如何判断吉布斯采样器是否具有不可约性？

Great question! I totally get why you might feel stuck—most intro examples focus on transition matrices for simple discrete chains, but Gibbs samplers (especially for continuous variables) don’t play by that exact rule. Let’s walk through how to check irreducibility here without building a full transition matrix.

First, let’s recap what irreducibility means in this context: For the Markov chain generated by Gibbs sampling, we can get from any starting state $(x_1, x_2)$ to any target state $(y_1, y_2)$ in a finite number of steps, with positive probability. That’s the core idea—no "isolated" states the chain can’t escape or reach.

How to Check Irreducibility for Bivariate Gibbs Sampling #

You don’t need a transition matrix here. Instead, focus on the support of your conditional distributions (the set of values each variable can take given the other):

• Let’s start with the standard bivariate Gibbs step:

  1. Given current state $(X_1^{(i)}, X_2^{(i)})$, sample $X_1^{(i+1)} \sim p(X_1 | X_2 = x_2^{(i)})$
  2. Then sample $X_2^{(i+1)} \sim p(X_2 | X_1 = x_1^{(i+1)})$

• For irreducibility, you just need two key properties:

  1. For any value $x_2$ that $X_2$ can take, the conditional distribution $p(X_1 | X_2 = x_2)$ has a support that includes all possible values of $X_1$. In plain terms: no matter what $x_2$ you’re starting with, you can sample any $y_1$ for $X_1$ with positive probability.
  2. Similarly, for any value $x_1$ that $X_1$ can take, $p(X_2 | X_1 = x_1)$ has a support that includes all possible values of $X_2$.

Example 1: Continuous Case (Bivariate Normal)

If $X_1$ and $X_2$ follow a bivariate normal distribution, each conditional distribution is a univariate normal with support over the entire real line. So:

• Starting from $(x_1, x_2)$, you can first sample $X_1$ to any $y_1$ (since the conditional normal has positive density everywhere), then sample $X_2$ to any $y_2$ using the new $y_1$. That means you can reach any $(y_1, y_2)$ in two steps with positive probability—so the chain is irreducible.

Example 2: Discrete Case (Two Dice)

Suppose $X_1$ and $X_2$ each take values 1-6, with a joint distribution where every conditional distribution $p(X_1 | X_2=k)$ has positive probability for all 1-6, and same for $p(X_2 | X_1=m)$. From any starting state like (1,1), you can sample $X_1$ to 3 in step 1, then $X_2$ to 5 in step 2—reaching (3,5) with positive probability. All states are reachable, so irreducible.

When Irreducibility Fails #

Irreducibility breaks if your conditional distributions have restricted supports. For example:

• Suppose $X_1$ and $X_2$ are non-negative integers, and if $X_2=0$, $X_1$ can only be 0; if $X_1=0$, $X_2$ can only be 0. The state (0,0) is a trap—you can never get to (1,1) from here, so the chain isn’t irreducible.

Why We Skip the Transition Matrix #

• For continuous variables, a transition matrix doesn’t even exist (there are uncountably many states).
• For large discrete state spaces, building a full transition matrix is impractical and unnecessary. Checking conditional supports is a far more efficient way to verify irreducibility.

内容的提问来源于stack exchange，提问作者mavavilj