Markov Chain Matrix Analyzer

Math Linear Algebra • Applications and Advanced Linear Algebra (capstone)

Written by STEM Calculators Team

Analyze a transition matrix \(P\) (row-stochastic). Compute a steady state \(\pi\) by solving \(\pi P=\pi\) with \(\sum_i \pi_i=1\), detect absorbing states, compute \(n\)-step transitions \(P^n\), and visualize state probabilities over time.

Number of states

Tolerance

Display decimals

Compute \(n\)-step transitions (\(P^n\))

Graph steps

Transition matrix \(P\)

Row \(i\) should sum to 1. Entry \(p_{ij}\) is the probability to move from state \(i\) to state \(j\).

Inputs accept 0.3, 2e-4, fractions 7/10, and constants pi, e.

Initial distribution \(v_0\)

Used for the graph \(v_{t+1}=v_tP\) and for \(v_k=v_0P^k\).

Preset

Tip: a valid distribution has nonnegative entries and sums to 1 (within tolerance).

Ready

Results

Row-stochastic check

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Absorbing states

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Steady state \(\pi\)

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Steady-state residual \(\|\pi P-\pi\|\)

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\(P^n\) (selected \(n\))

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Distribution after \(n\) steps: \(v_n = v_0P^n\)

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Long-term note

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Graph

Plots \(v_t\) versus time step \(t\). Includes axis tick values. Zoom with wheel/trackpad, drag to pan, double-click to reset. Play animates drawing the trajectories.

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Step-by-step

Enter \(P\) and click “Calculate”.

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Theory

Transition matrices and state distributions

A (finite) Markov chain is a stochastic process with a set of states \(\{1,2,\dots,n\}\) and the Markov property: the next state depends only on the current state. The chain is described by a transition matrix \(P\in\mathbb{R}^{n\times n}\), where \[ p_{ij}=\Pr(X_{t+1}=j\mid X_t=i). \] Because each row lists probabilities of moving out of state \(i\), \(P\) should be row-stochastic: \(p_{ij}\ge 0\) and \(\sum_j p_{ij}=1\) for every row \(i\).

If we store the state probabilities at time \(t\) as a row vector \(v_t=[\Pr(X_t=1),\dots,\Pr(X_t=n)]\), then the evolution is \[ v_{t+1}=v_tP. \] Repeating gives \(v_t=v_0P^t\). The matrix \(P^t\) is the \(t\)-step transition matrix: its entry \((i,j)\) equals the probability of going from state \(i\) to state \(j\) in exactly \(t\) steps.

Steady states and long-term behavior

A stationary (steady-state) distribution is a probability vector \(\pi\) that remains unchanged after applying \(P\): \[ \pi P=\pi,\qquad \sum_i \pi_i=1,\qquad \pi_i\ge 0. \] This is a left-eigenvector problem for eigenvalue 1. A common computational method rewrites it as \[ (P^{\mathsf T}-I)\pi^{\mathsf T}=0, \] and then replaces one of the equations with the normalization \(\sum_i \pi_i=1\) to form a square linear system.

Whether the chain converges to a unique steady state depends on structural properties. If the chain is regular (some power \(P^k\) has all strictly positive entries), then there is a unique stationary distribution and, for any initial distribution, \(v_t\to\pi\) as \(t\to\infty\). More generally, for ergodic chains (irreducible and aperiodic), the same conclusion holds. If the chain is reducible or periodic, you can have multiple stationary distributions or oscillatory behavior, and the limit may depend on \(v_0\).

Absorbing states and absorbing classes

A state \(i\) is absorbing if once you enter it, you stay there forever: \[ p_{ii}=1 \quad\text{and}\quad p_{ij}=0\ \text{for}\ j\ne i. \] Absorbing states force long-term mass to concentrate inside absorbing states or absorbing classes (closed sets of states). In such cases, the steady state may not be unique, and the eventual behavior depends on where the chain can end up from the starting state.

How to interpret the graph

The graph plots the components of \(v_t\) over time \(t\). If the chain is mixing/regular, the curves stabilize near a constant vector (the steady state \(\pi\)). If there are absorbing states, one or more components may approach 1 as the chain gets absorbed. If the chain is periodic, the curves may oscillate rather than converge.

Frequently Asked Questions

What does it mean for P to be row-stochastic?

Every entry is nonnegative and each row sums to 1. Row i represents a probability distribution over the next state given the current state i.

How is the steady state computed?

The calculator solves πP=π with sum(π)=1 by converting it to (P^T - I)π^T=0 and replacing one equation with the normalization constraint; if the system is ill-conditioned, it falls back to an iterative approximation.

What is an absorbing state?

A state i is absorbing if p_ii=1 and p_ij=0 for all j≠i. Once entered, the chain stays there forever.

Why might the chain not converge to a single steady state?

If the chain is reducible, periodic, or has multiple closed classes, it may have multiple stationary distributions or oscillate. Regular/ergodic chains converge to a unique π.

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Frequently Asked Questions

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