Normalizing a Finite Invariant Measure

From Invariant Weights to Probabilities

An invariant measure records weights that are unchanged by one transition of a Markov chain. Those weights need not initially sum to one, so they need not themselves be probabilities. When their total mass is positive and finite, dividing each weight by that total produces a probability distribution without changing the balance between incoming and outgoing mass. This is a basic conversion: the invariance calculation is already done, and normalization puts its weights on the probability scale.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with finite or countable state space \(S\). Its transition matrix is \(P=(p_{ij})_{i,j\in S}\), where \(p_{ij}\ge0\) and \(\sum_{j\in S}p_{ij}=1\) for every \(i\in S\). We use row vectors for measures and distributions. For a nonnegative row vector \(\nu=(\nu_i)_{i\in S}\), its total mass is \(Z=\sum_{i\in S}\nu_i\).

For countable \(S\), the sums in these definitions are sums of nonnegative terms. The invariance condition says that the total weight arriving at each state \(j\), computed from all possible preceding states, equals the weight assigned to \(j\). It does not require that the original weights sum to one. Normalization concerns that total separately.

Proof. Since every \(\nu_i\ge0\) and \(Z\) is positive and finite, \(\pi_i=\nu_i/Z\ge0\) for each \(i\), and \(\sum_i\pi_i=(1/Z)\sum_i\nu_i=Z/Z=1\). For every \(j\in S\), invariance of \(\nu\) gives \(\sum_i\nu_i p_{ij}=\nu_j\). Dividing this equality by \(Z\) yields \(\sum_i\pi_i p_{ij}=\nu_j/Z=\pi_j\). Thus \(\pi P=\pi\), and \(\pi\) is a stationary distribution. \(\square\)

The proof uses two separate facts: positive finite mass makes the division well-defined and gives a probability vector; invariance is preserved because every coordinate is divided by the same constant. No irreducibility, aperiodicity, or uniqueness assumption is needed for this normalization result.

Worked Normalizations

What Normalization Does—and Does Not—Show

Proof. For \(c>0\), nonnegativity is preserved and \((c\nu)P=c(\nu P)=c\nu\), so \(c\nu\) is invariant. Its total mass is \(cZ_\nu\), which is positive and finite, and its normalization at state \(i\) is \(c\nu_i/(cZ_\nu)=\nu_i/Z_\nu\). For the converse, write \(Z_\nu=\sum_i\nu_i\) and \(Z_\eta=\sum_i\eta_i\). Equality of the normalized distributions means \(\nu_i/Z_\nu=\eta_i/Z_\eta\) for every \(i\). Multiplying by \(Z_\eta\) gives \(\eta_i=(Z_\eta/Z_\nu)\nu_i\) for every \(i\). Since both masses are positive and finite, \(Z_\eta/Z_\nu>0\), proving the claim. \(\square\)

This proposition explains why invariant measures are often described as weights: multiplying all weights by the same positive constant changes their total but not the stationary distribution obtained from them. Normalization alone does not show that a stationary distribution is unique. It supplies a stationary distribution from the particular invariant measure given; deciding whether other stationary distributions exist is a separate question.

Common mistake. In the nonuniform example, dividing \((1,2,3)\) by its largest coordinate \(3\) gives \((\tfrac13,\tfrac23,1)\). These entries sum to \(\tfrac13+\tfrac23+1=2\), not \(1\), so this is not a probability distribution. The correct divisor is the total mass \(1+2+3=6\), which gives \((\tfrac16,\tfrac13,\tfrac12)\). Another avoidable error is multiplying the column of weights by \(P\): with the row-vector convention, invariance is \(\nu P=\nu\), and its \(j\)th coordinate is \(\sum_i\nu_i p_{ij}\).

The two conditions on total mass have different roles. If \(Z=0\), nonnegativity forces every \(\nu_i=0\), and division by \(Z\) is undefined. If \(Z=\infty\), there is no finite total by which to divide to obtain a probability vector. For example, on \(S=\{1,2,\ldots\}\), let \(p_{ij}=1\) when \(i=j\) and \(p_{ij}=0\) otherwise. Each row sums to \(1\), and \(\nu_i=1\) for every \(i\) is invariant, since \(\sum_i\nu_i p_{ij}=1=\nu_j\) for each \(j\). But \(\sum_i\nu_i=\infty\), so the finite-mass normalization theorem does not apply. This example does not contradict the theorem: its finite positive mass hypothesis is absent.

In applications, an invariant measure may be easier to find in a convenient scale than a probability vector. Once its weights satisfy the balance equations, the normalization step is mechanical: sum all coordinates, confirm that the result is positive and finite, divide every coordinate by that same sum, and check both the total and the invariance equation. Keeping these checks separate helps catch arithmetic errors and makes clear exactly what the result guarantees.