research27 min studyLesson 764 in Physics

Sample the target measure and measure the correlations

Separate stationarity, convergence, autocorrelation, and regulator systematics.

On this page

What you’ll understand

  • Check balance for a finite transition matrix.
  • State additional conditions needed for convergence from an initial state.
  • Inflate statistical errors using an explicit autocorrelation convention.

Importance sampling estimates a normalized lattice expectation by visiting configurations according to their target weight. It avoids spending most effort in regions that contribute almost nothing, but successive Markov configurations are correlated. Correct stationary weights, adequate exploration, and reliable uncertainty estimates are separate requirements. A simulation that satisfies a balance identity can still be trapped or poorly equilibrated, while a large sample count can hide a small effective number of independent measurements. None of these statistical diagnostics removes finite-spacing or finite-volume bias.

Balance identifies a stationary distribution

At a finite lattice regulator with nonnegative target weight, normalize π(U) proportional to exp(−Sg) times the included fermion determinant. Integrals over configurations use the specified product Haar measure. A transition kernel preserves π if integrating over the starting configuration produces π again. Detailed balance equates probability flow in each direction and is a sufficient construction for stationarity, although globally balanced algorithms need not satisfy it pair by pair.

The familiar Metropolis rule min[1,exp(−ΔS)] assumes a proposal symmetric under reversal with respect to the same measure. Asymmetric proposals require the appropriate proposal-density ratio. Rejected moves count as repeated configurations and must remain in the chain. Keeping accepted proposals only generally changes the sampled distribution, because states with different rejection rates would then be weighted incorrectly. This bookkeeping issue is independent of whether the action evaluation itself is accurate.

πiPij=πjPji⟹∑iπiPij=πj,Pij=P(i→j)\pi_iP_{ij}=\pi_jP_{ji}\quad\Longrightarrow\quad\sum_i\pi_iP_{ij}=\pi_j,\qquad P_{ij}=P(i\to j)
This finite row-stochastic convention is explicit: every row sums to one. Continuum configuration indices become Haar-measure integrals, and detailed balance is sufficient rather than necessary.↔ Scroll wide equations horizontally.

A stationary chain still needs to explore

Convergence from an arbitrary start requires suitable irreducibility and aperiodicity, or the corresponding ergodic conditions in the continuous state space. A deterministic alternating chain can have a stationary distribution but fail to approach it from one starting state. Near a continuum limit, rare transitions between topological sectors may create very long relaxation times. A short apparently stable record of common observables does not establish that every slow mode has equilibrated.

Discarding a chosen warmup period addresses initialization only if that period is long compared with relevant relaxation scales. Independent starts, sector histories, and observable-specific autocorrelations provide complementary checks. Thinning can reduce stored correlations but discards information and does not magically create more independent data than were generated. The Monte Carlo update index is algorithmic time; it is not Euclidean propagation time used to extract a hadron energy.

ρO(t)=Cov⁡(Ok,Ok+t)Var⁡(O),τint,O=12+∑t=1∞ρO(t),Neff≃N2τint,O\rho_O(t)=\frac{\operatorname{Cov}(O_k,O_{k+t})}{\operatorname{Var}(O)},\qquad \tau_{\rm int,O}=\frac12+\sum_{t=1}^{\infty}\rho_O(t),\qquad N_{\rm eff}\simeq\frac{N}{2\tau_{\rm int,O}}
The convention gives τint=1/2 for independent samples. N is a long stationary record and Neff is an asymptotic variance-equivalent count for observable O, not a universal property of the ensemble.↔ Scroll wide equations horizontally.

Statistical and systematic errors answer different questions

For a sufficiently mixing chain with summable correlations, the sample-mean variance is approximately the single-configuration variance divided by the effective count. Estimating the correlation sum itself requires a window choice and enough data to see its tail. Blocking or correlated resampling should respect those scales. A reported standard error with an unobserved slow tail can be optimistic even when its numerical computation is exact.

The Markov lab makes finite convergence visible. Reset and set p=0.2 and q=0.4, then Step twice; predict state-1 probability 0.28 and stationary value 1/3. Reset and set p=0.4 and q=0.2, then Step twice; predict 0.56 and stationary value 2/3. The lab evolves exact probability distributions from state 0 rather than generating a random chain. It has no gauge configurations, measured autocorrelation error bars, or continuum extrapolation.

Var⁡(Oˉ)≃2τint,ONVar⁡(O)\operatorname{Var}(\bar O)\simeq\frac{2\tau_{\rm int,O}}{N}\operatorname{Var}(O)
This large-N relation assumes stationarity and sufficient correlation decay. Cutoff artifacts, finite-volume shifts, mistuned quark masses, and fit-model bias are additional systematics, not terms removed by increasing N.↔ Scroll wide equations horizontally.

Check a two-state equilibrium

A row-stochastic chain has p=P(0→1)=0.2 and q=P(1→0)=0.4, initially in state 0. Find its stationary law, verify detailed balance, and compute state-1 probabilities after one and two transitions.

  1. Stationarity gives π₁=p/(p+q)=1/3 and π₀=2/3. The two opposing stationary flows are π₀p=(2/3)0.2 and π₁q=(1/3)0.4, both 2/15.

  2. The first-step state-1 probability is 0.2 because the initial state is certainly 0. For a current probability z, the next one is p(1−z)+(1−q)z=0.2+0.4z.

  3. The second-step value is 0.2+0.4×0.2=0.28. Its deviation from 1/3 decreases by a factor 0.4 per transition; the stationary identity did not mean the initial state already had the target law.

The result

The stationary law is (2/3,1/3), with matching flows 2/15 and probabilities 0.2 then 0.28 during relaxation.

Explore the model

Interactive lab

Interactive lab · markov

Inspect finite-step equilibration

Reset and set p=0.2,q=0.4, then Step twice; predict P₁=0.28 and stationary 1/3. Reset and set p=0.4,q=0.2, then Step twice; predict P₁=0.56 and stationary 2/3.

Inspect finite-step equilibrationExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe probabilities are exact and unsampled. This instrument does not generate gauge ensembles or estimate measurement autocorrelations.

Check your understanding

Does detailed balance alone guarantee convergence from every initial state?

Try it yourself

A stationary observable has variance 4 and normalized autocorrelation ρ(t)=(1/2)^t for positive integer lags. For N=900 measurements, estimate τint, Neff, and the standard error of the mean.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Sum the geometric tail using the convention τint=1/2+Σt≥1ρ(t).

Compare with the solution

The tail sums to 1, so τint=1.5. The effective count is 900/(2×1.5)=300, giving standard error √(4/300)≈0.1155. Treating all 900 measurements as independent would give 0.0667 and underestimate the uncertainty. The estimate presumes the stated stationary correlation law and a record long compared with its decay.

Take it with you

  • Balance fixes the stationary target but does not prove adequate exploration.
  • Autocorrelation inflation depends on the observable and the time convention.
  • More configurations reduce statistical error without automatically removing lattice systematics.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.