Solution Manual for Stochastic Processes with R: An Introduction, 2nd Edition by Olga Korosteleva — Complete Verified Solutions

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edition-by-korosteleva-2024-all-9-chapters-covered
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ALL 9 CHAPTER COVERED
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SOLUTIONS MANUAL dg

, 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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9

, CHAPTER 1 dg




0.3 0.4 0.3 dg d g dg d g




EXERCISE 1.1. Fora Markov chain witha one-step transitionprobabilitymatrix�0.2 0.3 0.5� dg d g dg dg dg dg dg dg dg dg dg dg gd dg d g dg d g gd




0.8 0.1 0.1
we compute:
dg d g dg d g



dg




(a) 𝑃𝑃(𝑋𝑋3 =2 |𝑋𝑋0 =1, 𝑋𝑋1 =2, 𝑋𝑋2 =3)=𝑃𝑃(𝑋𝑋3 =2|𝑋𝑋2 =3)
dg
dg
gd (by the Markov property)
gd
dg
gd dg
dg
gd dg
dg
gd gd dg
dg
gd gd gd
dg
gd d g dg d g




= 𝑃𝑃32 = 0.1. dg
d g
dg




(b) 𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋0 =2, 𝑋𝑋3 =1)=𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋3 =1)
dg
dg
(by the Markov property)
gd gd
dg
gd dg
dg
gd gd gd
dg
gd gd gd
dg
gd d g dg dg




= 𝑃𝑃13 = 0.3. dg
d g
dg




(c) 𝑃𝑃(𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋2 = 3, 𝑋𝑋3 = 1) = 𝑃𝑃(𝑋𝑋3 = 1 | 𝑋𝑋0 = 1,𝑋𝑋1 = 2, 𝑋𝑋2 = 3)𝑃𝑃(𝑋𝑋2 = 3 |𝑋𝑋0 = 1,
d g
d g
dg gd
d g
dg dg
d g
dg dg
d g
dg dg dg
d g
dg dg dg
d g
dg gd
d g
dg dg
d g
d g dg
dg
dg dg
d g
dg




𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by conditioning)
dg
dg dg
d g
dg dg dg
d g
dg dg
d g
dg d g dg




= 𝑃𝑃(𝑋𝑋3 = 1 | 𝑋𝑋2 = 3) 𝑃𝑃(𝑋𝑋2 = 3 | 𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by the Markov property)
d g
d g
d g dg dg
d g
d g dg
d g
d g dg dg
d g
d g dg
d g
d g dg dg
d g
d g dg
dg d g
dg dg dg dg dg




=𝑃𝑃31 𝑃𝑃23 𝑃𝑃12 𝑃𝑃(𝑋𝑋0 =1)=(0.8)(0.5)(0.4)(1)=0.16.
gd
dg dg dg dg
dg dg gd gd gd




(d) We first compute the two-step transition probability matrix. We obtain
d g dg dg dg dg dg dg dg dg dg




0.3
0.3 0.4 dg d g 0.4 dg d g 0.3 dg d g dg 0.41 0.27 0.32
0.3 d g




(2)
𝐏𝐏 dg

= �0.2 0.3 0.5��0.2 0.3 0.5� =�
dg gd d g d g gd gd gd d g d g gd d g dg 0.52 0.22 0.26�.
Now we write dg dg 0.8 dg d g 0.1 dg 0.8 dg d g 0.1 dg 0.34 0.36 0.30

d g 0.1 d g 0.1
𝑃𝑃(𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋3 = 3,𝑋𝑋5 = 1) = 𝑃𝑃(𝑋𝑋5 = 1 | 𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋3 = 3) 𝑃𝑃(𝑋𝑋3 = 3 |𝑋𝑋0 = 1,
dg
dg gd
d g
dg gd
d g
dg gd
d g
dg dg dg
d g
dg dg dg
d g
dg gd
d g
dg gd
d g
d g dg
d g
dg dg
d g
dg




𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by conditioning)
dg
dg dg
d g
dg dg dg
d g
dg dg
d g
dg d g dg




= 𝑃𝑃(𝑋𝑋5 = 1 | 𝑋𝑋3 = 3) 𝑃𝑃(𝑋𝑋3 = 3 | 𝑋𝑋1 = 2) 𝑃𝑃(𝑋𝑋1 = 2 | 𝑋𝑋0 = 1) 𝑃𝑃(𝑋𝑋0 = 1) (by the Markov property)
𝑃𝑃(𝑋𝑋 = 1) = (0.34)(0.26)(0.4)(1) = 0.03536.
d g d g dg dg d g dg d g dg dg d g dg d g dg dg d g dg dg dg dg dg dg


(2) (2)
d g d g d g d g d g d g dg d g

dg dg dg dg dg dg



𝑃𝑃
dg




= 𝑃𝑃31 dg
𝑃𝑃23 12 0


EXERCISE 1.2. dg dg d g (a) We plot a diagram of the Markov chain.
dg dg dg dg dg dg dg dg




#specifying transition probability matrix dg dg dg




tm<- matrix(c(1, 0, 0, 0, 0, 0.5, 0, 0, 0, 0.5, 0.2, 0, 0, 0, 0.8,
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg




0, 0, 1, 0, 0, 0, 0, 0, 1, 0), nrow=5, ncol=5, byrow=TRUE)
dg dg dg dg dg dg dg dg dg dg dg dg




#transposing transition probability matrix tm.tr<- dg dg dg dg




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




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




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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g dg dg




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