Stochastic Series - Pt 7 - Detailed Balance

16 Oct, 2026

In the last chapter we covered the stationary distribution for a Markov chain. As you might recall, this means that the distribution q is stationary, which means it does not change.

We use π (pi) to denote the stationary distribution.

π = π P

The above expression means that the stationary distribution is equal to the stationary distribution multiplied by the transition matrix.

Detailed Balance vs. Stationary Distribution

Stationary Distribution

Detailed balance is a stricter condition than stationarity, which means that if a chain satisfies detailed balance with respect to a distribution, that distribution must also be stationary.

A stationary distribution does not "care" how the distribution stays stationary, as long as the probability of each state stays the same.

Imagine we have 100 machines:

State	Machines	Probability
A	50	0.50
B	25	0.25
C	25	0.25
Total	100	1.00

If 10 machines from state A transition into B, it's okay for C to make up for the loss in A.

For example, imagine the following expected transitions:

From	To	Machines
A	B	10
B	C	10
C	A	10

We can represent these transitions with the following transition matrix:

P = (\begin{matrix} 0.8 & 0.2 & 0 \\ 0 & 0.6 & 0.4 \\ 0.4 & 0 & 0.6 \end{matrix})

This illustrates global balance. Each state receives as much as it loses, even though the machines move in a cycle.

At the end of the step, the distribution remains unchanged in expectation.

But for a chain to satisfy detailed balance, we need something stricter: the flow of probability must balance between every pair of states.

Probability Flow?

Probability flow is the probability of being in a specific state multiplied by the probability of transitioning to another state.

Let's say our stationary distribution looks like this:

π = (0.3, 0.2, 0.5)

To denote the stationary probability of state A, we use:

π (A) = 0.3

So imagine that we are at step n in the chain and the distribution is stationary. We know that π(A) is 0.3, so we can expect 30% of all machines to be in state A.

We also have this transition matrix ℗:

P = (\begin{matrix} 0.7 & 0.2 & 0.1 \\ 0.2 & 0.7 & 0.1 \\ 0.1 & 0 & 0.9 \end{matrix})

Now we can see that the probability of transitioning from state A to B is 0.2.

Then the probability flow becomes:

π (A) \times P (A, B) = 0.3 \times 0.2 = 0.06

So we took the stationary probability of state A, which was 0.3, and multiplied it by the probability of transitioning from A to B, which was 0.2.

This means that 6% of all machines are expected to transition from A to B during a single step.

Detailed Balance

Now that we have covered the concept of probability flow, we can take a closer look at what detailed balance means.

π (A) \times P (A, B) = π (B) \times P (B, A)

The above expression means that the probability flow from A to B matches the probability flow from B to A.

This condition must hold for every pair of states.

Our previous examples satisfied stationarity, but not detailed balance. So let's use a transition matrix that satisfies detailed balance.

We return to our original distribution of 100 machines:

π = (0.50, 0.25, 0.25)

And use this transition matrix:

P = (\begin{matrix} 0.6 & 0.2 & 0.2 \\ 0.4 & 0.2 & 0.4 \\ 0.4 & 0.4 & 0.2 \end{matrix})

In the first example, machines moved in a one-way cycle.

In the new matrix, the expected flow is equal in both directions for every pair:

A ↔ B: 10 machines each way
B ↔ C: 10 machines each way
A ↔ C: 10 machines each way

Let's check that every pair satisfies the detailed balance condition.

A ↔ B

0.50 \times 0.2 = 0.10

0.25 \times 0.4 = 0.10

A ↔ C

0.50 \times 0.2 = 0.10

0.25 \times 0.4 = 0.10

B ↔ C

0.25 \times 0.4 = 0.10

0.25 \times 0.4 = 0.10

All three pairs satisfy detailed balance.

The distribution remains stationary, but unlike our previous examples, the probability flow between every pair of states is now balanced in both directions.