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Markov chains model systems that move between states with fixed probabilities at each step. This reference helps students organize transition diagrams, matrices, state vectors, and long-run predictions. It is useful for probability, statistics, discrete math, and applied modeling problems.

The main idea is that the next state depends only on the current state, not the full past history.

A discrete-time Markov chain uses a transition matrix PP and a state vector xn\mathbf{x}_n to describe probabilities after nn steps. Matrix powers such as PnP^n predict long-run behavior, while steady states satisfy πP=π\boldsymbol{\pi}P=\boldsymbol{\pi}. Absorbing chains use canonical form to find absorption probabilities and expected time to absorption.

The most important skills are setting up PP, multiplying in the correct order, and interpreting each probability carefully.

Key Facts

  • A Markov chain has the memoryless property P(Xn+1=jXn=i,Xn1,,X0)=P(Xn+1=jXn=i)P(X_{n+1}=j\mid X_n=i, X_{n-1},\ldots,X_0)=P(X_{n+1}=j\mid X_n=i).
  • A transition matrix PP is stochastic, so every entry satisfies 0pij10\le p_{ij}\le 1 and each row sum is jpij=1\sum_j p_{ij}=1.
  • If xn\mathbf{x}_n is a row state vector, then the next state vector is xn+1=xnP\mathbf{x}_{n+1}=\mathbf{x}_nP.
  • The state vector after nn steps is xn=x0Pn\mathbf{x}_n=\mathbf{x}_0P^n when using row vectors.
  • A steady-state distribution π\boldsymbol{\pi} satisfies πP=π\boldsymbol{\pi}P=\boldsymbol{\pi} and iπi=1\sum_i \pi_i=1.
  • For many regular chains, the rows of PnP^n approach the same steady-state vector π\boldsymbol{\pi} as nn\to\infty.
  • In an absorbing chain written as P=[QR0I]P=\begin{bmatrix}Q&R\\0&I\end{bmatrix}, the fundamental matrix is N=(IQ)1N=(I-Q)^{-1}.
  • For an absorbing chain, absorption probabilities are given by B=NRB=NR and expected steps to absorption are given by t=N1\mathbf{t}=N\mathbf{1}.

Vocabulary

State
A state is one possible condition or category that the system can occupy at a given step.
Transition Probability
A transition probability pijp_{ij} is the probability of moving from state ii to state jj in one step.
Transition Matrix
A transition matrix PP stores all one-step transition probabilities for a Markov chain.
State Vector
A state vector xn\mathbf{x}_n lists the probabilities of being in each state after nn steps.
Steady-State Distribution
A steady-state distribution π\boldsymbol{\pi} is a probability vector that does not change after multiplying by PP.
Absorbing State
An absorbing state is a state that cannot be left once entered, so its self-transition probability is 11.

Common Mistakes to Avoid

  • Using columns when the problem uses row vectors is wrong because the update rule changes from xn+1=xnP\mathbf{x}_{n+1}=\mathbf{x}_nP to xn+1=Pxn\mathbf{x}_{n+1}=P\mathbf{x}_n. Match the vector convention before multiplying.
  • Forgetting that each row of PP must sum to 11 is wrong because transition probabilities from one current state must cover all possible next states.
  • Treating PnP^n as multiplying every entry of PP by nn is wrong because PnP^n means repeated matrix multiplication, not scalar multiplication.
  • Solving only πP=π\boldsymbol{\pi}P=\boldsymbol{\pi} without iπi=1\sum_i \pi_i=1 is incomplete because a steady-state distribution must be normalized as a probability vector.
  • Using N=(IP)1N=(I-P)^{-1} for an absorbing chain is wrong because the fundamental matrix uses only the transient-to-transient block, so N=(IQ)1N=(I-Q)^{-1}.

Practice Questions

  1. 1 A two-state chain has P=[0.70.30.20.8]P=\begin{bmatrix}0.7&0.3\\0.2&0.8\end{bmatrix} and x0=[10]\mathbf{x}_0=\begin{bmatrix}1&0\end{bmatrix}. Find x1\mathbf{x}_1 and x2\mathbf{x}_2.
  2. 2 Find the steady-state distribution π\boldsymbol{\pi} for P=[0.60.40.10.9]P=\begin{bmatrix}0.6&0.4\\0.1&0.9\end{bmatrix} using πP=π\boldsymbol{\pi}P=\boldsymbol{\pi} and π1+π2=1\pi_1+\pi_2=1.
  3. 3 For an absorbing chain with Q=[0.50.20.10.4]Q=\begin{bmatrix}0.5&0.2\\0.1&0.4\end{bmatrix} and R=[0.30.5]R=\begin{bmatrix}0.3\\0.5\end{bmatrix}, compute N=(IQ)1N=(I-Q)^{-1} and B=NRB=NR.
  4. 4 Explain why a transition matrix with a row sum of 1.21.2 cannot represent a valid Markov chain, even if all entries are positive.