Mixing Times and Convergence to Stationarity

You build a Markov chain model of a stock's regime (bull, bear, choppy) and want to simulate 10,000 draws from its long-run regime distribution to feed a Monte Carlo backtest. You start the simulation in the "bull" state (today's actual regime) and run it forward. But the first several hundred simulated steps aren't really draws from the long-run distribution — they're still colored by the fact that you deliberately started in "bull." Using those early steps as if representative subtly biases the simulation toward the starting state. Mixing time is the number of steps a chain needs before its distribution over states gets close enough to its true stationary distribution that starting-point memory has effectively faded — the number telling you how much of a simulation to discard before trusting it.

An analogy: a shuffled deck forgetting its starting order

A brand-new deck is perfectly ordered by suit and rank. After one riffle shuffle, it's still noticeably closer to the original order than to a truly random arrangement — an expert could still exploit leftover structure. After about seven riffle shuffles, mathematically, the deck is close enough to uniformly random that no practical test can tell it apart from a truly random deck. "Seven shuffles" is the deck's mixing time: not the point it becomes exactly random (that takes infinitely many shuffles), but the point it's random enough that starting order no longer detectably influences anything you'd care about.

The formulas, one piece at a time

Let $\pi$ be a Markov chain's stationary distribution — the long-run probability of being in each state, unchanging once reached. Let $P^t(x, \cdot)$ be the distribution over states after $t$ steps, starting from state $x$. The total variation distance between two distributions measures how far apart they are:

$$d_{TV}\big(P^t(x, \cdot), \pi\big) = \frac{1}{2} \sum_{y} \big| P^t(x, y) - \pi(y) \big|.$$

In plain English: this sums up, across every state $y$, how much more (or less) likely the chain is to be in state $y$ after $t$ steps compared to its true long-run frequency, and halves the total to keep the distance between 0 (identical) and 1 (completely different). The mixing time $t_{\text{mix}}(\epsilon)$ is the smallest $t$ such that this distance drops below a small threshold $\epsilon$ (commonly $\epsilon = 0.25$ or $1/4$) for the worst-case starting state $x$. In plain English: it's how many steps you need to run the chain before its current-state distribution is within a small, agreed-upon tolerance of the true long-run distribution, no matter where you started.

Worked example 1: a two-state regime chain

Consider a simple two-state regime chain: bull today stays bull with probability 0.9, switches to bear with probability 0.1; bear stays bear with probability 0.8, switches to bull with probability 0.2. Solving $\pi P = \pi$ gives the stationary distribution $\pi_{\text{bull}} = 0.2/(0.2+0.1) \approx 0.667$, $\pi_{\text{bear}} \approx 0.333$. Starting deterministically in "bull" ($P^0 = (1, 0)$), after one step the distribution is $(0.9, 0.1)$ — total variation distance from $\pi$ is $\frac{1}{2}(|0.9-0.667| + |0.1-0.333|) = \frac{1}{2}(0.233+0.233) = 0.233$. After a few more steps this distance keeps shrinking geometrically (roughly by a factor related to the chain's second-largest eigenvalue, here about 0.7 per step), reaching below $0.05$ by around step 8–9 — meaning after about 9 simulated steps, the "started in bull" bias has become negligible for practical purposes.

Worked example 2: why burn-in length matters for the backtest

Suppose you run 10,000 simulated regime paths, each starting from today's actual "bull" state, and want the long-run regime frequencies in your Monte Carlo sample to match the true $\pi = (0.667, 0.333)$. If you use all 10,000 steps of every path including the first 9, you're injecting extra "bull" weight from the biased early steps — a small effect per path, but if paths are short (say, only 20 steps total), those first 9 biased steps are nearly half the sample, materially skewing the regime frequencies toward bull relative to the true stationary mix. The standard fix, called "burn-in," discards the first $t_{\text{mix}}$ steps of each simulated path (here, roughly the first 9) before recording any statistics, ensuring what's kept is a fair sample of $\pi$ rather than a sample still colored by the deliberately chosen bull starting point.

What this means in practice

Mixing time governs the burn-in period in any Markov-chain Monte Carlo (MCMC) simulation, regime-switching model simulation, or Markov chain-based scenario generator: too little burn-in leaves starting-point bias in your results, while excessive burn-in just wastes computation. Chains that mix slowly (near-absorbing states, or states that rarely communicate) need much longer burn-in and are also a red flag that the chain's long-run behavior may be poorly identified from short historical data in the first place.