Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 7 / b

Since $Y\ge 1$, its finite expected value satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative andThus these values are a probability mass function. They define the size-biased distribution of $Y$: larger values receive proportionally more weight.

For $0\le s<1$, differentiate the probability generating function term by term and multiply by $s/\mu$:This is the size-biased probability generating function identity. At $s=1$, the finite-mean assumption gives the limiting value one on both sides; at zero, positive support gives value zero.

For $Y\sim \mathrm{Geom}(p)$, its mean is $1/p$, so the size-biased probability generating function isChoose $X_1,X_2$ independent with distribution $\mathrm{Geom}(p)$. Their shifted sum has generating function, initially for $s>0$,with both sides continuous and zero at $s=0$. Equality of probability generating functions provesEquivalently, $X_1+X_2=k+1$ has $k$ possible positive summand pairs, each of probability $p^2(1-p{)}^{k-1}$, matching the size-biased geometric distribution probabilities.