Path-Reversal Symmetry Under Detailed Balance

From Edge Balance to Path Balance

Detailed balance compares the probability flow across one transition in each direction. Path-reversal symmetry is the corresponding statement for an entire finite sequence of transitions: in a stationary reversible chain, a path and the same path read backwards have equal probabilities. This is a statement about the probability of specified paths, not a claim that every individual transition has the same probability as its reverse.

Let $(X_n)_{n\ge0}$ be a discrete-time Markov chain on a finite or countable state space $S$, defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Write $P=(p_{ij})$ for its row-stochastic transition matrix, so $\sum_{j\in S}p_{ij}=1$ for each $i$. Suppose the initial distribution is the row vector $\pi$, with $\sum_i\pi_i=1$, and $\pi P=\pi$. The Markov property gives the probability of any specified finite path by multiplying its initial probability by its successive transition probabilities.

Definition: The chain is reversible with respect to $\pi$ if it satisfies detailed balance, $\pi_i p_{ij}=\pi_j p_{ji}$ for every $i,j\in S$. For a path $x_0,\ldots,x_n$, its reversal is $x_n,\ldots,x_0$. The probability of the path means the joint probability $\mathbb{P}(X_0=x_0,\ldots,X_n=x_n)$, not a probability conditional on its endpoints.

For a chain started with distribution $\pi$, the path probability is $\pi_{x_0}\prod_{r=0}^{n-1}p_{x_r x_{r+1}}$. Reversal symmetry therefore compares this expression with $\pi_{x_n}\prod_{r=0}^{n-1}p_{x_{r+1}x_r}$. The initial distribution is part of the comparison: detailed balance alone is a condition on the transition matrix and does not make paths symmetric under every possible initial distribution.

Theorem (Path-reversal symmetry): Let $(X_n)_{n\ge0}$ be a discrete-time Markov chain on a finite or countable state space $S$, with initial distribution $\pi$ and transition matrix $P$. If $\pi$ is stationary and detailed balance holds with respect to $\pi$, then, for every integer $n\ge0$ and every $x_0,\ldots,x_n\in S$, $$ \mathbb{P}(X_0=x_0,\ldots,X_n=x_n) = \mathbb{P}(X_0=x_n,\ldots,X_n=x_0). $$

Proof. The Markov property gives the two probabilities as $\pi_{x_0}\prod_{r=0}^{n-1}p_{x_r x_{r+1}}$ and $\pi_{x_n}\prod_{r=0}^{n-1}p_{x_{r+1}x_r}$, respectively. First suppose the forward path has positive probability. Then $\pi_{x_0}>0$, and every forward transition on the path has positive probability. For each $r$, detailed balance gives $\pi_{x_r}p_{x_r x_{r+1}}=\pi_{x_{r+1}}p_{x_{r+1}x_r}>0$. In particular, every $\pi_{x_{r+1}}$ is positive, and division gives $p_{x_r x_{r+1}}=(\pi_{x_{r+1}}/\pi_{x_r})p_{x_{r+1}x_r}$. Multiplying these identities over $r=0,\ldots,n-1$, then multiplying by $\pi_{x_0}$, makes the ratios telescope: $\pi_{x_0}\prod_r p_{x_r x_{r+1}}=\pi_{x_n}\prod_r p_{x_{r+1}x_r}$. If the forward probability is zero but the reversed probability is positive, applying the positive-probability argument to the reversed path would imply that the forward probability is positive, a contradiction. Thus in this case both probabilities are zero. When $n=0$, the path and its reversal are the same one-state path and the equality holds directly. This covers all cases. $\square$

The proof uses detailed balance along each edge of the path. On a path with positive probability, the ratios of stationary probabilities cancel between consecutive edges. This cancellation explains why the initial factor $\pi_{x_0}$ must be included: it is what turns the product of transition ratios into the terminal factor $\pi_{x_n}$.

Worked Path Comparisons

Worked Example: A Path with a Repeated State

Consider $S=\{1,2\}$ and $$ P=\begin{bmatrix}\tfrac34&\tfrac14\\[2pt]\tfrac12&\tfrac12\end{bmatrix}. $$ Each row sums to $1$: $\tfrac34+\tfrac14=1$ and $\tfrac12+\tfrac12=1$. The detailed-balance equation across states 1 and 2 is $\pi_1\tfrac14=\pi_2\tfrac12$, so $\pi_1=2\pi_2$. Normalizing gives $\pi=(\tfrac23,\tfrac13)$. Check stationarity directly: $(\tfrac23,\tfrac13)P=(\tfrac12+\tfrac16,\tfrac16+\tfrac16)=(\tfrac23,\tfrac13)$. The self-transition balance equations hold automatically.

Compare the path $1,1,2,1$ with its reversal $1,2,1,1$. The forward probability is $\pi_1p_{11}p_{12}p_{21}=\tfrac23\cdot\tfrac34\cdot\tfrac14\cdot\tfrac12$. In sequence, $\tfrac23\cdot\tfrac34=\tfrac12$, then $\tfrac12\cdot\tfrac14=\tfrac18$, and then $\tfrac18\cdot\tfrac12=\tfrac1{16}$.

The reversed path probability is $\pi_1p_{12}p_{21}p_{11}=\tfrac23\cdot\tfrac14\cdot\tfrac12\cdot\tfrac34$. Here $\tfrac23\cdot\tfrac14=\tfrac16$, $\tfrac16\cdot\tfrac12=\tfrac1{12}$, and $\tfrac1{12}\cdot\tfrac34=\tfrac1{16}$. The probabilities agree. As a check using the edge-balance ratios, the forward-to-reverse transition product ratio is $(p_{11}/p_{11})(p_{12}/p_{21})(p_{21}/p_{12})=1\cdot\tfrac12\cdot2=1$.

Worked Example: A Path on Three States

Let $S=\{1,2,3\}$, $\pi=(\tfrac12,\tfrac13,\tfrac16)$, and $$ P=\begin{bmatrix} \tfrac56&\tfrac16&0\\[2pt] \tfrac14&\tfrac12&\tfrac14\\[2pt] 0&\tfrac12&\tfrac12 \end{bmatrix}. $$ The row sums are $\tfrac56+\tfrac16+0=1$, $\tfrac14+\tfrac12+\tfrac14=1$, and $0+\tfrac12+\tfrac12=1$. For the edge between 1 and 2, both flows are $\pi_1p_{12}=\tfrac12\cdot\tfrac16=\tfrac1{12}$ and $\pi_2p_{21}=\tfrac13\cdot\tfrac14=\tfrac1{12}$. For the edge between 2 and 3, they are $\pi_2p_{23}=\tfrac13\cdot\tfrac14=\tfrac1{12}$ and $\pi_3p_{32}=\tfrac16\cdot\tfrac12=\tfrac1{12}$. For 1 and 3 both flows are zero, and the diagonal balance equations compare each quantity with itself. Thus detailed balance holds.

Stationarity can also be checked directly: \[ (\tfrac12,\tfrac13,\tfrac16)P = (\tfrac5{12}+\tfrac1{12},\ \tfrac1{12}+\tfrac16+\tfrac1{12},\ \tfrac1{12}+\tfrac1{12}) =(\tfrac12,\tfrac13,\tfrac16). \] Now compare $1,2,3,2,1$ with its reversal $1,2,3,2,1$; this path happens to be its own reversal. Its probability is $\tfrac12\cdot\tfrac16\cdot\tfrac14\cdot\tfrac12\cdot\tfrac14$. The successive products are $\tfrac1{12}$, $\tfrac1{48}$, $\tfrac1{96}$, and $\tfrac1{384}$. Since the reversed sequence is identical, its probability is also $\tfrac1{384}$. This self-reversing example also checks that a path need not have distinct states.

Worked Example: Detailed Balance Without a Stationary Start

Use the two-state transition matrix from the first example, whose rows sum to $1$ and which satisfies detailed balance with $\pi=(\tfrac23,\tfrac13)$. Start the chain instead with $\mu=(1,0)$, concentrated at state 1. Compare the one-step path $1,2$ with $2,1$.

Under this initial distribution, $\mathbb{P}(X_0=1,X_1=2)=\mu_1p_{12}=1\cdot\tfrac14=\tfrac14$. In the reverse order, $\mathbb{P}(X_0=2,X_1=1)=\mu_2p_{21}=0\cdot\tfrac12=0$. They are not equal. The initial distribution is not stationary, since $(1,0)P=(\tfrac34,\tfrac14)\ne(1,0)$. If the initial distribution is changed to $\pi$, the two probabilities become $\tfrac23\cdot\tfrac14=\tfrac16$ and $\tfrac13\cdot\tfrac12=\tfrac16$. The comparison confirms that the theorem concerns the stationary chain, not merely a transition matrix that satisfies detailed balance for some distribution.

What the Symmetry Does—and Does Not—Say

Because $\pi$ is stationary, the same path formula applies to a path segment beginning at any fixed time $m\ge0$. In fact, $\mathbb{P}(X_m=x_0,\ldots,X_{m+n}=x_n)=\pi_{x_0}\prod_{r=0}^{n-1}p_{x_r x_{r+1}}$. Applying the theorem therefore shows that the joint probabilities for this segment and its reversal agree. This conclusion concerns a fixed finite block of states; it does not require conditioning on a future state or on a random time.

In particular, path reversal does not say $p_{ij}=p_{ji}$ for every pair of states. The first example has $p_{12}=\tfrac14$ and $p_{21}=\tfrac12$, which are different, while the stationary flows are equal: $\pi_1p_{12}=\tfrac23\cdot\tfrac14=\tfrac16$ and $\pi_2p_{21}=\tfrac13\cdot\tfrac12=\tfrac16$. It is the stationary weights together with the transition probabilities that balance, and the stationary factors telescope across a path.

Common mistake. One might compare only the products of transition probabilities and conclude that reversing a path changes nothing. For the path $1,1,2$ in the first example, the forward transition product is $p_{11}p_{12}=\tfrac34\cdot\tfrac14=\tfrac3{16}$, whereas the reversed transition product for $2,1,1$ is $p_{21}p_{11}=\tfrac12\cdot\tfrac34=\tfrac38$. These are not equal. The full path probabilities include different initial factors: $\pi_1\cdot\tfrac3{16}=\tfrac23\cdot\tfrac3{16}=\tfrac18$, and $\pi_2\cdot\tfrac38=\tfrac13\cdot\tfrac38=\tfrac18$. The correction is to compare joint path probabilities, including the initial stationary probability, rather than transition products alone.

A second error is to infer that a zero-probability path has a positive-probability reversal. The theorem’s proof rules this out: if either direction had positive probability, detailed balance along its edges would imply positive probability in the other direction as well. Thus the equality includes paths that are impossible, without requiring division by a zero stationary probability.

Check Your Understanding

For each question, distinguish the initial distribution from the transition matrix and use the full joint path probability.

For $P=\begin{bmatrix}\tfrac34&\tfrac14\\ \tfrac12&\tfrac12\end{bmatrix}$ and $\pi=(\tfrac23,\tfrac13)$, compute the probabilities of $2,1,1,2$ and its reversal.
Why is it not valid to divide the detailed-balance equation by $\pi_i$ when $\pi_i=0$? How does the theorem handle this case?
For the three-state matrix in the second worked example, verify $\pi_2p_{22}=\pi_2p_{22}$ and explain why this is a detailed-balance equation.
Can a reversible transition matrix have $p_{ij}\ne p_{ji}$? Give the values from the two-state example and compare the corresponding stationary flows.
State precisely which assumption fails in the third worked example, and compute the two one-step path probabilities under its initial distribution.

Path-Reversal Symmetry Under Detailed Balance

What You'll Learn