# hw7sol - 36-226 Summer 2010Homework 7Solutions1(a)FY(y...

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Unformatted text preview: 36-226 Summer 2010Homework 7Solutions1.(a)FY(y) =P(Y < y) =P(FX(X)< y) =P(X < F-1X(y))=FX(F-1X(y)) =yBut this is the CDF of a U(0,1) random variable, soY∼U(0,1).(b) We can use the CDF ofX,FX(x|θ) = 1-e-x/θas a pivotal quantity. Thus,Q= 1-e-X/θ∼U(0,1).To find a 100(1-α)% CIα=P(1-α/2<1-e-X/θ< α/2)=P(1-α/2< e-X/θ< α/2)=P(log(1-α/2)<-X/θ <log(α/2))=P-Xlog(1-α/2)< θ <-Xlogα/2.So a 100(1-α)% CI is-Xlog(1-α/2),-Xlogα/2.2.(a) We said in class that ifX1,..,Xnare iid and∼exp(θ), then2θnXi=1Xi∼χ2(2n)Using the hint, in this case we have:-lnX1,-lnX2,...,-lnXniid∼exp(1θ)⇒T=21θnXi=1(-lnXi) = 2θnXi=1(-lnXi)∼χ2(2n)This makes T a pivotal quantity forθin theBeta(θ,1) distribution. Using this pivot,we can find the confidence interval by setting the quantile equation and solving forθ.⇒1-α=P"χ21-α/2(2n)≤ -2θnXi=1lnXi≤χ2α/2(2n)#=P"χ21-α/2(2n)-2∑ni=1lnXi≤θ≤χ2α/2(2n)-2∑ni=1lnXi#So a 100(1-α)% C.I. forθis:"χ21-α/2(2n)-2∑ni=1lnXi,χ2α/2(2n)-2∑ni=1lnXi#1(b) I found that-2∑8i=1logXi= 10.4,χ2.975(2n) = 1.72, andχ2.025(2n) = 7.21. Pluggingin gives me a 95% CI forθ∈[0.16,.69].(c) The approximate CI using the MLE involvesbθMLE=-n∑ni=1logXi(shown in class) andthe CRLBθ2/n. The form of the approximate CI is thenbθMLE±zα/2qCRLB|θ=bθMLE=-n∑ni=1logXi±zα/2n√n∑ni=1logXi= 1.54±2×.54So the approximate 95% CI is [0.45,2.63].(d) Clearly the first is much narrower. It is also much more appropriate sincen= 8 is notvery big. The approximate CI is for large samples.3.(a) We discussed the pivot for this interval in class.nXi=1Xi∼Ga(2n,θ)⇒2θnXi=1Xi|{z}∼χ24npivotal forθ⇒1-α=P"χ24n,1-α/2≤2θnXi=1Xi≤χ24n,α/2#Thus, a (1-α) confidence interval forθis"2∑ni=1Xiχ24n,α/2,2∑ni=1Xiχ24n,1-α/2#(b) From the data, we can compute the following quantities:n= 10,∑...
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hw7sol - 36-226 Summer 2010Homework 7Solutions1(a)FY(y...

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