Self-Loops and Aperiodicity

Why a Self-Loop Matters

Period is determined by the lengths of possible returns to a state. A self-loop is a direct return in one step, so even when longer return paths are restricted to particular lengths, that one-step return changes the greatest common divisor. This gives a useful sufficient condition for aperiodicity. It is not a necessary condition: a chain can have period one even when every diagonal transition probability is zero.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a finite or countable state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with transition matrix \(P=(p_{ij})\). The matrix is row-stochastic, so \(\sum_{j\in S}p_{ij}=1\) for each \(i\). Write \(\mathbb{P}_i\) for probability when \(X_0=i\), and \(p_{ii}^{(n)}=\mathbb{P}_i(X_n=i)\) for the \(n\)-step return probability.

The set in the greatest common divisor uses positive times \(n\ge1\); time zero is not counted. Also, \(p_{ii}^{(n)}>0\) means there is at least one length-\(n\) path from \(i\) back to \(i\) whose transition probabilities are all positive. The probability of that particular path is the product of its transition probabilities, so its positivity verifies that \(n\) belongs to the return-time set.

The Self-Loop Criterion

Proof. Since \(p_{ii}>0\), the one-step return probability \(p_{ii}^{(1)}=p_{ii}\) is positive. Thus \(1\) belongs to the set of positive return times, and the greatest common divisor of a set containing \(1\) is \(1\). This proves that \(d(i)=1\). For the final statement, use the standard period-invariance result for communicating states, proved just below: every state communicating with \(i\) has the same period as \(i\). Irreducibility means every pair of states communicates, so every state has period one. \(\square\)

The first part of the theorem is local: it establishes the period of the state with the self-loop without requiring irreducibility. The second part needs irreducibility if we want to conclude that all states have period one. More generally, the conclusion extends to the communicating class of the state with the loop, not automatically to states in other classes.

Proof. Choose a positive-probability path from \(i\) to \(j\) of length \(a\ge0\) and a positive-probability path from \(j\) to \(i\) of length \(b\ge0\). If \(a+b=0\), then \(i=j\) and the conclusion is immediate. Otherwise, concatenating the paths gives a positive-probability return to \(i\) of length \(a+b\). Now take any \(n\ge1\) such that \(p_{jj}^{(n)}>0\). Concatenate the path from \(i\) to \(j\), a length-\(n\) return from \(j\) to \(j\), and the path from \(j\) to \(i\). The result is a positive-probability return to \(i\) of length \(a+n+b\). Therefore \(d(i)\) divides both \(a+b\) and \(a+n+b\), and hence divides their difference \(n\). This holds for every positive return time \(n\) at \(j\), so \(d(i)\) divides \(d(j)\). Interchanging \(i\) and \(j\) gives that \(d(j)\) divides \(d(i)\); both periods are positive integers, so they are equal. \(\square\)

A related elementary observation is often useful when there is no self-loop: if two consecutive positive return lengths \(r\) and \(r+1\) occur at a state, then its period is one. Indeed, its period divides both lengths and therefore divides their difference, \(1\). This gives a direct way to verify aperiodicity without finding every possible return time.

Aperiodicity Without Self-Loops

The self-loop condition is sufficient, not necessary. If there is no one-step return, the return-time set may still contain lengths with greatest common divisor one. In particular, return paths of two consecutive lengths prove period one. The next example has no self-loops at all, yet every state is aperiodic.

Here the extra routes between states make return paths of different lengths possible. The definition depends on the gcd of all possible positive return lengths, not on whether the shortest return happens to be one. A pair of coprime return lengths is enough; two consecutive lengths are a particularly convenient pair.

No Self-Loop Does Not Mean Aperiodic

Scope and Common Pitfalls

A positive-probability self-loop settles the period of its own state, and period invariance settles the periods of states that communicate with it. It does not settle the periods of states in other communicating classes. For example, consider a chain on \(\{1,2,3\}\) with transition matrix \(\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}\). Its row sums are \(1+0+0=1\), \(0+0+1=1\), and \(0+1+0=1\). State \(1\) has a self-loop and period one, while states \(2\) and \(3\) alternate and each has period two. The chain is reducible, so the self-loop at \(1\) gives no conclusion about the other class.

A common mistake is to reason, “There is no self-loop, so the state has period two.” The three-state chain in the preceding example shows why that calculation fails: at state \(1\), the returns of lengths \(2\) and \(3\) have probabilities \(\tfrac14\) and \(\tfrac18\), so the period divides both \(2\) and \(3\), and is therefore \(1\), not \(2\). Absence of a one-step return only says that \(1\) is not in the return-time set; it does not imply that every return time is even.

Another error is to conclude that one self-loop makes every state in any chain aperiodic. The reducible example above corrects this: a loop at a state forces period one there, and the conclusion transfers only to states in its communicating class. Before using the class-wide conclusion, check irreducibility or identify the relevant class.