Solution of Assignment #2

Transcription

1 Soluion of Assignmen #2 Insrucor: A. Simchi Quesion #1: a r 1 c i 7, and λ n c i n i An approximae 95% confidence inerval for λ is given by ˆλ ± 1.96 ˆλ r.189 ± Noe ha he above confidence inerval is an approximae confidence inerval while for exponenial we can find an exac confidence inerval. In addiion, he confidence inerval is based on large sample while we have only 1 observaions. Therefore, i is beer o use he following formula o find a 95% confidence inerval for λ. ˆλ χ 2 2r,.975 2r ˆλ χ 2 2r,.25 2r ˆλ χ 2 14, ˆλ χ 2 14, b For exponenial disribuion S4 e 4λ. A 95% confidence inerval for λ, from par a, is.73,.337. So a 95% confidence inerval for S4 e 4λ is e 4.337, e ,.7468 Anoher way o find a 95% confidence inerval for gλ S4 e 4λ is o use gˆλ ± 1.96 S.E. gˆλ where Var gˆλ 2 d dλ gλ Varˆλ 4e 4λ 2 λ 2 r S.E. gˆλ 4e ˆλ 4ˆλ r So a 95% confidence inerval for gλ S4 e 4λ is gˆλ ± 1.96 S.E. gˆλ e ± e ± ,.745 1

2 ] exp[±1.96 S.Eˆν] Anoher 95% confidence inerval for S4 is [Ŝ4 where So Varˆν Var gˆλ ν ln [ ln S4] ln 4 + ln λ gλ 2 d dλ gλ Varˆλ Therefore a 95% confidence inerval for S4 is 2 1 λ 2 λ r 1 r 1 7 S.Eˆν 1 7 ] exp[±1.96 S.Eˆν] [Ŝ4 [e 4.189] exp[±1.96/ 7].219,.78 I should be menioned ha he las wo confidence inervals are approximae confidence inervals and hey are based on large samples while we have only 1 observaions in his problem.so we should use he firs confidence inerval. Quesion #2: a For Weibull disribuion, ˆγ, Weibull shape, and is sandard deviaion are given by SAS. To find ˆλ and is sandard deviaion, we have λ e β gβ. So Var ˆλ 2 d gβ Var dβ ˆβ e β 2 Var ˆβ S.E.ˆλ e ˆβ S.E. ˆβ Therefore, for group 1, we have ˆγ , S.E.ˆγ 1.323, and ˆλ 1 e ˆβ e and S.E.ˆλ 1 e Similarly, for group 2, we have ˆγ 2.542, S.E.ˆγ , and ˆλ 2 e ˆβ e and S.E.ˆλ 2 e b The log-likelihoods for Weibull disribuion for wo groups 1 and 2 are and , respecively. If γ 1, he Weibull disribuion reduces o an exponenial disribuion. So The log-likelihoods for Exponenial disribuion for wo groups 1 and 2 are and , respecively. So for esing H : γ 1 γ 2 1 versus H 1 : γ 1 1 or γ 2 1, he es saisic is T S 2 {[ ] [ ]} The es saisic, under null hypohesis, has a chi-square disribuion wih degrees of freedom. Therefore, P -value P T S > Since p-value is less han α.5, we rejec H. 2

3 c For Weibull disribuion we have S exp [ λ γ ]. So In addiion, For Group 1 : Ŝ15 exp { [.415] }.9232 For Group 2 : Ŝ25 exp { [.455].542}.8792 ν ln [ ln S5] γ ln5λ γ ln λ + ln 5 1 σ β + ln 5 Hence ν β 1 σ, ν σ 1 σ β 2 + ln 5, and ν ν Var ˆβ Cov Varˆν ˆβ, ˆσ β σ Cov ˆβ, ˆσ Varˆσ Therefore, for group 1, we have ν β , ν σ ν β ν σ ln , and Varˆν S.E.ˆν Similarly, for group 2, we have ν β , ν σ ln , and Varˆν S.E.ˆν Finally, 95% confidence inervals for S 1 5 and S 2 5, respecively, are given by ] exp[±1.96 S.Eˆν1] [Ŝ1 5 [.9232] exp[± ].823,.9683 ] exp[±1.96 S.Eˆν2] [Ŝ2 5 [.8792] exp[± ].7887,.9325 The produc-limi esimae of S 1 5 is Ŝ15 Ŝ A 95% confidence inerval for S 1 5 is.75322, Similarly, he produc-limi esimae of S 2 5 is Ŝ25 Ŝ A 95% confidence inerval for S 2 5 is.76335, Confidence inervals obained by produc limi esimaes are lile wider han confidence inervals obained from Weibull disribuions. 3

4 Quesion #3: a We find momen generaing funcion of Y 2nˆµ/µ 2nλT Ee uy E e u2nλt E e u2λ n Ti n E e 2λuTi E e 2λuTi Since T i are independen. The momen generaing funcion of T i is Ee uti 1 u/λ 1. Hence Therefore, Ee 2λuTi 1 2λu/λ 1 1 2u 1 Ee uy 1 2u 1 1 2u n Noe ha 1 2u n is he momen generaing funcion of a Chi-Square disribuion wih degrees of freedom 2n. Since momen generaing funcion uniquely deermine he disribuion, Y has a Chi-Square disribuion wih degrees of freedom 2n and his complees he proof. b Suppose T 1, T 2,, T n is a random sample from an exponenial disribuion wih parameer λ. We will solve he problem when he disribuion is woparameer exponenial afer ha. The join disribuion of T 1, T 2,, T n is g n [ ] T 1, T 2,, T n n! ft i n! λe λt i n! λ n exp λ T i We will find he join disribuion of W 1 nt 1 and W i n i + 1T i T i 1 for i 2,, n Noe ha n T i n W i and he Jacobian of he he ransformaion is J 1/n!. Therefore, he join disribuion of W 1, W 2, W n is f W 1, W 2,, W n n!g W 1, W 2,, W n J [ ] λ n exp λ W i λ e λw i This means ha W 1, W 2, W n are independen and W i has an exponenial disribuion wih parameer λ. On he oher hand, Y n Ti T 1 T i nt 1 T i nt 1 4 W i W 1 i2 W i

5 The momen generaing funcion of Y is [ ] n Ee uy E exp u W i E e uwi i2 1 u/λ n 1 12 Ee uwi 12 1 u/λ 1 This is he momen generaing funcion of a gamma disribuion wih parameers n 1 and λ. Since momen generaing funcion uniquely deermine he disribuion, Y has a gamma disribuion wih parameers n 1 and λ. Now, if T 1, T 2,, T n is a random sample from wo-parameer disribuion wih parameer λ and G, hen T 1 G, T 2 G,, T n G is a random sample from an exponenial disribuion wih parameer λ. Therefore, T 1 G, T 2 G,, T n G is he order saisics in a random sample of size nfrom an exponenial disribuion wih parameer λ. Hence, by he above resuls, W 1 n T 1 G and W i n i + 1 [ T i G T i 1 G ] n i + 1T i T i 1, i 2,, n, are independen and have a exponenial disribuion wih parameer λ. In addiion n Y Ti T 1 i2 W i has a gamma disribuion wih parameers n 1 and λ. Quesion #4: a We proved, in par b of quesion #3, ha W 1 n T 1 G and W i n i + 1T i T i 1 for i 2,, n are independen and have a exponenial disribuion wih parameer λ. I will use PROC LIfETEST o check wheher W 2,, W n is a random sample from an exponenial disribuion You can also use Cox-Snell residuals plo o check wheher W 2,, W n is a random sample from an exponenial disribuion. The plos on ln Ŝ versus for Types A and B are given in he las page. Boh plos show no serious deviaion from sraigh line and herefore we canno rejec ha he daa does no have wo-parameer exponenial disribuion. b As I showed in he class, since T 1, T 2,, T n are independen and have wo parameer exponenial disribuion, he maximum likelihood esimae of λ is ˆλ n n n i2 W n. i i2 Ti T 1 12 In addiion, by using par b of quesion #3, We know ha W 1 n T 1 G and W i n i + 1T i T i 1 for i 2,, n 5

6 are independen and have a exponenial disribuion wih parameer λ. In addiion Y n i2 W i has a gamma disribuion wih parameers n 1 and λ. Hence, by par a of quesion #3, 2λY has a Chi-square wih 2n 1 degrees of freedom. Therefore, a 11 α% confidence inerval for λ is given by ˆλ χ 2 2n 1,.975 2n ˆλ χ 2 2n 1,.25 2n For ype A, we have ˆλ 12/ and a 95% confidence inerval for λ is ˆλ χ ,.975 ˆλ χ , Similarly, for ype B, we have ˆλ 12/ and a 95% confidence inerval for λ is ˆλ χ ,.975 ˆλ χ , Quesion #5: a By using he Fundamenal Theorem of Calculus, we have d Sx dx Sx d In addiion Therefore, lim lim m lim n S lim n Sx dx lim S lim f Sx dx S lim S S f lim lim f f lim S f lim f f By using L Hospial s Rule S f 1 lim By using L Hospial s Rule dd ln f 1 6

7 b The p.d.f. of a gamma disribuion is f λ Γγ λγ 1 e λ So λ d ln f ln f ln + γ 1 lnλ λ γ 1 λ Γγ d Therefore, by using par a, we have lim m lim d 1 d ln f lim λ γ 1 1 λ 1 1 λ f The p.d.f. of a log-normal disribuion is [ 1 f σ 2π exp 1 ] ln µ2 2σ2 So 1 ln f ln + ln σ 2π 1 ln µ2 2σ2 d ln f 1 d + 1 ln µ σ ln σ 2 µ, as, since, by using L Hospial s Rule, Therefore, by using par a, we have ln 1/ lim lim 1 lim 1. lim m lim d 1 d ln f, as. Quesion #6: [ H hxdx α + β ] dx [αx + β lnx + γ] o x + γ + γ + γ α + β ln + γ β lnγ α + β ln α + ln γ γ β 7

8 { [ ]} β + γ S exp { H} exp α + ln γ f hs [ α + β ] β γ e α + γ + γ You can also find fx by using, for example, fx S x. β γ e α + γ Quesion #7: So f, λ f λgλ λe λ λ k 1 e λ/α f f, λ dλ α k Γk Le u [ + 1/α]λ du [ + 1/α] dλ. Then λk α k Γk e [+1/α]λ λ k α k Γk e [+1/α]λ dλ f 1 α k Γk 1 α k [ + 1/α] k+1 k u e u du + 1/α + 1/α u k e u du Γk α Γk + 1 αk α + 1 k+1 Γk α + 1 k+1 Therefore and S f fxdx Since h f/s, we have h kα 1 + α k+1 >, k >, α > kα 1 + αx k+1 dx [ 1 + αx k] α k kα/1 + αk+1 1/1 + α k kα 1 + α 8

9 Log- Sur vi val pl o f or ype A Ti mea Log- Sur vi val pl o f or ype B Ti meb