Steady State and Equilibrium (AI HL)
Run a Markov chain forward for long enough and, for most transition matrices, the state vector stops changing between steps - it settles at a fixed distribution called the steady state. This page covers the equilibrium equation \(T\pi=\pi\), how to solve it by hand or by matrix power on your GDC, and the mistakes that most often cost marks. It's part of the broader Transition Matrices & Markov Chains topic.
18 questions on this sub-topic.
Finding the steady state
Covered under IB syllabus reference AHL4.19: calculation of steady-state and long-term probabilities by repeated multiplication of the transition matrix, or by solving a system of linear equations, with awareness that the solution is the eigenvector corresponding to eigenvalue 1.
Equilibrium equation
\(T\pi = \pi,\ \sum \pi_i = 1\)
Not in the formula booklet - it's derived from \(s_n=T^ns_0\) by requiring that one more transition leaves the distribution unchanged. The normalisation condition \(\sum\pi_i=1\) is what pins down an exact answer.
State vector after \(n\) transitions
\(s_n = T^n s_0\)
In the formula booklet. Raising \(T\) to a high power and multiplying by \(s_0\) on the GDC is often quicker than solving \(T\pi=\pi\) algebraically, especially for a \(3\times3\) system.
Need the full syllabus wording and formula-booklet reference table? See Transition Matrices & Markov Chains. For calculator-specific steps, see the parent topic's GDC guidance.
Worked examples
Each year 10% of city residents move to suburbs and 5% of suburb residents move to the city. With city C and suburb S, write the transition matrix and find the long-run proportion living in the city.
Worked solution
\(T=\begin{pmatrix}0.9&0.05\\0.1&0.95\end{pmatrix}.\) M1
Steady: \(0.9p+0.05(1-p)=p\Rightarrow0.05=0.15p.\) M1
\(p=\tfrac13\approx33.3\%.\) A1
Find the steady-state distribution for \(T=\begin{pmatrix}0.8&0.4\\0.2&0.6\end{pmatrix}.\)
Worked solution
Let \(\pi=\begin{pmatrix}p\\1-p\end{pmatrix}\) with \(T\pi=\pi\): \(0.8p+0.4(1-p)=p\Rightarrow0.4=0.6p\Rightarrow p=\tfrac23.\) M1
Steady state \(\begin{pmatrix}2/3\\1/3\end{pmatrix}.\) A1
Common mistakes
- Forgetting the normalisation condition. The equation \(T\pi=\pi\) alone only fixes the ratio between the state probabilities - you still need \(\sum \pi_i = 1\) to pin down exact values, not just a proportional relationship.
- Multiplying in the wrong order. The transition matrix acts on the state vector as \(Ts_0\), not \(s_0T\) - matrix multiplication isn't commutative, so getting the order backwards gives nonsense (or an error) on the GDC.
- Stopping at an approximate power instead of the exact steady state. \(T^{20}s_0\) will look converged to a few decimal places, but when a question asks to "find" or "solve for" the steady state, an exact fraction from \(T\pi=\pi\) is expected rather than a rounded decimal read off a high matrix power.
Ready to practise properly?
18 steady-state questions, marked instantly like the real exam.
Quick answers
What is the steady-state distribution of a Markov chain?
The state vector \(\pi\) that satisfies \(T\pi=\pi\) and sums to 1 - once the system reaches it, further transitions leave the distribution unchanged.
How do you find a steady-state distribution by hand?
Set \(T\pi=\pi\), write the resulting equations, add the normalisation condition that the entries of \(\pi\) sum to 1, then solve the system - or raise \(T\) to a high power on the GDC and read off the converged columns.