# Detailed Balance: What is the continuous analogy of the transition matrix?

I am having trouble understanding the definition of detailed balance in the case of a continuous state space. The definition of detailed balance that I am working from is:

A pmf $$\pi$$ on $$\mathcal{X}$$ satisfies detailed balance with respect to $$T$$ if $$\pi_a T_{ab} = \pi_b T_{ba} \ \ \forall a,b \in \mathcal{X}$$

As I understand it, $$T$$ is the transition matrix and $$T_{ab}$$ represents the probability of moving from state $$a$$ to state $$b$$. I am struggling with the notion of $$T$$ in the case where $$\mathcal{X}$$ is continuous. Say for example that $$\mathcal{X} = \mathbb{R}$$, then wouldn't $$T_{ab}=0$$ as a single value in $$\mathbb{R}$$ has measure 0? Also, what is the structure of the transition matrix $$T$$? Does it have an uncountably infinite number of rows and columns? How is this possible?

• Perhaps stats.stackexchange.com/questions/46389 helps.
– whuber
Commented May 26, 2021 at 18:55
• Thanks @whuber . I think this is getting towards what I am struggling with although some of it is over my head. For an example, if $\pi = \mathcal{N}(\mu, \Sigma)$ what would the elements of $T$ be? I'm not sure I can answer this question based on your example? Commented May 26, 2021 at 19:23

Detailed balance for a continuous Markov chain with transition kernel $$K(\cdot,\cdot)$$ and stationary density $$f(\cdot)$$ writes as $$f(x)K(x,y)=f(y)K(y,x)$$ It means that, in a stationary regime, the joint distribution of $$(x_t,x_{t+1})$$ is the same as the joint distribution of $$(x_{t+1},x_{t})$$. This implies that the chain $$(x_t)$$ is (time-)reversible.
• I see. So am I understanding correctly that the transition kernel $K$ is the continuous analogue to the transition matrix $T$? Commented May 26, 2021 at 20:01
• The transition kernel $K$ is used to generate the following value of the Markov chain conditional on the current value (first argument). It thus acts as a conditional density. Commented May 26, 2021 at 20:29