Long-term behaviour and steady states | Year 12

01 / LEARN

The idea, step by step

Equilibrium balances the flows

A steady state $s$ satisfies $T s = s$ . Individuals may still move between categories, but arrivals and departures balance in each category. In a two-state model, use one independent balance equation and the total. The second balance equation is dependent because the model conserves the total.

For rows (0.8,0.3),(0.2,0.7), balance in A is $a = 0.8 a + 0.3 b$ , so $0.2 a = 0.3 b$ and $a : b = 3 : 2$ . With total 100 the fixed distribution is (60,40), not necessarily an equal split.

Fixed does not always mean attracting

Many mixing models approach one stable distribution, which can be investigated by repeated powers. But a stable-looking rounded calculator output is not a proof about every transition matrix or starting state.

For the swap matrix, (50,50) is fixed but a start (100,0) alternates forever. For the identity matrix, every starting distribution is fixed. Solve the equilibrium equations, then separately ask whether the initial state approaches that equilibrium and whether the transition assumptions remain sensible.

A common trap

Assuming that every initial state approaches a steady state just because one exists.

Instead: Check the dynamics: swapping and disconnected models are counterexamples.

02 / WORKED EXAMPLES

See the thinking

Example 1

Commuting choices

Rows are (0.7,0.2),(0.3,0.8). Find steady counts for 250 commuters.

Balance gives $0.3 a = 0.2 b$ , hence $a : b = 2 : 3$ .
Use $a + b = 250$ to obtain $a = 100, b = 150$ .

(100,150).

Example 2

Three-state balance

Find a steady state totalling 200 for the displayed matrix.

Columns=current, rows=next; order A,B,C.
Next / current	A	B	C
A	0.5	0.25	0
B	0.5	0.5	0.5
C	0	0.25	0.5

A balance gives $a = 0.5 a + 0.25 b$ , hence $a = 0.5 b$ .
C balance gives $c = 0.25 b + 0.5 c$ , hence $c = 0.5 b$ .
The total is $2 b = 200$ .

(50,100,50).

Check your understanding

How many fixed distributions total 100 under the identity matrix?

Show answer

Infinitely many.

Any nonnegative $(a, 100 - a)$ is unchanged.

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For rows (0.8,0.3),(0.2,0.7), find steady proportions.
3 marks
Use the same model for 350 members. Find steady counts.
3 marks
For rows (0.9,0.4),(0.1,0.6), find steady proportions.
3 marks
For the swap matrix, find steady proportions and explain the behaviour from (1,0).
3 marks
Find steady proportions for rows (0.5,0.25,0),(0.5,0.5,0.5),(0,0.25,0.5).
3 marks
A closed model has steady counts (80,120) out of 200.
4 marks
1. (a)
  State the steady proportions.
  1 mark
2. (b)
  Predict steady counts if the total doubles and the transition rule is unchanged.
  2 marks
3. (c)
  Give one reason observed long-run counts might differ.
  1 mark

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For rows (0.8,0.3),(0.2,0.7), find steady proportions.
3 marks
$0.2 a = 0.3 b$ and $a + b = 1$ .
(0.6,0.4).
Use the same model for 350 members. Find steady counts.
3 marks
Multiply (0.6,0.4) by 350.
(210,140).
For rows (0.9,0.4),(0.1,0.6), find steady proportions.
3 marks
$0.1 a = 0.4 b$ and $a + b = 1$ .
(0.8,0.2).
For the swap matrix, find steady proportions and explain the behaviour from (1,0).
3 marks
Balance gives $a = b$ .
Applying the swap repeatedly never approaches that balance from (1,0).
Steady (0.5,0.5); the specified start alternates (1,0),(0,1).
Find steady proportions for rows (0.5,0.25,0),(0.5,0.5,0.5),(0,0.25,0.5).
3 marks
The first and third equations give $a = c = 0.5 b$ .
Normalise using $2 b = 1$ .
(0.25,0.5,0.25).
A closed model has steady counts (80,120) out of 200.
4 marks
a) Divide by 200.
b) Multiply the proportions by 400.
c) The equilibrium prediction is conditional on its assumptions.
1. (a)
  State the steady proportions.
  1 mark
  Divide by 200.
  (0.4,0.6).
2. (b)
  Predict steady counts if the total doubles and the transition rule is unchanged.
  2 marks
  Multiply the proportions by 400.
  (160,240).
3. (c)
  Give one reason observed long-run counts might differ.
  1 mark
  The equilibrium prediction is conditional on its assumptions.
  The system may open or transition behaviour may change.
a) (0.4,0.6).; b) (160,240).; c) The system may open or transition behaviour may change.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics matrix-models option. This is the offered fifth-topic option, alongside the four compulsory topics; a school-designed Open Topic is an alternative, not prescribed content in this chapter.

Official curriculum source ↗

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