Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / S (Simulation)

For $0<r<1$, start at any $x\in \{0,\dots ,n\}$. Propose $Y$ uniformly from this same finite set and accept it with probabilityOtherwise retain $x$. The uniform proposal is symmetric, so this is the Metropolis–Hastings algorithm with target $\pi (\cdot \mid r)$. Detailed balance follows because accepted probability flow is proportional to $\min \{\pi (x),\pi (Y)\}$. The chain is irreducible and has positive self-transition probability, hence converges to that binomial equilibrium law. At the degenerate parameters zero or one, the target is simply a point mass and can be simulated as a constant chain.

Use the prior itself as an independence proposal: draw $J\ge 1$ with $\mathbb{P}(J=j)=2^{-j}$ and propose $s=2^{-J}$. This indeed samples $p(s)=s$, since the geometric series sums to one. Starting from any allowed current value $r$, accept the proposal with probabilityThe Hastings proposal ratio cancels the prior ratio in the posterior, leaving exactly the binomial likelihood ratio; its binomial coefficient also cancels. All allowed states communicate, and proposing the current state gives a self-loop. Detailed balance therefore gives the specified posterior as the stationary law, and the chain converges to it. Its normalizing constant is finite and positive because it is a prior average of a nonnegative, nonzero likelihood.