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# hw5sol - 36-226 Summer 2010Homework...

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Unformatted text preview: 36-226 Summer 2010Homework 5Solutions1. (a)Xi∼N(μi,1)(independent)L(μ1,...,μn) =nYi=11√2πe-12(Xi-μi)2= (2π)-n2e-12∑ni=1(Xi-μi)2logL(μ1,...,μn) =-n2log 2π-12nXi=1(Xi-μi)2(b) To find the MLE for eachμ, differentiate with respect to eachμiseparately because theXiare independent, soidoes not interact withjfori6=jin the likelihood function:∂logL(μ1,...,μn)∂μi=-(Xi-μi)set= 0⇒Xi=μi(critical point)∂2logL(μ1,...,μn)∂μ2i=-1⇒max⇒bμMLEi=XiTherefore, the MLE forμisbμMLE= (X1,...,Xn).(c) To find the MSE forbμMLEi, we start from the definitionMSE(bμMLEi) =E[(bμMLEi-μi)2] =E[(Xi-μi)2] =V ar[Xi] = 1.Now summing these overigives∑ni=11 =n.(d) To find the MSE of the new estimator, we consider this case by case. Supposei∈S.Then,MSE(bμSi) =E[(bμSi-μi)2] =E[(Xi-μi)2] =V ar[Xi] = 1just like for the MLE. Now supposei6∈S. Then,MSE(bμSi) =E[(bμSi-μi)2] =E[(0-μi)2] =E[μ2i] =μ2i....
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hw5sol - 36-226 Summer 2010Homework...

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