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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## This document was uploaded on 07/14/2011.

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online