Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 4

The joint probability mass function isIt is nonnegative and its sum over $S\times S$ is one. Summing over one coordinate gives the other marginal probability mass function.

For $\mathbb{P}(X=x)>0$, the conditional probability mass function isThe denominator is the sum of the numerator over all possible $y$, so these conditional probabilities sum to one.

A partition of a sample space is a finite or countable family $(A_j)$ of pairwise disjoint events whose union is the whole sample space. If $X$ is integrable, the law of total expectation statesThis follows by summing $X{1}_{A_j}$ over the partition of a sample space; absolute integrability justifies the countable sum. For nonnegative $X$, the identity also holds with infinite expectations. Zero-probability parts contribute zero without requiring a conditional value on them.

The first swap necessarily removes one red ball from urn 1 and replaces it with a blue ball. Thus $\boxed{R_1=n-1}$ with probability one.

Conditional on $R_k=r$, urn 1 has $r$ red and $n-r$ blue balls, while urn 2 has $n-r$ red and $r$ blue balls. The independent selections therefore giveEvery other conditional probability is zero. The out-of-range terms at $r=0$ or $r=n$ already have zero probability. These are the transition probabilities of the Bernoulli-Laplace diffusion model.

Only upward and downward moves change the value, so the conditional expectation isThus $\boxed{\alpha =1,\beta =1-2/n}$.

Apply the law of total expectation to the partition of a sample space given by the possible values of $R_k$. Since the conditional mean is affine,Here $r_0=n$. This uses linearity of expectation and does not assert independence of successive urn counts.

The equilibrium of the first-order linear recurrence is $n/2$. Subtracting it givesConsequentlyFor $n=2$, interpret the formula at $k=0$ through the initial condition: $r_0=2$, while $r_k=1$ for $k\ge 1$. For every $n\ge 2$, $r_k\to n/2$.