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hwch8 - MC convergence MH Ch 8 Not to turn in Hw Ch 8...

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MC convergence MH. Ch 8. Not to turn in. Hw Ch 8. Convergence Explain clearly what you do and why you are making the (convergence) decision. Ex 1. According to exercise 1 in MH. Hw 6 p ( μ | α ) = 1 B ( a μ , a μ ) α a μ μ a μ - 1 (1 + αμ ) a μ + a μ , and the integrated likelihood function is given by p ( k | α ) = integraldisplay 0 p ( k | α, μ ) p ( μ | α ) (1) = bracketleftBigg productdisplay j 1 B ( k j + 1 , α - 1 )( k j + α - 1 ) bracketrightBigg B ( a μ + j k j , ( J + a μ ) ) B ( a μ , a μ ) , where k denotes the sample k = ( k 1 , . . . , k J ) NegBin( μ, α ), and p ( k | α, μ ) = producttext j p ( k j | α, μ ) is the likelihood function. Assume that α exponential( mean = 0 . 5) and that you observed k = c (16 , 24 , 4 , 0 , 13 , 2 , 6 , 5 , 2 , 1 , 34 , 1 , 4 , 12 , 2 , 18 , 8 , 15 , 12 , 18) Considering q ( y | x ) = exponential( y | mean = 0 . 838). 1. We ran three chains with corresponding starting values α (1) 1 = 0 . 1 , α (1) 2 = 1 , α (1) 3 = 2 of length T = 10 4 . Figure 1 shows the output of indices=100:T acf(alphas[indices,1],main="") acf(alphas[indices,2],main="") acf(alphas[indices,3],main="") What does this figure tell you?, If you wanted a quasi-independent chain, what would be the batch size that you would use?, why? 2. Using alphas[indices=100:T,], I compute the KS statistic comparing the first half of the chain vs its second half and the KS comparing chains 1 and 2. Figure 2 1 of 8

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MC convergence MH. Ch 8. Not to turn in. 0 10 20 30 40 0.0 0.2 0.4 0.6 0.8 1.0 Lag ACF 0 10 20 30 40 0.0 0.2 0.4 0.6 0.8 1.0 Lag ACF 0 10 20 30 40 0.0 0.2 0.4 0.6 0.8 1.0 Lag ACF Figure 1: acf(alphas[indices,1],main=””) for each one of the chains 0 2000 4000 6000 8000 10000 0.0 0.2 0.4 0.6
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