Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 37 / 5 / d

The algorithm is a Random-scan Gibbs sampler. Start at a point where the target is positive, with a proper normalizing integral and well-defined full conditional distributions. Its state is the entire coordinate vector. Choosing the initial point fixes the initial law; it need not already be the invariant law. The transition rule below preserves the target irrespective of this choice.

At each transition, choose a coordinate uniformly and independently of the previous coordinate choices. Draw a fresh value from that coordinate's full conditional distribution given the current remaining coordinates, and keep all other coordinates fixed. This supplies the random-selection and conditional-draw operations lost in the TeX transcription. The update depends only on the current vector and fresh randomness, which proves the Markov property. The printed mathematical requests are the kernel and its detailed balance, answered in the following two subsections.

For the Random-scan Gibbs sampler, the correct transition kernel is a measure, since a one-coordinate update is singular with respect to full-dimensional Lebesgue measure when $p>1$. Write $x_{-i}$ for all coordinates except $i$. ThenEach Dirac measure fixes an unupdated coordinate. Equivalently, $P(x,A)=p^{-1}{\sum }_i\int 1_A(x_{-i},z){\pi }_i(z\mid x_{-i})dz$, with $z$ placed in coordinate $i$. Replacing this kernel by an ordinary density on the whole product space would omit those fixed-coordinate constraints.

Let $\bar{\pi }$ be the normalized target and ${\bar{\pi }}_{-i}$ its marginal distribution. A coordinate-$i$ kernel $P_i$ leaves $x_{-i}=x_{-i}^{\prime }$ and has the joint old-new measureThis is symmetric in the old and new state: the common remaining coordinates are fixed and the two conditional factors exchange places. Thus each coordinate kernel satisfies detailed balance. Averaging them with the fixed weights $1/p$ provesIntegrating out the old state proves invariance of $\bar{\pi }$. The common normalization cancels, so detailed balance can also be written with the unnormalized target. This establishes detailed balance of a random-scan Gibbs sampler and a reversible Markov chain. Fresh independent random choices make the update a Markov chain, as explained above; invariance is not an assertion that an arbitrarily initialized chain starts in stationarity.