I'm learning about Markov Chains Monte Carlo methods and mixing times, and could use some help understanding the concept of the Spectral Gap and why / how it relates to the mixing times.

Thus far what I have learned is this: given an ergodic (irreducible, aperiodic) and reversible (e.g. $\pi(x)P(x,y) = \pi(y)P(y,x)$ for all $x,y \in \Omega=$state space) finite Markov Chain and its transition matrix $P$, the spectral gap is defined as:

$$\xi = 1 - \lambda_\max$$

where $\lambda_\max = \max\{|\lambda_2|, |\lambda_n|\}$. Alternatively, the spectral gap can be understood as the smallest positive nonzero eigenvalue (I think?)

I vaguely get that this is somehow related to the mixing time of the Markov chain (that the bigger the spectral gap, the faster the mixing time), but I lack an intuitive fundamental understanding of why this is the case. My linear algebra knowledge is sketchy at best, so I'd very much appreciate an explanation as if I were a 5-year-old of why the eigenvalues are bounded the way they are (between 1 and -1), why the negative eigenvalues don't really concern us for purposes of MCMC, why we care about $\lambda_\max$, and how it is that the spectral gap is indicative of the mixing time.

  • $\begingroup$ Did my answer help, or is there too much math and not enough intuition? $\endgroup$
    – Ian
    Oct 9, 2017 at 22:54
  • $\begingroup$ Thanks for your response! I think I understand that, given the diagonalization equation you provided, larger the spectral gap implies faster convergence. But my understanding of matrix diagonalization is very shoddy so I'm unsure how it came about. Also, by "modulus" do you mean absolute value? I haven't yet gotten to the part where you describe the relationship between the spectral gap and mixing time but will put in some time trying to understand it. $\endgroup$ Oct 10, 2017 at 0:03
  • $\begingroup$ Modulus is the complex version of absolute value. In the reversible case all eigenvalues are actually real, so with that assumption you can replace all instances of "modulus" by "absolute value" in my answer. As for diagonalization, honestly I think you're best off looking in a linear algebra book. For example, David Lay's book in Chapter 5, or Gilbert Strang's book also in chapter 5. $\endgroup$
    – Ian
    Oct 10, 2017 at 0:14
  • $\begingroup$ To just give the basic idea of diagonalization, in the notation of my answer, $q_i P = \lambda_i P$. Iterating that gives $q_i P^t=\lambda_i^t q_i$. By linearity, if $p_0=\sum_{i=1}^n c_i q_i$ then $p_0 P^t$ is given by the formula in my answer. The interesting thing is that there are "enough" eigenvectors to write $p_0$ that way in the first place...and that takes some significant work to prove. $\endgroup$
    – Ian
    Oct 10, 2017 at 0:16

1 Answer 1


Strictly speaking, MCMC means constructing a Markov chain which has some desired property (usually either a prespecified equilibrium distribution or an equilibrium distribution with a prespecified expectation) and then numerically drawing samples from it. If you are given a Markov chain it is really just called Monte Carlo simulation.

Anyway, at each fixed time $t$, the value of the chain will have some given distribution, given by $p_0 P^t$, where $p_0$ is the initial distribution (written as a row vector) and $P$ is the transition probability matrix (in the row-stochastic convention i.e. $\sum_{j=1}^n P_{ij}=1$).

So the long term behavior of the distributions at each fixed $t$ is determined by the behavior of $P^t$ as $t \to \infty$. Any stochastic matrix has all eigenvalues of modulus at most $1$. It also has $1$ as an eigenvalue. The ergodicity assumption additionally tells us that the eigenvalue $1$ has multiplicity $1$ and that all of its other eigenvalues will have modulus strictly less than $1$. (Without irreducibility it is possible that the eigenvalue $1$ has a higher multiplicity; without aperiodicity it is possible to have an eigenvalue like $-1$ or $e^{2\pi i/3}$.) The reversibility assumption tells you (among other things) that $P$ is diagonalizable with real eigenvalues (because reversibility furnishes a similarity transformation which turns $P$ into a symmetric matrix).

As a result of these three facts you can write

$$p_0 P^t = \pi + \sum_{i=2}^n c_i \lambda_i^t q_i$$

where $\lambda_i$ are eigenvalues all less than $1$ in absolute value, $q_i$ are the left eigenvectors, and $c_i$ are coefficients depending on $p_0$. This is just what you get from diagonalizing $P^t$ (without going through and computing the $c_i$ since they do not really matter for this presentation).

You can see that as $t \to \infty$, the remaining eigenvalues will decay because of the power of $t$. Assuming the $c_i$ are all significant (the typical situation for a "random" initial distribution), the slowest decay is by $\lambda_{\max}$, since it is closer to $1$ in absolute value. So the larger the spectral gap $\xi$, the faster the convergence. Note that the negative eigenvalues do matter, which is why you wrote it as $\lambda_{\max}=\max \{ |\lambda_2|,|\lambda_n| \}$.

The precise relationship between the spectral gap and the dynamical properties of the chain is more subtle than the rest of what I've said here. Loosely speaking when the spectral gap is small, the eigenvector corresponding to $\lambda_{\max}$, which I might call a "metastable eigenvector", will be significant and positive for one set of states and significant and negative for another set of states. It might also be smaller on a third set of states.

The "metastable eigenvector" then corresponds to the slow exchange of probability between the first two sets of states (because the decay of its contribution to the distribution results in a flow from one set to the other, with which set is which depending on the sign of the corresponding $c_i$). You can look up metastability in Markov chains for more about this. To get a feel for it, consider a matrix like

$$P=\begin{bmatrix} 1/3 & 2/3 & 0 & 0 \\ 1/4-\epsilon/2 & 3/4-\epsilon/2 & \epsilon & 0 \\ 0 & \epsilon & 1/5-\epsilon/2 & 4/5-\epsilon/2 \\ 0 & 0 & 1/6 & 5/6 \end{bmatrix}$$

where $0<\epsilon \ll 1$ is a small parameter. For $\epsilon=10^{-6}$ for example, this has relatively rapid equilibration within the sets $\{ 1,2 \}$ and $\{ 3,4 \}$ (driven by eigenvalues of about $1/12$ and $1/30$ respectively) and slow equilibration between the two of them (driven by an eigenvalue of about $1-10^{-6}$).

  • $\begingroup$ Thank you, @Ian! Can you provide a good resource for learning about these ideas more carefully? $\endgroup$ Oct 22, 2021 at 19:18
  • $\begingroup$ @curious_dan Unfortunately I learned the overwhelming majority of these concepts on an informal basis. If you want details about metastability I would refer you to Anton Bovier's work but this is somewhat subtle and may be hard to understand if you aren't acquainted with the (seemingly somewhat unrelated) subject of potential theory. $\endgroup$
    – Ian
    Oct 22, 2021 at 19:19
  • $\begingroup$ ok I see. In that case I hope you don't mind if I pose another question. Happy to write it as a separate post if you're interested in answering it there for credit. How do I go about constructing the transition matrix $P$ for an algorithm such as Metropolis-Hastings or MALA? $\endgroup$ Oct 22, 2021 at 19:26
  • $\begingroup$ @curious_dan I'd put that as its own question (although at that level of generality there is absolutely no way to answer it, you need to know something about the problem to propose a matrix for decent runtime). $\endgroup$
    – Ian
    Oct 22, 2021 at 19:27

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