Question Solved1 Answer 3) Given a Markov chain with one-step transition matrix: 1 0.4 0.6 P = 1 0.2 0.8 And initial probability distribution, P(Xo-i) as follows: 1 P(X0=i) 2/3 1/3 Compute a. P(X4 01X2 1) (10 points) b. PCX's= 0 Xo-1, X1- 1 , X2- 0 , X3 - 0 , X4= 1) (10 points)

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Transcribed Image Text: 3) Given a Markov chain with one-step transition matrix: 1 0.4 0.6 P = 1 0.2 0.8 And initial probability distribution, P(Xo-i) as follows: 1 P(X0=i) 2/3 1/3 Compute a. P(X4 01X2 1) (10 points) b. PCX's= 0 Xo-1, X1- 1 , X2- 0 , X3 - 0 , X4= 1) (10 points)
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Transcribed Image Text: 3) Given a Markov chain with one-step transition matrix: 1 0.4 0.6 P = 1 0.2 0.8 And initial probability distribution, P(Xo-i) as follows: 1 P(X0=i) 2/3 1/3 Compute a. P(X4 01X2 1) (10 points) b. PCX's= 0 Xo-1, X1- 1 , X2- 0 , X3 - 0 , X4= 1) (10 points)
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{:[P={:[0,0,1],[1,0.4,0.6],[0.2,0.8]:}],[P(x_(0)=0)=1//3],[P(x_(0)=1)=2//3]:}a) P(x_(4)=0∣x_(2)=1)we have to calculate p_(10)^((2))for that we find {:p^(2))=P**p{:[{:[[[0.4,0.6],[0.2,0.8]][[0.4,0.6],[0.2,0.8]]=[[0.16+0.12,0.24+0.48],[0.08+0.16,0.12+0.64]]],[=[[0.28,0.72],[0.24,0.76]]],[P_(10)^((2))=0.24[(i,1.0 ... See the full answer