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

Put $q=1-p$. Since $X$ takes positive integer values, $\mathbb{P}(X>0)=1=q^0$. Taking differences of the given tail probabilities therefore yields, also for $k=1$,This is the probability mass function of the geometric distribution on $\{1,2,\dots \}$.

Using the geometric series, the probability generating function isDifferentiation gives

The mean is the derivative at one of the probability generating function. ThereforeEvaluation at one is valid here because the probability generating function has radius of convergence $1/(1-p)>1$.

The event $W>k$ occurs exactly when both $X_1>k$ and $X_2>k$. By independence,ThusThe new success probability $2p-p^2$ is the chance that at least one of the two trials succeeds.

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 $k$.