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How do we predict the future state of a system that moves between categories at fixed rates?

Use transition matrices and an initial state vector to predict future states and to find the long-term steady state of a system.

How to set up a transition matrix from movement percentages, apply it to an initial state to predict future states, and find the long-term steady state of a system.

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  1. What this dot point is asking
  2. Setting up a transition matrix
  3. Predicting future states
  4. The long-term steady state
  5. Why a steady state exists
  6. Interpreting the result in context

What this dot point is asking

You must build a transition matrix from movement data, predict the state after one or more steps, and find or interpret the long-term steady state.

Setting up a transition matrix

A system moves objects (customers, animals, voters) between a fixed set of categories at constant rates each period. The transition matrix TT records these rates. Using the convention Sn+1=TSnS_{n+1} = T S_n, the entry in row ii, column jj is the proportion moving from category jj into category ii. Each column therefore sums to 1, since everything currently in a category must go somewhere.

Predicting future states

To advance one week, multiply the current state by TT. Repeat, or use Sn=TnS0S_n = T^n S_0 to jump ahead.

The long-term steady state

As nn grows, the state vector usually settles to a fixed steady state SS that satisfies S=TSS = TS and stops changing from week to week. You can find it by computing high powers TnS0T^n S_0 until the numbers stabilise, or by solving S=TSS = TS with the constraint that the categories add to the total.

Why a steady state exists

The reason the system settles is that the transition matrix gently pulls the state toward a balance point. Whenever more leave a category than join it, that category shrinks; whenever more join than leave, it grows. The steady state is precisely where these flows balance for every category at once, so TS=STS = S. For a regular transition matrix - one where some power has all positive entries - this balance point is unique and is reached from any starting distribution, which is why the long-run market share does not depend on where the shoppers began. This is the key insight examiners want you to articulate when interpreting a steady state.

Interpreting the result in context

The steady state predicts the eventual market share regardless of where shoppers started, provided the transition rates stay constant. Examiners reward a sentence interpreting the numbers, for example "store B eventually holds about two-thirds of the market."

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20235 marksCalculator-assumed. Each week 80% of shoppers at store A return to A and 20% switch to B; 90% at B return to B and 10% switch to A. Initially 1000 shop at A and 1000 at B. (a) Write the transition matrix TT (order A then B). (b) Find the distribution after one week. (c) Find the long-term steady-state distribution of the 2000 shoppers.
Show worked answer β†’

(a) Columns describe where shoppers start: T=[0.80.10.20.9]T = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}, each column summing to 1. (1 mark)

(b) S1=TS0=[0.80.10.20.9][10001000]=[9001100]S_1 = T S_0 = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}\begin{bmatrix} 1000 \\ 1000 \end{bmatrix} = \begin{bmatrix} 900 \\ 1100 \end{bmatrix}: 900 at A, 1100 at B. (2 marks)

(c) Steady state [ab]\begin{bmatrix} a \\ b \end{bmatrix} with a+b=2000a + b = 2000 and TS=STS = S. The first row gives 0.8a+0.1b=a0.8a + 0.1b = a, so 0.1b=0.2a0.1b = 0.2a, i.e. b=2ab = 2a. Then 3a=20003a = 2000, so aβ‰ˆ667a \approx 667 and bβ‰ˆ1333b \approx 1333. (2 marks)

Marks: one for TT with valid columns, two for the one-week product, two for solving the steady state.

SACE 20223 marksCalculator-assumed. A transition matrix is T=[0.70.40.30.6]T = \begin{bmatrix} 0.7 & 0.4 \\ 0.3 & 0.6 \end{bmatrix} with initial state S0=[500500]S_0 = \begin{bmatrix} 500 \\ 500 \end{bmatrix}. Find the state after two steps, S2S_2.
Show worked answer β†’

First step: S1=TS0=[0.7(500)+0.4(500)0.3(500)+0.6(500)]=[550450]S_1 = T S_0 = \begin{bmatrix} 0.7(500) + 0.4(500) \\ 0.3(500) + 0.6(500) \end{bmatrix} = \begin{bmatrix} 550 \\ 450 \end{bmatrix}. [1 mark]

Second step: S2=TS1=[0.7(550)+0.4(450)0.3(550)+0.6(450)]=[565435]S_2 = T S_1 = \begin{bmatrix} 0.7(550) + 0.4(450) \\ 0.3(550) + 0.6(450) \end{bmatrix} = \begin{bmatrix} 565 \\ 435 \end{bmatrix}. [2 marks]

The total stays 1000 at each step, confirming no loss.

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