hw4sol - 36-226 Summer 2010Homework 4Solutions1(a Note that...

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Unformatted text preview: 36-226 Summer 2010Homework 4Solutions1. (a) Note that this is aBeta(1,θ) distribution.⇒E[X] =11 +θ.We therefore solve:[E[X] =X=⇒11 +bθ=X=⇒bθ=1-XX(b) Compute the mle using calculus as done before.L(θ) =θn"nYi=1(1-xi)#θ-1lnL(θ) =nlnθ+ (θ-1)nXi=1ln(1-xi)∂lnL(θ)∂θ=nθ+nXi=1ln(1-xi) = 0⇒θ=-n∑ni=1ln(1-xi)is a critical point∂2lnL(θ)∂2θ=-nθ2<⇒bθ=-n∑ni=1ln(1-Xi)is the MLE(c) Based on the output,x= 0.11. The estimate forθis thereforebθ=1-.11.11≈8.47.Code for 1(c)---------------------(1-mean(defects))/mean(defects)(d) Using the data,nXi=1ln(1-xi) =-21.25⇒bθ=-100-21.25≈4.7Code for 1(d)----------------------1/mean(log(1-defects))(e) This probability can be computed asP(X >.025) =Z1.025θ(1-x)θ-1=-(1-x)θ1.025= (1-.025)θNow, just replaceθbybθto get an estimate of the probability.Usingbθ= 8.47,\P(X >.025)≈.81.Usingbθ= 107.74,\P(X >.025)≈.89.12. (a)E[bθ1] =E[4X3] =43E[X] =43E[X] = [E[X] is given] =43·34θ=θV ar[bθ1] =V ar[43X] =169V ar[X] =169nV ar[X] = [V ar[X] is given] =169n380θ2=θ215n(b) To findE[bθ2] =E[X(n)], we first need to findfX(n)(x).FX(x) =Zx3t2θ3dt=x3θ3fX(n)(x) =nx3θ3n-13x2θ3=3nθ3nx3n-1,< x < θTherefore,E[X(n)] =Zθx3nθ3nx3n-1dx=3nθ3nx3n+13n+ 1θ=3n3n+ 1θ⇒bias(bθ2) =E[X(n)]-θ=3n3n+ 1θ-θ=-13n+ 1θ(c)MSE(bθ1) =Bias2(bθ1) +V ar[bθ1] = 0 +θ215nby part (a)MSE(bθ2) =Bias2(bθ2) +V ar[bθ2=3n(3n+ 2)(3n+ 1)2θ2+1(3n+ 1)2θ2by part (b)=2(3n+ 2)(3n+ 1)θ2after some algebraHence,MSE(bθ2)< MSE(bθ1) =V ar[bθ1] forn >2....
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hw4sol - 36-226 Summer 2010Homework 4Solutions1(a Note that...

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