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Unformatted text preview: 620-202 Tutorial/Computing Labo-ratory 2. Answers.1. (a) .(b) The likelihood isL(θ) =θnnYi=1xθ-1i.Hence the log-likelihood is‘(θ) =nlnθ+ (θ-1)nXi=1ln(xi),and the score function iss(θ) =nθ+nXi=1ln(xi),so that settings(θ) = 0 yieldsˆθ=-n∑ni=1ln(Xi)=-nln(Qni=1Xi)(c) HenceˆθX= 0.549,ˆθY= 2.210,ˆθZ= 0.959. To find the method of momentsestimators solve ¯x=θ/(θ+ 1) which yields˜θ= ¯x/(1-¯x) and hence˜θX=.598,˜θY= 2.400 and˜θZ= 0.865.2. Recall that for a random sampleX1,···,Xn,E(¯X) =1nnXi=1E(Xi) =E(X1)andVar(¯X) =1n2nXi=1Var(Xi) =Var(X1)n1(a) Now, theXiare i.i.d. exponential random variables with meanθ. HenceE(¯X) =E(Xi) =θand¯Xis unbiased.(b)σ2= Var(Xi) =θ2. Hence Var(¯X) =σ2/n=θ2/nas required. A goodestimator ofθisˆθ= ¯x= 3.48. (This is a good estimator as it is unbiasedand its variance tends to zero asn→ ∞.)3.S2=nXi=1(Xi-¯X)2n-1=1n-1nXi=1(X2i-2Xi¯X+¯X2)=1n-1nXi=1X2i-2n¯X2+n¯X2!...
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This note was uploaded on 10/05/2009 for the course STATS 620-202 taught by Professor R during the One '09 term at University of Melbourne.
- One '09