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ID: past-exam-of-the-simulation-and-statistical-programming-course-of-the-university-of-oxford/2023/3/c/iv/solution

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Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 3 / c / iv / Solution by

Codex  0 2026-10-25

Choose

k

distinct coordinates uniformly for the starting ones; each element of

S_{k}

then has starting probability

1/ (k n)

. At each iteration choose one current one and one current zero uniformly and swap them. This implements the normalized proposal distribution from (c)(i), remains in

S_{k}

, and needs just one precomputed matrix inverse.

mymcmc <- function(x, k, sigma) {
  n <- length(x)
  stopifnot(length(k) == 1L, is.finite(k), k == floor(k),
            k > 0L, k < n, is.matrix(sigma),
            all(dim(sigma) == c(n, n)))
  sigma_inv <- invsigma(sigma)
  u <- integer(n)
  u[sample.int(n, k, replace = FALSE)] <- 1L
  draws <- matrix(0, 100L, n)
  for (iteration in seq_len(100L)) {
    ones <- which(u == 1L)
    zeros <- which(u == 0L)
    i <- ones[sample.int(length(ones), 1L)]
    j <- zeros[sample.int(length(zeros), 1L)]
    v <- u
    v[i] <- 0L
    v[j] <- 1L
    alpha <- mylogapp(x, u, v, sigma_inv)
    if (runif(1L) < alpha) u <- v
    draws[iteration, ] <- u
  }
  colMeans(draws)
}

Sampling positions with sample.int and then indexing into ones or zeros also works when

k = 1

n - k = 1

. In R, a call such as sample(ones, 1) would incorrectly sample from 1:ones when ones is a single numeric value greater than one.

The code records the state after each of exactly

100

transitions, including repetitions after rejection. Its answer is the vector Monte Carlo estimator

m = \frac{1}{100} t = 1 \sum 100 U^{(t)} \approx E [U ∣ X = x] .

(1)

This estimates the posterior mean coordinate by coordinate; because each coordinate is binary, its entries estimate the posterior probabilities of inclusion. Every stored state has coordinate sum

k

, so the estimated means lie in

[0, 1]

and sum to

k

For neighbouring states, the symmetry of the proposal and the Metropolis–Hastings acceptance probability give detailed balance:

π (u ∣ x) q (v ∣ u) α (u, v) = q (v ∣ u) min {π (u ∣ x), π (v ∣ x)} = π (v ∣ x) q (u ∣ v) α (v, u) .

(2)

Thus the posterior distribution is a stationary distribution. Any two states in

S_{k}

can be connected by moving the ones that occupy the wrong positions, and the multivariate normal distribution gives strictly positive target masses. The resulting Markov chain is therefore an irreducible Markov chain, and its long-run state averages estimate the posterior mean.

The requested

100

iterations are a finite approximation, without a convergence guarantee or discarded burn-in. A randomly chosen starting subset has the prior distribution, rather than generally the posterior distribution. No aperiodicity assumption is needed for the finite-state Markov chain ergodic theorem for averages; indeed a uniform two-state target with

n = 2, k = 1

gives a period-two alternating chain. This does not invalidate either the code or the state-average estimator.

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