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

The variable has positive-integer support, so $\mathbb{P}(X>0)=1$. Subtract consecutive values of the survival function:This is the positive-support convention for the geometric distribution.

Summing a geometric series gives the probability generating functionThe series converges absolutely for $|s|<1/(1-p)$. Differentiating the rational expression yields

The expected value is the derivative of the probability generating function at one. The expression from part (ii) givesIts finiteness also follows directly from the exponentially decaying geometric probabilities.

By independence, for $k\ge 0$,This is the geometric minimum law, givingThe support remains the positive integers; this parameter is a success probability, not a mean.

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.