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Incoming links: Burn-in
Markov chain Monte Carlo Created 2026-09-24 Updated 2026-11-05
Markov chain Monte Carlo estimates expectations under a target probability distribution by simulating a Markov chain that preserves . For an integrable observable , the assumptions of a Markov-chain law of large numbers justifyalmost surely. A Gibbs sampler preserves the target through full conditional distributions; a Metropolis–Hastings algorithm proposes a move and accepts or rejects it using a ratio of target and proposal densities. The unknown normalization constant cancels from that ratio, which makes unnormalized Bayesian posterior densities usable.
Successive states are generally dependent, so Monte Carlo precision depends on the observable's integrated autocorrelation time, rather than the raw iteration count alone. Burn-in can reduce initial-state effects but is not a proof of convergence. A correct stationary distribution also does not ensure rapid exploration: isolated modes or strong dependence can produce a long mixing time. Markov chain Monte Carlo convergence diagnostics investigate these limitations.
A single randomly chosen internal-site flip is a symmetric proposal: the reverse flip has the same probability . Therefore the Metropolis–Hastings algorithm accepts withComparing logarithms avoids overflow. On rejection the old state must be retained, including a repeated copy in the stored Markov chain; retaining only accepted states would sample a different distribution. This implementation also stores the actual initial state at row zero:
def sample_methylation(observed, nsteps=5000, seed=2024, score=methylation_score):
rng=np.random.default_rng(seed)
state=rng.integers(0,2,size=len(observed));state[[0,-1]]=1
states=np.empty((nsteps,len(state)),dtype=int);values=np.empty(nsteps)
states[0]=state;values[0]=score(state,observed)
accepted=0
for t in range(1,nsteps):
candidate=state.copy();site=rng.integers(1,len(state)-1)
candidate[site]=1-candidate[site]
trial=score(candidate,observed)
if np.log(rng.random())<min(0.,trial-values[t-1]):
state=candidate;values[t]=trial;accepted+=1
else:values[t]=values[t-1]
states[t]=state
return values,states,accepted/(nsteps-1)
lls, meths, acceptance = sample_methylation(obs, nsteps=5000, seed=2024)
import matplotlib.pyplot as plt
plt.plot(lls)
plt.xlabel("Stored state index")
plt.ylabel("Conditional-product log score")
plt.show()For this random-number convention, the acceptance fraction is approximately . The trace climbs substantially during its first several thousand updates, so a small fixed burn-in such as 100 iterations would discard too little of the initial transient. Five thousand single-site proposals are only about fifty updates per site on average.
For a binary state, its expected value is its probability of being one. Hence the Monte Carlo method estimate of the posterior methylation profile under the chosen target isThe averaging axis must be zero: average over sampled patterns while keeping the site index. A provisional choice for this particular trace is , retaining 2000 states:
def mean_methylation(meths, burnin):
if not 0 <= burnin < len(meths):
raise ValueError("burn-in must leave at least one state")
return np.asarray(meths)[burnin:].mean(axis=0)
burnin = 3000
profile_estimate = mean_methylation(meths, burnin)
print(lls[burnin:].mean())
plt.plot(profile_estimate, label="retained-sample mean")
plt.scatter(np.arange(len(obs)), .2*obs, s=10, label="0.2 × observation")
plt.xlabel("Site index")
plt.ylabel("Methylation probability")
plt.legend()
plt.show()The retained mean log score is , and the profile is shown below the trace in part (b). The fixed endpoints have estimate exactly one; adjacent sites are often smoothed into methylated or unmethylated stretches despite individual conflicting observations. The rescaled observations are only a visual comparison and are not another estimate of the posterior probability.
This is a finite-chain estimate for the conditional-product target defined in part (a). A burn-in chosen by inspection does not establish convergence, and the long excursions in this short Markov chain leave substantial uncertainty in individual-site estimates; several longer runs would be needed for precise inference. To sample the coherent joint alternative instead, pass
score=joint_methylation_score and inspect that new trace before choosing its own burn-in.Use several complementary Markov chain Monte Carlo convergence diagnostics, monitoring both and rather than judging the chain solely by its length:
- Trace plots of each parameter and the log-posterior against iteration, including early iterations and later windows. Look for initial drift, long flat stretches, abrupt switches and sustained changes in spread. Several independent chains started at dispersed points should explore the same region after burn-in.
- Plots of cumulative sample means and selected sample quantiles against iteration. Compare chains and successive long blocks; persistent movement indicates that the estimated posterior moments or tails have not stabilized.
- Autocorrelation function plots for both parameters after tentative burn-in. Slow decay reveals strong serial dependence and a small effective sample size of a Markov chain, even when iterations have been generated.
- Overlaid histograms or kernel density estimation curves from successive retained blocks and from different chains. Their agreement provides evidence of stable marginal posterior distributions.
- A scatter plot of retained pairs, colored by time block if helpful. Strong dependence or separate clusters can expose slow joint exploration that marginal plots conceal.
A million iterations and stable-looking plots do not prove convergence. Agreement across dispersed chains, exploration of all visible modes and adequate effective sample size of a Markov chain provide stronger evidence. Estimate the Monte Carlo standard error using the chain's autocorrelation; the independent random variables formula from 3(b)(ii) does not apply directly to these dependent draws.
Choose distinct coordinates uniformly for the starting ones; each element of then has starting probability . 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 , 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 or . 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 transitions, including repetitions after rejection. Its answer is the vector Monte Carlo estimatorThis 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 , so the estimated means lie in and sum to .
For neighbouring states, the symmetry of the proposal and the Metropolis–Hastings acceptance probability give detailed balance:Thus the posterior distribution is a stationary distribution. Any two states in 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 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 gives a period-two alternating chain. This does not invalidate either the code or the state-average estimator.
Past exam of the systems biology course of the University of Cambridge 2024 Paper 3 2 e Solution 2026-10-20
TODO: Complete the data-specific numerical results and diagnostics when the original
Iris_data.csv is available.Plot all 10,001 stored states, including the initial value, and indicate the provisional burn-in boundary. Also display the retained chain's autocorrelation function:
png("q2-trace.png", width = 1000, height = 700)
par(mfrow = c(2, 1), mar = c(4, 4.5, 2.5, 1))
plot(0:B, mu, type = "l", xlab = "Iteration", ylab = expression(mu),
main = "Complete Metropolis trace")
abline(v = burn, col = "firebrick", lty = 2)
abline(h = m, col = "steelblue", lty = 3)
acf(post, main = "Autocorrelation after provisional burn-in")
dev.off()
# Check sensitivity to the chosen discard and to the retained time window.
for (discard in c(500L, 1000L, 2000L)) {
kept <- mu[(discard + 2L):(B + 1L)]
cat("Discard", discard, ": posterior mean", mean(kept), "\n")
}
cat("Means of retained halves:", mean(post[1:4500]),
mean(post[4501:9000]), "\n")A satisfactory trace plot should show an initial transient, if any, followed by repeated exploration of one stable range with no continuing drift. Repeated values are expected because rejections are stored. If is large, the prescribed proposal standard deviation one can be much wider than the posterior scale , giving low acceptance and conspicuous plateaus. A flat-looking trace is therefore not by itself evidence of good exploration.
Compare the retained histogram with the exact posterior density, inspect autocorrelation, compare separate retained windows, and assess sensitivity to the discarded count. Agreement supports the adequacy of this run; neither 10,000 iterations nor a chosen burn-in of 1,000 proves convergence. The effective sample size of a Markov chain can be much smaller than the number of stored states.
The actual trace cannot be diagnosed without
Iris_data.csv. The code produces the required trace when the measurements are available; the comments here specify what to examine rather than asserting an unseen chain has converged.