Lagmental Vicfred

Chapman--Kolmogorov Is Matrix Multiplication by Vicfred

Summing over an intermediate state composes transition probabilities. I want the notation, the mechanism, and the failure mode visible at the same time.

The mathematical object

A discrete-time Markov chain has transition matrix \(P=(p_{ij})\) and forgets the past after conditioning on the present. Matrix powers \(P^n\) give multi-step transition probabilities.

$$ p_{ij}^{(n)}=\mathbf P(X_n=j\mid X_0=i) $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ P^{m+n}=P^mP^n $$

One explicit computation

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ p_{ij}^{(m+n)}=\sum_kp_{ik}^{(m)}p_{kj}^{(n)} $$

Why the identity matters

The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.

$$ \begin{aligned} \mathsf{D}\;&:\quad p_{ij}^{(n)}=\mathbf P(X_n=j\mid X_0=i),\\[5pt] \mathsf{C}\;&:\quad P^{m+n}=P^mP^n. \end{aligned} $$

Where it can fail

A stationary distribution need not be unique without irreducibility, and convergence to it can fail without aperiodicity.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] P^{m+n}=P^mP^n \end{gathered}} $$

I would use the boxed line as a reference later, while returning to the full display whenever a hypothesis becomes uncertain. That division keeps compression from becoming ambiguity.

This article was posted on Wed 29 April 2009. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.