****INSTANT DOWNLOAD***PDF***Solutions Manual for Stochastic Processes with R An Introduction 1st Edition By Olga Korosteleva

ALL 9 CHAPTER COVERED
c c c




SOLUTIONS MANUAL c

, TABLEOFCONTENTS
c c




CHAPTER 1 ……………………………………………………………………………………. 3
c c c




CHAPTER 2 ……………………………………………………………………………………. 31
c c c




CHAPTER 3 ……………………………………………………………………………………. 41
c c c




CHAPTER 4 ……………………………………………………………………………………. 48
c c c




CHAPTER 5 ……………………………………………………………………………………. 60
c c c




CHAPTER 6 ……………………………………………………………………………………. 67
c c c




CHAPTER 7 ……………………………………………………………………………………. 74
c c c




CHAPTER 8 ……………………………………………………………………………………. 81
c c c




CHAPTER 9 ……………………………………………………………………………………. 87
c c c




2

, CHAPTER 1 c




0.3 0.4 0.3 c c c c




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




0.8 0.1 0.1 c c c c



we compute: c




(a) 𝑃𝑃(𝑋𝑋3 =2|𝑋𝑋0 =1, 𝑋𝑋1 =2, 𝑋𝑋2 =3)=𝑃𝑃(𝑋𝑋3 =2|𝑋𝑋2 =3)
c
c
c c
c
c c
c
c c
c
c c c
c
c c c
c
c (by the Markov property)
c c c




= 𝑃𝑃32 = 0.1. c
c
c




(b)𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋0 =2, 𝑋𝑋3 =1)=𝑃𝑃(𝑋𝑋4 =3|𝑋𝑋3 =1)
c
c
c c
c
c c
c
c c c
c
c c c
c
c (by the Markov property)
c c c




= 𝑃𝑃13 = 0.3. c
c
c




(c) 𝑃𝑃(𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋2 = 3,𝑋𝑋3 = 1) = 𝑃𝑃(𝑋𝑋3 = 1 | 𝑋𝑋0 = 1,𝑋𝑋1 = 2,𝑋𝑋2 = 3)𝑃𝑃(𝑋𝑋2 = 3|𝑋𝑋0 = 1,
c
c
c c
c
c c
c
c c
c
c c c
c
c c c
c
c c
c
c c
c
c c
c
c c
c
c




𝑋𝑋1 = 2)𝑃𝑃(𝑋𝑋1 = 2|𝑋𝑋0 = 1)𝑃𝑃(𝑋𝑋0 = 1) (byconditioning)
c
c c
c
c c c
c
c c
c
c c c




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




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




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




0.3 0.4 0.3 c c c c 0.3 0.4 0.3 c c c c 0.41 0.27 0.32
𝐏𝐏(2) =�0.2 0.3 0.5��0.2 0.3 0.5� =� c

c c c c c c c c c c c c 0.52 0.22 0.26�.
Now we write c c 0.8 0.1 0.1 c c c c 0.8 0.1 0.1 c c c c 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,
c
c c
c
c c
c
c c
c
c c c
c
c c c
c
c c
c
c c
c
c c
c
c c
c
c




𝑋𝑋1 = 2)𝑃𝑃(𝑋𝑋1 = 2|𝑋𝑋0 = 1)𝑃𝑃(𝑋𝑋0 = 1) (byconditioning)
c
c c
c
c c c
c
c c
c
c c c




= 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) =
c c c c c c c c c c c c c c c c c c c c c


(2) (2)
c c c c c c c c

cc c c c c c




= 𝑃𝑃31 c
𝑃𝑃23 12 0

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




#specifying transition probability matrix c c c




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




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




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




t(tm)
c




#plotting diagram library(diagram) c c




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




box.prop=0.5, box.size=0.12, box.type="circle", cex.txt=0.8, lwd=1,
c c c c c




self.cex=0.3,self.shiftx=0.01, self.shifty=0.09)
c c c




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