hw5sol - 36-226 Summer 2010Homework 5Solutions1.

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Unformatted text preview: 36-226 Summer 2010Homework 5Solutions1. (a)XiN(i,1)(independent)L(1,...,n) =nYi=112e-12(Xi-i)2= (2)-n2e-12ni=1(Xi-i)2logL(1,...,n) =-n2log 2-12nXi=1(Xi-i)2(b) To find the MLE for each, differentiate with respect to eachiseparately because theXiare independent, soidoes not interact withjfori6=jin the likelihood function:logL(1,...,n)i=-(Xi-i)set= 0Xi=i(critical point)2logL(1,...,n)2i=-1maxbMLEi=XiTherefore, the MLE forisbMLE= (X1,...,Xn).(c) To find the MSE forbMLEi, we start from the definitionMSE(bMLEi) =E[(bMLEi-i)2] =E[(Xi-i)2] =V ar[Xi] = 1.Now summing these overigivesni=11 =n.(d) To find the MSE of the new estimator, we consider this case by case. SupposeiS.Then,MSE(bSi) =E[(bSi-i)2] =E[(Xi-i)2] =V ar[Xi] = 1just like for the MLE. Now supposei6S. Then,MSE(bSi) =E[(bSi-i)2] =E[(0-i)2] =E[2i] =2i....
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hw5sol - 36-226 Summer 2010Homework 5Solutions1.

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