What Symmetry Means for a Transition Matrix
When a transition matrix is symmetric, every one-step transition from state \(i\) to state \(j\) has the same probability as the reverse transition from \(j\) to \(i\). This entrywise condition has an immediate consequence: if the chain is in its uniform distribution, one transition leaves that distribution unchanged. The key is to distinguish symmetry of the matrix from a diagram that merely looks symmetric or has undirected edges.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\ge0}\) be a discrete-time Markov chain with nonempty finite state space \(S\), where \(|S|=m\). Its transition matrix is \(P=(p_{ij})_{i,j\in S}\), with \(p_{ij}=\mathbb{P}(X_{n+1}=j\mid X_n=i)\). Each entry is nonnegative and each row sums to one. A probability distribution is written as a row vector \(\mu\); after one step it is \(\mu P\). A stationary distribution is a probability row vector \(\pi\) satisfying \(\pi P=\pi\).
The uniform distribution on \(S\) assigns probability \(1/m\) to each state. The result below uses only finiteness, row-stochasticity, and matrix symmetry; it does not require the chain to be irreducible or aperiodic.
Proof. Fix a state \(j\). By symmetry and the row-sum condition for row \(j\), the sum of column \(j\) is \(\sum_{i\in S}p_{ij}=\sum_{i\in S}p_{ji}=1\). Therefore the \(j\)th coordinate of \(\pi P\) is \((\pi P)_j=\sum_{i\in S}\pi_i p_{ij}=(1/m)\sum_{i\in S}p_{ij}=1/m=\pi_j\). This holds for every \(j\), so \(\pi P=\pi\). The vector is a probability distribution because it has \(m\) entries, each \(1/m\), and their sum is one. The argument also applies when \(m=1\). \(\square\)
The proof has a useful intermediate conclusion: every column of a symmetric row-stochastic matrix sums to one. Thus such a matrix is doubly stochastic. The earlier material on doubly stochastic chains establishes that this column-sum condition also suffices for uniform stationarity; symmetry is one convenient way to guarantee it. Symmetry is sufficient, but it is not necessary: a matrix can have column sums one without having matching off-diagonal entries.
Proof. Since \(\pi_i=\pi_j=1/m\) and \(p_{ij}=p_{ji}\), we have \(\pi_i p_{ij}=(1/m)p_{ij}=(1/m)p_{ji}=\pi_j p_{ji}\). This also holds when \(i=j\), so it holds for every pair of states. \(\square\)
Worked Matrix Calculations
Worked Example: A Symmetric Three-State Chain
Let \(S=\{1,2,3\}\), and consider a chain on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with discrete time \(n\ge0\) and transition matrix
Before using the matrix, check its row sums: row 1 sums to \(\tfrac12+\tfrac14+\tfrac14=1\); row 2 sums to \(\tfrac14+\tfrac12+\tfrac14=1\); and row 3 sums to \(\tfrac14+\tfrac14+\tfrac12=1\). The entries across the diagonal match, so \(P\) is symmetric. The theorem gives the candidate \(\pi=(\tfrac13,\tfrac13,\tfrac13)\).
To check by direct multiplication, each column sums to \(\tfrac12+\tfrac14+\tfrac14=1\). Hence each coordinate of \(\pi P\) is \(\tfrac13\) times that column sum:
The distribution is normalized since \(\tfrac13+\tfrac13+\tfrac13=1\). This calculation illustrates the mechanism: uniform mass is preserved because the total probability entering each state, summed over all possible source states, is one.
Worked Example: A Symmetric Path with Unequal Self-Transition Probabilities
Let \(S=\{1,2,3,4\}\), with discrete-time chain \((X_n)_{n\ge0}\) on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and take
The four row sums are, in order, \(\tfrac34+\tfrac14=1\), \(\tfrac14+\tfrac12+\tfrac14=1\), \(0+\tfrac14+\tfrac12+\tfrac14=1\), and \(0+0+\tfrac14+\tfrac34=1\). Matching entries on opposite sides of the diagonal show that the matrix is symmetric. In particular, its diagonal entries need not all be equal: symmetry requires \(p_{ij}=p_{ji}\), not that every entry have the same value.
The proposed uniform law is \(\pi=(\tfrac14,\tfrac14,\tfrac14,\tfrac14)\). The column sums are \(\tfrac34+\tfrac14=1\), \(\tfrac14+\tfrac12+\tfrac14=1\), \(0+\tfrac14+\tfrac12+\tfrac14=1\), and \(0+0+\tfrac14+\tfrac34=1\). Therefore direct multiplication gives
The entries of \(\pi\) sum to \(4(\tfrac14)=1\), and substitution confirms stationarity. The chain can have different self-transition probabilities at different states while still preserving a uniform stationary law; what matters here is the equality of each transition probability with its reverse.
Diagram Symmetry Is Not Matrix Symmetry
A transition diagram often records which transitions are possible. If it draws an undirected edge between \(i\) and \(j\), this can mean only that both \(p_{ij}>0\) and \(p_{ji}>0\). It does not say that those probabilities are equal. Likewise, a drawing may look symmetric because of where the states are placed, even when its transition probabilities do not have that symmetry. The matrix condition is exact and numerical: \(p_{ij}=p_{ji}\) for every pair.
Worked Example: An Undirected Diagram That Does Not Give Uniform Stationarity
Let \(S=\{1,2,3\}\), and consider a discrete-time Markov chain on \((\Omega,\mathcal{F},\mathbb{P})\) with
Each row sums to one: row 1 has sum \(0+1+0=1\), row 2 has sum \(\tfrac12+0+\tfrac12=1\), and row 3 has sum \(0+1+0=1\). The possible transitions are \(1\leftrightarrow2\) and \(2\leftrightarrow3\), so the support can be drawn as an undirected path. But the matrix is not symmetric: \(p_{12}=1\) whereas \(p_{21}=\tfrac12\). The uniform distribution is not stationary, since
For a further check, solve for a stationary law \((a,b,c)\). Multiplication gives \((a,b,c)P=(b/2,a+c,b/2)\). Equating coordinates requires \(a=b/2\), \(b=a+c\), and \(c=b/2\). The first and third equations give \(a=c=b/2\); normalization gives \(a+b+c=2b=1\), so \(b=\tfrac12\) and \(a=c=\tfrac14\). Substitution verifies the result: \((\tfrac14,\tfrac12,\tfrac14)P=(\tfrac14,\tfrac12,\tfrac14)\). Thus an undirected support graph does not imply a uniform stationary law.
Common mistake. A tempting but incorrect calculation in the last example is to see reciprocal edges and infer that every column sums to one, then claim uniform stationarity. For instance, column 1 sums to \(0+\tfrac12+0=\tfrac12\), not one; column 2 sums to \(1+0+1=2\). The correction is to use the actual entries, not just the presence of edges. Uniform stationarity requires every column sum to be one, and symmetry guarantees that property only when the matrix entries themselves match in opposite directions.
What the Result Does—and Does Not—Say
The theorem proves stationarity: if \(X_0\) has the uniform distribution, then \(X_1\) has that distribution as well. Repeated application gives the same marginal distribution at each discrete time, since \(\pi P=\pi\) implies \(\pi P^n=\pi\) for every \(n\ge0\). This does not by itself say that the chain approaches the uniform law from an arbitrary initial distribution, nor does it assert that the stationary law is unique. Those are distinct questions requiring additional hypotheses or analysis.
In particular, symmetry does not force uniqueness. For a direct check, on a two-state space let \(P\) be the identity matrix. Its rows sum to one and it is symmetric, so the uniform law is stationary by the theorem. But for any probability row vector \(\mu=(a,1-a)\), multiplication gives \(\mu P=(a,1-a)=\mu\). There are therefore stationary laws other than the uniform one. This example does not change the theorem: it shows why its conclusion is existence of the uniform stationary law, not uniqueness.
When assessing a proposed symmetric-chain argument, check the following in order:
- Are the rows nonnegative and do they each sum to one?
- Does every pair of opposite entries satisfy \(p_{ij}=p_{ji}\)?
- Is the candidate the uniform row vector with one entry \(1/|S|\) per state?
- Does direct multiplication, or equivalently checking column sums, confirm \(\pi P=\pi\)?
These checks keep three ideas separate: a valid transition matrix, a symmetric transition matrix, and a stationary distribution. A picture can help identify possible transitions, but only the numerical transition matrix determines whether the symmetry argument applies.
Check Your Understanding
For each question, distinguish the row-sum condition from the symmetry condition and verify any proposed stationary vector by multiplication.
- For \(P=\begin{bmatrix}\tfrac12&\tfrac12\\ \tfrac12&\tfrac12\end{bmatrix}\), check row sums and symmetry, then compute the uniform stationary distribution.
- Give the logical step that shows every column of a symmetric row-stochastic matrix sums to one.
- In the three-state path example, compute the sum of each column and explain why the uniform distribution fails to be stationary.
- Does a symmetric transition matrix necessarily have a unique stationary distribution? Use the identity matrix on two states to justify your answer.
- State the numerical condition on \(p_{ij}\) and \(p_{ji}\) that an undirected transition diagram alone does not guarantee.