SME
Linear Algebra · Problem type
The steady state of a Markov chain
Use this for “Each year 10% of city dwellers move to the suburbs and 30% of suburb dwellers move to the city. What share lives in each in the long run?”
Diagram
A is 2 × 2 and B is 2 × 1, so AB is 2 × 1.
Row 1 of A times column 1 of B gives entry (1, 1): c₁₁ = 9/10 × 3/4 + 3/10 × 1/4 = 3/4.
Tap an entry of AB to see its row and column.
This is an example. Type your own values to start a new problem.
P₁₁
p
q
P₂₂
s₁
s₂
=
s₁
s₂
λ₂
Other eigenvalue of P
worked out (not typed)
Formulas
P₁₁
=
1
−
p
0.9 = 1 − 0.1
P₂₂
=
1
−
q
0.7 = 1 − 0.3
s₁
=
q
p
+
q
0.75
=
0.3
0.1 + 0.3
s₂
=
p
p
+
q
0.25
=
0.1
0.1 + 0.3
λ₂
=
1
−
p
−
q
0.6 = 1 − 0.1 − 0.3
Assumptions
•
Column j of P holds the chances of going from state j to each state in one step, so each column adds to 1.
•
The steady state s solves Ps = s: it is an eigenvector of P for λ = 1, with s₁ + s₂ = 1.
•
The other eigenvalue of P is 1 − p − q; when it is between −1 and 1, every start tends to s.
•
The chances and shares are decimals: 10% is 0.1. State 1 is the city and state 2 the suburbs in the example.
Step-by-step
WE KNOW
p = 0.1, q = 0.3
FIND
chance of staying in state 1 (P₁₁), chance of staying in state 2 (P₂₂), long-run share in state 1 (s₁), long-run share in state 2 (s₂), other eigenvalue of P (λ₂)
Step 1 · Find chance of staying in state 1 (P₁₁)
Use P₁₁ = 1 − p
Column 1 adds to 1: whoever doesn’t leave state 1 stays.
P₁₁
=
1
−
0.1
P₁₁
=
0.9
Step 2 · Find chance of staying in state 2 (P₂₂)
Use P₂₂ = 1 − q
Column 2 adds to 1 the same way.
P₂₂
=
1
−
0.3
P₂₂
=
0.7
Step 3 · Find long-run share in state 1 (s₁)
Use s₁ = q ÷ (p + q)
Row
1
of
Ps
=
s
gives
(1
−
p
)
s₁
+
qs₂
=
s₁
,
so
ps₁
=
qs₂.
With
s₁
+
s₂
=
1,
s₁
=
q
p
+
q
.
s₁
=
0.3
0.1 + 0.3
s₁
=
0.3
0.4
s₁
=
0.75
Step 4 · Find long-run share in state 2 (s₂)
Use s₂ = p ÷ (p + q)
The
rest
of
the
shares:
s₂
=
1
−
s₁
=
p
p
+
q
.
s₂
=
0.1
0.1 + 0.3
Check: [[0.9, 0.3], [0.1, 0.7]] × ⟨0.75, 0.25⟩ = ⟨0.75, 0.25⟩
s₂
=
0.25
Step 5 · Find other eigenvalue of P (λ₂)
Use λ₂ = 1 − p − q
The
trace
of
P
is
λ₁
+
λ₂
=
(1
−
p
)
+
(1
−
q
)
with
λ₁
=
1.
Each
step
shrinks
the
gap
from
s
by
the
factor
λ₂
.
λ₂
=
1
−
0.1
−
0.3
λ₂
=
0.6
Check
Put the numbers back in. Both sides should match.
0.9 = 1 − 0.1 ✓
0.7 = 1 − 0.3 ✓
0.75
=
0.3
0.1 + 0.3
✓
0.25
=
0.1
0.1 + 0.3
✓
0.6 = 1 − 0.1 − 0.3 ✓
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Powers of a matrix by its eigenvectors
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Complex eigenvalues: a turn and a stretch
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