f X (x i ;θ) = n ( n logx i = 0 = θml = n/ n logx i 1 θ +1 n n 2 < 0 for all θ (θ +1) n logx i 1 ESTIMATOR: = logx i θ n for all θ θ 2 < 0 2θ 2 x 3

Transcription

1 MATH EXERCISES SOLUTIONS 1 The lkelhoo the orgal parameterzato s ( ( 1 L (θ 1,θ x 1,x = θ x 1 1 x (1 θ 1 1 x 1 θ x 1 x (1 θ x If φ = θ /(1 θ, the θ = φ/(1+φ θ 1 = (φψ/(1+φψ. Ether by wrtg out the lkelhoo full the ew parameterzato, a maxmzg the usual way, or by usg varace propertes of maxmum lkelhoo estmates, we coclue that ψ(x 1,x = θ 1 /(1 θ 1 θ /(1 θ = (x 1/ 1 /(1 x 1 / 1 (x / /(1 x / = x 1/( 1 x 1 x /( x X 1,...,X Gamma(α,β so that X [X ;α,β] = α/β a Var X [X ;α,β] = α/β so that = α(α+1 X [ X ;α,β ] = Var X [X ;α,β]+{ X [X ;α,β]} = α β + ( α β β Hece for the metho of momets estmators α MM a β MM, ee to solve the followg: FIRST MOMENT SECOND MOMENT Solve Solve 1 1 x = x = α β x = (x +S = α(α+1 β wheres = 1 (x x. Elemetary algebra gves α MM = (x S a β MM = x S. 3 Wrtg L (θ for L(x;θ throughout: ( Forθ > STEP L (θ = f X (x ;θ = θx θ 1 log L (θ = logθ+(θ 1 logx θ {log L (θ} = θ + ( θ 1 = θ x logx = = θml = / logx Hece θ {log L (θ} = θ < ESTIMATE : θ ML = logx for all θ ESTIMATOR: logx MATH 557 EXERCISES SOLUTIONS Page 1 of 9

2 ( Forθ > STEP L (θ = f X (x ;θ = (θ+1x (θ+ log L (θ = log(θ+1 (θ + logx θ {log L (θ} = θ +1 θ {log L (θ} = < for all θ (θ +1 ( (θ+ = (θ+1 x logx = = θml = / logx 1 Hece ESTIMATE : θ ML = 1 ESTIMATOR: = logx 1 logx ( Forθ > STEP L (θ = f X (x ;θ = ( θ x exp{ θx } = θ log L (θ = logθ+ logx θ θ {log L (θ} = θ x x = = θml = x { x exp θ } x Hece θ {log L (θ} = θ < ESTIMATE : θ ML = x for all θ ESTIMATOR: = X (v Because of the costrat the pf that x θ L (θ ( θ x 3 = θ x 3 θ x 1,...,x f X (x ;θ == otherwse STEP log L (θ = log+logθ 3 logx At ths pot we ote that the lkelhoo s mootocally creasg θ, a hece the lkelhoo s maxmze wheθ s as large as possble but so that the costratθ x 1,...,x s stll satsfe, hece ESTIMATE : θ ML = m{x 1,...,x } ESTIMATOR: m{x 1,...,X } MATH 557 EXERCISES SOLUTIONS Page of 9

3 (v Forθ > STEP L (θ = f X (x ;θ = log L (θ = log+logθ θ x θ {log L (θ} = θ ( { } θ exp{ θ x } = θ exp θ x x = = θml = x Hece θ {log L (θ} = θ < ESTIMATE : θ ML = x for all θ (v Because of the costrat the pf that θ 1 x θ L (θ 1,θ ESTIMATOR : x 1 f X (x ;θ = (θ θ 1 = 1 (θ θ 1 θ 1 x 1,...,x θ otherwse STEP log L (θ 1,θ = log(θ θ 1 At ths pot we ote that the lkelhoo s mootocally creasg θ 1 a mootocally ecreasg θ, a hece the lkelhoo s maxmze wheθ 1 s as large as possble a whe θ s as small as possble, but so that the costrat θ 1 x 1,...,x θ (a θ θ 1 s stll satsfe, hece ESTIMATE : θ1ml = m{x 1,...,x } θml = max{x 1,...,x } ESTIMATOR: θ 1 m{x 1,...,X } θ max{x 1,...,X } (v Notg the costrat the pf thatx θ, we have L (θ 1,θ = f X (x ;θ = θ 1 θ θ 1 x (θ 1+1 = θ 1θ θ 1 ( (θ1 +1 x STEP x 1,...,x θ log L (θ 1,θ = logθ 1 +θ 1 logθ (θ 1 +1 logx {log L (θ 1,θ } = +logθ θ 1 θ 1 = θ 1ML = θ {log L (θ 1,θ } = θ 1 θ logx = logx log θ ML MATH 557 EXERCISES SOLUTIONS Page 3 of 9

4 The seco of the partal ervatve equatos cates aga that the maxmum of the lkelhoo occurs wheθ s as large as possble, that s, whe θ ML = m{x 1,...,x }.Hece ESTIMATES θ1ml = [ ] logx log{m{x 1,...,x }} θml = m{x 1,...,x } ESTIMATORS: θ 1 [ ] logx log{m{x 1,...,X }} θ m{x 1,...,X } 4 For theposso(λ case STEP x L(λ = f X (x λ = e λ λ x = e λ λ ( x! x! ( log L(λ = logx! λ+ x logλ ( 1 {log L(λ} = + λ x λ = = λml = 1 ( 1 {log L(λ} = x λ λ < for all λ x = x Therefore We see that a T 1 s ubase. Now, f ESTIMATE : λ ML = x T 1 = X = 1 ESTIMATOR:X X = T1 [T 1 ;λ] = λ τ = τ(λ = e λ so that λ = logτ a we ca reformulate the lkelhoo terms of τ, gvg ( log L(τ = logx!+logτ + x log( logτ Dfferetatg wrtτ the usual way a equatg to zero, we f that τ ML = exp{ x}, so that τ ML (λ = τ( λ ML. Usg a Taylor approxmato g( λ g(λ+ġ(λ( λ λ+ g(λ( λ λ/ MATH 557 EXERCISES SOLUTIONS Page 4 of 9

5 wth g(t = e t a λ = x, a takg expectatos, otg that X [X] = λ, T1 [exp{ X};λ] e λ +e λ Var T1 [X;λ]/ = e λ + 1 e λ λ/ so the bas s approxmately 1 e λ λ/. 5 Notg the costrat the pf thatx η, we have STEP L(λ,η = f X (x ;θ = { λexp{ λ(x η} = λ exp λ log L(λ,η = logλ λ (x η = logλ λ x +λη } (x η η x 1,...,x, zero otherwse λ {log L(λ,η} = λ + {log L(λ,η} = λ η x η = = λml = x η The seco of the partal ervatve equatos cates aga that the maxmum of the lkelhoo occurs wheη s as large as possble, that s, whe η = m{x 1,...,x }.Hece ESTIMATES λ ML = = x m{x 1,...,x } η ML = m{x 1,...,x } wth the obvous correspog estmators. (x m{x 1,...,x } 6 X 1,...,X Expoetal(1/θ so that X [X ;θ] = θ a hece, usg staar mgf techques, we have X = X Gamma(,1/θ = E fx [X] = /(1/θ = θ so that f T 1 = X = 1 X the T1 [T1;θ] = 1 θ = θ a hecet 1 s a ubase estmator ofθ. Now f Y 1 = m{x 1,...,X }, the prevous orer statstcs results gve that { ( F Y1 (y = 1 {1 F X (y} = e y/θ} = 1 e y/θ y > so that Y 1 Expoetal(/θ. Hece f Z = Y 1 the Y1 [Y 1 ;θ] = θ = Z [Z;θ] = θ = θ a hecez s a ubase estmator ofθ. MATH 557 EXERCISES SOLUTIONS Page 5 of 9

6 7 We have f X (x = 1 θ 1 x θ+1 F X (x = x (θ 1 = x θ+1 θ 1 x θ+1 E X [X ;θ] = θ (by tegrato, or by otg that the pf s costat a hece symmetrc about θ a hece, usg staar expectato techques, we have that f T 1 = X T1 [T 1 ;θ] = 1 X [X ;θ] = 1 θ = 1 θ = θ a hecet 1 s a ubase estmator ofθ. Now f Y 1 = m{x 1,...,X }a Y = max{x 1,...,X }, results o orer statstcs results gve that f Y1 (y = f X (y{1 F X (y} 1 = 1 { } y (θ = { } 1+θ y 1 θ 1 y θ+1 a f Y (y = f X (y{f X (y} 1 = 1 For the expectatos, Y1 [Y 1 ;θ] = θ+1 θ 1y { y (θ 1 { } 1+θ y 1 y } 1 = { } 1 θ +y 1 θ 1 y θ+1 = ((1+θ tt 1 t settgt = (1+θ y/ = y = (1+θ t = (1+θ t 1 t t t = (1+θ +1 a Y [Y ;θ] = θ+1 { } 1 θ +y 1 y θ 1y = (t (1 θt 1 t settgt = (1 θ+y/ = y = t (1 θ = t t (1 θ t 1 t = +1 (1 θ so that f M = (Y 1 +Y / the by propertes of expectatos M [M;θ] = 1 Y 1 [Y 1 ]+ 1 [ 1 Y [Y ] = (1+θ ] [ ] (1 θ = θ a hecem s a ubase estmator forθ. MATH 557 EXERCISES SOLUTIONS Page 6 of 9

7 8 ( Forλ > STEP L(λ = f X (x λ = log L(λ = logλ logγ(+ logx λ λ ( { } Γ( x exp{ λx } = λ {Γ(} x exp λ x x θ {log L(λ} = λ x = = λml = x {log L(λ} = θ λ < Hece the estmator s λ = atλ = λ ML ( X ( By the varace property of maxmum lkelhoo estmators, we must have that the ML estmator of τ = 1/λ st = X /. Now, usg mgfs, t s straghtforwar to show that so that X Gamma(,λ = T = 1 X Gamma(,λ T [T;λ] = λ = 1 λ Var ft [T;λ] = (λ = 1 [ λ T T ;λ ] = 1 ( 1 λ + = +1 λ λ 9 (a We have for the lkelhoo 1 1 L(x;θ = π 1+(x θ = 1 π l(x;θ = logπ log(1+(x θ. 1 1+(x θ yelg the followg score equato (whch caot be solve aalytcally l θ = (x θ 1+(x θ = (b The coe below evaluates the log-lkelhoo Ro a fe gr o the rage (-,: x.at<-c(7.36,5.14,3.71,3.15,6.,6.38,1.34,6.73;thvec<-seq(-,,legth=1 log.pf.fuc<-fucto(x,th{retur(-log(p-log(1+(x-th^} log.pf.mat<-outer(thvec,x.at,log.pf.fuc log.lke.vec<-apply(log.pf.mat,1,sum MATH 557 EXERCISES SOLUTIONS Page 7 of 9

8 Log lkelhoo θ plot(thvec,log.lke.vec,type= l,xlab=expresso(theta,ylab= Log-lkelhoo th.max<-thvec[whch.max(log.lke.vec] I ths coe, the varable th.max cotas the umercal value of the estmate of θ; to two ecmal places, we f that θ(x = The coe also prouces the followg plot. The followg coe uses theoptmze fucto to prouce the same result: the commas log.lke.fx<-fucto(x,x{retur(sum(log.pf.fuc(x,x[1]} optmze(f=log.lke.fx,x=x.at,terval=c(-,,maxmum=true yel $maxmum [1] $objectve [1] The followg R coe computes the N = 5 estmates. N<-5 <-8 ml.est<-me.est<-rep(,n theta<-5 for( 1:N{ } x<-rcauchy(+theta ml<-optmze(f=log.lke.fx,x=x,terval=c(-,,maxmum=true ml.est[]<-ml$maxmum me.est[]<-mea(x MATH 557 EXERCISES SOLUTIONS Page 8 of 9

9 The hstograms below epct the 5 maxmum lkelhoo a mea estmates: Max. Lke. Mea Frequecy Frequecy ml.est It ca be scere from these fgures that the varace of the mea estmator s slghtly larger tha for the maxmum lkelhoo estmator; oe smulato, I got that the varace for the sample of mea estmates was.468, a the varace for the sample of maxmum lkelhoo estmates was.387. me.est MATH 557 EXERCISES SOLUTIONS Page 9 of 9