# solution5 - 1 C + (1 b 1 ) X = b 1 ( C X ) + X . V ar ( X (...

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Homework 5 solution 1 Problem 1 For this problem, what we need to remember is how to update the weight for each chain we generated. The formula is w i, 0 = 1 ,w i,T = g ( Y T | μ T ) w i,T - 1 , where i denotes the i-th chain or sample, T = 1 , 2 ,..., 10 denotes the T-th position in the chain, and g ( Y T | μ T ) = dnorm ( Y T T , 1). We need to estimate E ( μ 1 | Y 1 ) ,E ( μ 2 | Y 1 ,Y 2 ) ,...,E ( μ 10 | Y 1 ,Y 2 ,...,Y 10 ). R code: mus<-matrix(0,10,50) mu<-numeric(10) y=c(0.2,-0.1,-0.4,-1.3,1.0,-1.2,1.1,-0.1,-2.4,0.9) w<-rep(1,50) for(t in 1:10) { for(m in 1:50) { u<-runif(1) if(u<0.1) { mus[t,m]<-rnorm(1) } else 1

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{ if(t>1) { mus[t,m]=mus[t-1,m] } else { mus[t,m]=0 } } w[m]=w[m]*dnorm(y[t],mus[t,m],1) } mu[t]<-sum(mus[t,]*w)/sum(w) } mu In my run, the estimates are μ 1 μ 2 μ 3 μ 4 μ 5 0.021986384 0.036023218 -0.005661837 -0.056319977 0.054143467 μ 6 μ 7 μ 8 μ 9 μ 10 -0.020782239 0.057099946 -0.027044208 -1.163091381 -0.233858758. 2
2 Problem 2 Let b 1 = 1 b , then X ( b ) = X ( b 1 ) = b
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Unformatted text preview: 1 C + (1 b 1 ) X = b 1 ( C X ) + X . V ar ( X ( b 1 )) = b 2 1 V ar ( C X ) + 2 b 1 Cov ( C X,X ) + V ar ( X ) = V ar ( C X )( b 1 + Cov ( C X,X ) V ar ( C X ) ) 2 + V ar ( X ) Cov ( C X,X ) 2 V ar ( C X ) = V ar ( C X )( b 1 + Cov ( X,C ) V ar ( X ) V ar ( C X ) ) 2 + V ar ( X ) ( Cov ( X,C ) V ar ( X )) 2 V ar ( C X ) Because V ar ( X ) = Cov ( X,C ), we have Cov ( X,C X ) = Cov ( X,C ) V ar ( X ) = 0, which implies V ar ( C X ) = 0. Therefore, if we choose b 1 = Cov ( X,C )-V ar ( X ) V ar ( C-X ) , then V ar ( X ( b )) = V ar ( X ( b 1 )) = V ar ( X ) ( Cov ( X,C )-V ar ( X )) 2 V ar ( C-X ) &lt; V ar ( X ). 3...
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## This note was uploaded on 10/24/2010 for the course STAT 428 taught by Professor Chen during the Spring '08 term at University of Illinois, Urbana Champaign.

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solution5 - 1 C + (1 b 1 ) X = b 1 ( C X ) + X . V ar ( X (...

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