Date | May Example question | Marks available | 5 | Reference code | EXM.1.AHL.TZ0.18 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Find | Question number | 18 | Adapted from | N/A |
Question
A 2×22×2 transition matrix for a Markov chain will have the form M=(a1−b1−ab),0<a<1,0<b<1M=(a1−b1−ab),0<a<1,0<b<1.
Show that λ=1λ=1 is always an eigenvalue for M and find the other eigenvalue in terms of aa and bb.
Find the steady state probability vector for M in terms of aa and bb.
Markscheme
|a−λ1−b1−ab−λ|=0⇒(a−λ)(b−λ)−(1−b)(1−a)=0∣∣∣a−λ1−b1−ab−λ∣∣∣=0⇒(a−λ)(b−λ)−(1−b)(1−a)=0 M1A1
⇒λ2−(a+b)λ+a+b−1=0⇒(λ−1)(λ+(1−a−b))=0⇒λ2−(a+b)λ+a+b−1=0⇒(λ−1)(λ+(1−a−b))=0 A1
⇒λ=1orλ=a+b−1⇒λ=1orλ=a+b−1 AGA1
[4 marks]
(a1−b1−ab)(p1−p)=(p1−p)⇒ap+1−b−p+bp=p(a1−b1−ab)(p1−p)=(p1−p)⇒ap+1−b−p+bp=p M1A1
⇒1−b=(2−a−b)p⇒p=1−b2−a−b⇒1−b=(2−a−b)p⇒p=1−b2−a−b M1
So vector is (1−b2−a−b1−a2−a−b)⎛⎝1−b2−a−b1−a2−a−b⎞⎠ A1A1
[5 marks]