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

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Unformatted text preview: 36-226 Summer 2010Homework 4Solutions1. (a) Note that this is aBeta(1,) distribution.E[X] =11 +.We therefore solve:[E[X] =X=11 +b=X=b=1-XX(b) Compute the mle using calculus as done before.L() =n"nYi=1(1-xi)#-1lnL() =nln+ (-1)nXi=1ln(1-xi)lnL()=n+nXi=1ln(1-xi) = 0=-nni=1ln(1-xi)is a critical point2lnL()2=-n2<b=-nni=1ln(1-Xi)is the MLE(c) Based on the output,x= 0.11. The estimate foris thereforeb=1-.11.118.47.Code for 1(c)---------------------(1-mean(defects))/mean(defects)(d) Using the data,nXi=1ln(1-xi) =-21.25b=-100-21.254.7Code for 1(d)----------------------1/mean(log(1-defects))(e) This probability can be computed asP(X >.025) =Z1.025(1-x)-1=-(1-x)1.025= (1-.025)Now, just replacebybto get an estimate of the probability.Usingb= 8.47,\P(X >.025).81.Usingb= 107.74,\P(X >.025).89.12. (a)E[b1] =E[4X3] =43E[X] =43E[X] = [E[X] is given] =4334=V ar[b1] =V ar[43X] =169V ar[X] =169nV ar[X] = [V ar[X] is given] =169n3802=215n(b) To findE[b2] =E[X(n)], we first need to findfX(n)(x).FX(x) =Zx3t23dt=x33fX(n)(x) =nx33n-13x23=3n3nx3n-1,< x < Therefore,E[X(n)] =Zx3n3nx3n-1dx=3n3nx3n+13n+ 1=3n3n+ 1bias(b2) =E[X(n)]-=3n3n+ 1-=-13n+ 1(c)MSE(b1) =Bias2(b1) +V ar[b1] = 0 +215nby part (a)MSE(b2) =Bias2(b2) +V ar[b2=3n(3n+ 2)(3n+ 1)22+1(3n+ 1)22by part (b)=2(3n+ 2)(3n+ 1)2after some algebraHence,MSE(b2)< MSE(b1) =V ar[b1] forn >2....
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hw4sol - 36-226 Summer 2010Homework 4Solutions1. (a) Note...

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