generating function,Termwise differentiation of the power series is valid for $|s|<1$. At $s=1$, the equality extends by the finite mean assumption, and both sides equal one.

For the geometric distribution, $\mu =1/p$, so the probability generating function of the size-biased distribution isIf $X_1,X_2$ are independent and identically distributed with geometric distribution of parameter $p$, the probability generating function of an independent sum givesAt zero, interpret the expression by its continuous extension. The probability generating functions agree, henceOne can also check the probability mass functions directly: for $k\ge 1$, there are $k$ positive integer pairs with sum $k+1$, each having probability $p^2{q}^{k-1}$. Thus both sides assign probability $k{p}^2{q}^{k-1}$ to $k$.