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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

2×22×2 transition matrix for a Markov chain will have the form M=(a1b1ab),0<a<1,0<b<1M=(a1b1ab),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.

[4]
a.

Find the steady state probability vector for M in terms of aa and bb.

[5]
b.

Markscheme

|aλ1b1abλ|=0(aλ)(bλ)(1b)(1a)=0aλ1b1abλ=0(aλ)(bλ)(1b)(1a)=0        M1A1

λ2(a+b)λ+a+b1=0(λ1)(λ+(1ab))=0λ2(a+b)λ+a+b1=0(λ1)(λ+(1ab))=0         A1

λ=1orλ=a+b1λ=1orλ=a+b1         AGA1

[4 marks]

a.

(a1b1ab)(p1p)=(p1p)ap+1bp+bp=p(a1b1ab)(p1p)=(p1p)ap+1bp+bp=p       M1A1

1b=(2ab)pp=1b2ab1b=(2ab)pp=1b2ab         M1

So vector is (1b2ab1a2ab)1b2ab1a2ab         A1A1

[5 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4—Statistics and probability » AHL 4.19—Transition matrices – Markov chains
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Topic 4—Statistics and probability

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