Solution Manual for Stochastic Processes with R: An Introduction (1st Edition by Olga Korosteleva) – Complete Step-by-Step Solutions

ALL9 CHAPTER COVERED
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SOLUTIONS MANUALl

, TABLEOFCONTENTS
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CHAPTER 1 ……………………………………………………………………………………. 3
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CHAPTER 2 ……………………………………………………………………………………. 31
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CHAPTER 3 ……………………………………………………………………………………. 41
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CHAPTER 4 ……………………………………………………………………………………. 48
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CHAPTER 5 ……………………………………………………………………………………. 60
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CHAPTER 6 ……………………………………………………………………………………. 67
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CHAPTER 7 ……………………………………………………………………………………. 74
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CHAPTER 8 ……………………………………………………………………………………. 81
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CHAPTER 9 ……………………………………………………………………………………. 87
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2

, CHAPTER 1 l




0.3 0.4 0.3 l l l l




EXERCISE1.1. ForaMarkovchainwithaone-steptransitionprobabilitymatrix�0.2 0.3 0.5� l l l l l l l l l l l l l l l l l l




0.8 0.1 0.1 l l l l



wecompute: l




(a) 𝑃𝑃(𝑋𝑋3 =2|𝑋𝑋0 =1,𝑋𝑋1 =2, 𝑋𝑋2 =3)=𝑃𝑃(𝑋𝑋3 =2|𝑋𝑋2 =3)
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l (by the Markov property)
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= 𝑃𝑃32 = 0.1. l
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(b)𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋0 =2, 𝑋𝑋3 =1)=𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋3 =1)
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l (by the Markov property)l l l




= 𝑃𝑃13 = 0.3. l
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(c) 𝑃𝑃(𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋2 = 3,𝑋𝑋3 = 1) = 𝑃𝑃(𝑋𝑋3 = 1 | 𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋2 = 3)𝑃𝑃(𝑋𝑋2 = 3|𝑋𝑋0 = 1,
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𝑋𝑋1 = 2)𝑃𝑃(𝑋𝑋1 = 2|𝑋𝑋0 = 1)𝑃𝑃(𝑋𝑋0 = 1) (byconditioning)
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= 𝑃𝑃(𝑋𝑋3 = 1 | 𝑋𝑋2 = 3) 𝑃𝑃(𝑋𝑋2 = 3 | 𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by the Markov property)
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=𝑃𝑃31 𝑃𝑃23 𝑃𝑃12 𝑃𝑃(𝑋𝑋0 =1)=(0.8)(0.5)(0.4)(1)=0.16.
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(d) We first compute the two-step transition probability matrix. We obtain
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0.3 0.4 0.3 l l l l 0.3 0.4 0.3 l l l l 0.41 0.27 0.32
𝐏𝐏(2) =�0.2 0.3 0.5��0.2 0.3 0.5� =� l

l l l l l l l l l l l l 0.52 0.22 0.26�.
Nowwewrite l l 0.8 0.1 0.1 l l l l 0.8 0.1 0.1 l l l l 0.34 0.36 0.30
𝑃𝑃(𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋3 = 3,𝑋𝑋5 = 1) = 𝑃𝑃(𝑋𝑋5 = 1|𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋3 = 3)𝑃𝑃(𝑋𝑋3 = 3|𝑋𝑋0 = 1,
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𝑋𝑋1 = 2)𝑃𝑃(𝑋𝑋1 = 2|𝑋𝑋0 = 1)𝑃𝑃(𝑋𝑋0 = 1) (byconditioning)
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= 𝑃𝑃(𝑋𝑋5 = 1 | 𝑋𝑋3 = 3) 𝑃𝑃(𝑋𝑋3 = 3 | 𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by the Markov property)
(0.34)(0.26)(0.4)(1) = 0.03536.
𝑃𝑃 𝑃𝑃(𝑋𝑋 = 1) =
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(2) (2)
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ll l l l l l




= 𝑃𝑃31 l
𝑃𝑃23 12 0

EXERCISE 1.2. (a) We plot a diagram of the Markov chain. l l l l l l l l l l l




#specifying transition probability matrix l l l




tm<- matrix(c(1, 0, 0, 0, 0, 0.5, 0, 0, 0, 0.5, 0.2, 0, 0, 0, 0.8,
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0, 0, 1, 0, 0, 0, 0, 0, 1, 0), nrow=5, ncol=5, byrow=TRUE)
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#transposing transition probability matrix tm.tr<- l l l l




t(tm)
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#plotting diagram library(diagram) l l




plotmat(tm.tr, arr.length=0.25, arr.width=0.1, box.col="light blue", box.lwd=1, l l l l l




box.prop=0.5, box.size=0.12, box.type="circle", cex.txt=0.8, lwd=1,
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self.cex=0.3,self.shiftx=0.01, self.shifty=0.09)
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