AMS Final Exam Spring 2018

Transcription

1 AMS57.1 Fal Exam Sprg 18 Name: ID: Sgature: Istructo: Ths s a close book exam. You are allowed two pages 8x11 formula sheet (-sded. No cellphoe or calculator or computer or smart watch s allowed. Cheatg shall result a course grade of F. Please provde complete solutos for full credt. The exam goes from 11:15am-1:45pm. Good luck! 1. Let 1,, be d Posso(λ, let λ be a gamma(α, dstrbuto, the cojugate famly for the Posso. (a Fd the posteror dstrbuto of λ, ad subsequetly the Bayesa estmator of λ. (b Derve the mode of the posteror dstrbuto of λ, ad the dscuss whe ad how we ca derve the 1(1-α% Bayesa HPD credble set for λ. Soluto: (a The Posso pdf s: Therefore the lkelhood fucto s: f(x λ = f(x 1,, x λ = e λ λ x f(x λ = e λ λ x, x =, 1, x! x! = [e λ λ x ] [ By the factorzato theorem, Y = s a suffcet statstc for λ. 1 x The mgf of Y λ s: M Y λ (t = e λ(et 1 = e λ(et 1 Therefore we kow Y λ ~ Posso(λ.! ] = g ( x, λ h(x 1,, x I. For the pror: The margal pdf of Y s m(y = f(y λπ(λ dλ = Thus the posteror dstrbuto s, 1 π(λ = Γ(α α λα 1 e (λy e λ 1 y! Γ(α α λα 1 e λ y dλ = y! Γ(α y y+a = Γ(y + α ( y! Γ(αα + 1 λ /(+1 λ λy+α 1 α π(λ y = f(y λπ(λ = λy+α 1 e y+a ~Gamma (y + α, m(y Γ(y + α ( The Bayesa estmator for λ s: E(λ y = (y + α + 1 = + 1 y + 1 (y + α α = (b λ l[π(λ y] = c + (y + α 1lλ /( + 1 e λ /(+1 dλ 1

2 l[π(λ y] (y + α 1 1 (y + α 1 = = λ =, for y + α 1 > λ λ Therefore whe y + α 1 >, the mode exsts, we ca obta the Bayesa HPD [a, b] as llustrated below by the Theorem of Optmal Iterval. II. For the pror: The posteror dstrbuto s, π(λ = α Γ(α λα 1 e λ π(λ y = f(y λπ(λ ~Gamma(y + α, +. m(y The Bayesa estmator for λ s: (y + α E(λ y = + The mode for the posteror of λ s: (y + α 1 λ =, for y + α 1 > +. Let 1,, be a radom sample from a expoetal dstrbuto exp (θ wth pdf: Please derve f (x; θ = 1 exp ( x, where θ > ad x >, θ θ (a The maxmum lkelhood estmator for θ. (b Is the above MLE ubased for θ? (c Please derve the dstrbuto of the frst order statstc (1 ad further show whether Y = (1 s a ubased estmator of θ or ot. (d Please calculate the MSE (mea squared error of Y, ad the MSE of the MLE for θ, whch oe s smaller? (e Is there a best estmator (UMVUE for θ? Please show the etre dervato. (f Is there a effcet estmator for θ? Please show the etre dervato. Ht: Cramér-Rao Iequalty: Let ˆ h 1,,, radom sample from a populato wth pdf f ; be ubased for, where, 1,,, s a x satsfyg all regularty codtos. The

3 Soluto: 1 1 ˆ l f x; l f x; Var E E (a The lkelhood fucto s L = f (x ; θ = 1 θ exp ( x θ = 1 exp ( x θ θ The log lkelhood fucto s ] = lθ x θ l = ll = l [ 1 exp ( x θ θ Solvg l θ = θ + x θ = We obta the MLE for θ: θ = (b Sce E[ ] = E[] = x 1 θ exp ( x θ dx = θ We kow the MLE θ = s a ubased estmator for θ. (c Now we derve the geeral formula for the pdf of the frst order statstc as follows: Therefore we have P( (1 > x = P( 1 > x,, > x = P( > x 1 F (1 (x = [1 F(x] = [1 F(x] Dfferetatg wth respect to x, ad the multplyg by (-1 at both sdes leads to: f (1 (x = f(x[1 F(x] 1 Now we ca derve the pdf of the frst order statstc for a radom sample from the gve expoetal famly. Frst, the populato cdf s: x x F(x = 1 θ exp ( u θ du = [ exp ( u θ ] = 1 exp ( x θ Now pluggg the populato pdf ad cdf we have: f (1 (x = 1 θ exp ( x θ [exp ( x θ ] 1 = x exp ( θ θ, whe θ > ad x >. Thus we kow that (1 ~exp ( θ, ad ts mea should be E[ (1 ] = θ 3

4 Therefore we have: E[Y] = E[ (1 ] = θ = θ That s, Y = (1 s a ubased estmator of θ. (d Sce both estmators are ubased, MSE = Varace. We ca easly derve the populato varace for ~exp (θ to be: Var[] = θ Thus we have: Smlar we kow Thus we have: Var[ ] = Var[] Var[ (1 ] = ( θ = θ = θ Var[Y] = Var[ (1 ] = θ = θ It s obvous that the varace, ad thus the MSE, of the MLE s smaller. (e The expoetal famly exp (θ wth pdf: f (x; θ = 1 θ I (, exp ( x == c(θh(x exp[w(θt(x] θ s a regular expoetal famly wth ad therefore T( = s a complete suffcet statstc for θ. Thus the MLE of θ: θ = T( = s a fucto of the complete suffcet statstc, ad s ubased for θ. By the Lehma Scheffe Theorem, the MLE s a best estmator (UMVUE for θ. (f Now we calculate the Cramer-Lower boud for the varace of a ubased estmator for θ: l[f (x; θ] = l [ 1 θ exp ( x θ ] = lθ x θ l[f (x; θ] = 1 θ θ + x θ l[f (x; θ] θ = 1 θ x θ 3 4

5 E [ l[f (x; θ] θ ] = E [ 1 θ θ 3] = [ 1 θ 1 θ 3 θ] = θ Thus the Cramer-Rao lower boud s: Therefore we clam that the MLE s a effcet estmator for θ. ***Note: You ca also frst prove (f the MLE s a effcet estmator for θ, ad subsequetly mmedately clam t s the (e best estmator (UMVUE for θ. 3. To test whether the brth rates of boys ad grls are equal, a radom sample of 3 famles wth exactly three chldre wth the age of 18 was take ad the dstrbuto of the chldre s geder s provded below. Please test whether the brth rates are equal or ot at the sgfcace level of α =.5. Please frst derve your test usg the pvotal quatty approach, cludg all steps for full pots. θ Soluto: 3 grls grls, 1 boy 1 grl, boys 3 boys No. of famles Ths problem ca be doe several ways cludg, a large sample -test for oe populato proporto, or a Ch-square goodess of ft test. Here we smply use the large sample -test. Part 1. Dervato of the geeral large sample -test for oe populato proporto. Samplg from the Beroull Dstrbuto Toss a co, ad get the result as followg: Head(H, H, Tal(T, T, T, H, Let th 1, f the toss s head. A proporto of p s head, the populato. th, f the toss s tal. x 1 x ~ Beroull( p P( x p (1 p, x =, 1; = 1,,,. (*Bomal dstrbuto wth = 1 Iferece o p the populato proporto: 1 Pot estmator: pˆ 1 Large sample ferece o p:, ˆp s the sample proporto ad also, the sample mea 3 Large sample (the orgal pvotal quatty for the ferece o p. 5

6 Alteratve P.Q. = p p p (1 p ~ N(,1 *** You ca use ether of the pvotal quatty. I the followg, we wll use the frst pvotal quatty, ad llustrate the procedure for a two-sded test. 4 Large Sample Test: H : p p Ha : p p Test statstc (large sample usg the orgal P.Q.: At the sgfcace level, we reject H favor of H a : p p f: P( c H P( c H P( c H P( c H P( c Reject H c H f Part. Now we apply the above geeral test to the gve problem. Here we have = 9. Let Y be the total umber of grls amog the 9 chldre, we have y = 453. Let p be the proporto of etrepreeurs wth domestc cars, we have p = 453, ad we are testg: H : the dstrbuto of the umber of daughters bor follows a bomal dstrbuto wth p.5 versus H : the dstrbuto of the umber of daughters bor follows a bomal dstrbuto wth p.5 a The test statstcs s: = p.5.5(1.5/9 =. 9 6

7 Sce =. < 1.96, we ca ot reject the ull hypothess of equal brth rates at the sgfcace level of Suppose we have a smple radom sample from a ormal populato as follows: 1,,, ~N(μ, σ, where μ ad σ are both ukow. (a At the sgfcace level α, please costruct a test usg the pvotal quatty approach to test H : σ = σ versus H a : σ > σ where σ s a postve costat (*Please clude the dervato of the pvotal quatty, the proof of ts dstrbuto, ad the dervato of the rejecto rego for full credt. (b At the sgfcace level α, please derve the lkelhood rato test for testg H : σ = σ versus H a : σ > σ. Subsequetly, please show whether ths test s equvalet to the oe derved part (a. (c Is the test part (b a uformly most powerful test? Please provde detaled justfcatos for your aswer. Soluto: ω = {(μ, σ : < μ <, σ = σ } Ω = {(μ, σ : < μ <, σ σ } Note: Now that we are ot just dealg wth the two-sded hypothess, t s mportat to kow that the more geeral defto of ω s the set of all ukow parameter values uder H, whle Ω s the set of all ukow parameter values uder the uo of H ad H a. Therefore we have: So, L ω = f(x 1, x,, x μ, σ = σ = f(x μ, σ = σ (x μ πσ e σ = 1 = (πσ e 1 σ (x μ L Ω = f(x 1, x,, x μ, σ σ = f(x μ, σ σ = 1 πσ e ll ω μ (x μ σ = (πσ e 1 σ (x μ = σ (x μ = μ ω = x sup(l ω = (πσ e 1 σ (x x { logl Ω logl Ω μ σ = σ + 1 σ 4 (x μ = σ (x μ = μ Ω = x = σ = 1 (x x logl Ω (σ = σ 4 1 σ 6 (x μ < 7

8 sup(l Ω = { (πσ e 1 LR = sup(l ω sup(l Ω = {( (πσ e, f σ > σ (x x σ σ (x x e 1, f σ σ σ (x x 1, f σ σ P(LR c H = α P (W = (x x Reject H favor of H a f W = (x x σ χ 1,α,upper σ, f σ > σ. c H = α. Furthermore, t s easy to show that for each σ 1 > σ, the lkelhood rato test of H : σ = σ versus H a : σ = σ 1 rejects the ull hypothess at the sgfcace level α for W = (x x χ 1,α,upper By the Neyma-Pearso Lemma, the lkelhood rato test s also the most powerful test. Now sce for each σ 1 > σ, the most powerful sze α test of H : σ = σ versus H a : σ = σ 1 rejects the ull hypothess for W = (x x χ 1,α,upper. Ths rejecto rego does ot deped o σ 1 -- t s the σ same rejecto rego, hece test, for each σ 1. Sce ths same test s most powerful for each σ 1 > σ, ths test s UMP for H : σ = σ versus H a : σ > σ, by the defto of the UMP test. σ 5. Suppose that the radom varables Y 1, Y,, Y satsfy Y = x + ε, = 1,,, where x 1, x,, x are fxed costats, ad ε 1, ε,, ε are depedet Gaussa radom varables wth mea ad varace c 1 σ, c σ,, c σ, respectvely (c >, = 1,,, are kow costats, whle σ s also a kow costat. Please derve: (a The (ordary least squares estmator (LSE for. (b The maxmum lkelhood estmator (MLE for. Is the MLE the same as the LSE? (c At the sgfcace level α, please derve a test for H : = versus H a :. Soluto: (a To mmze SS = x (y x SS = (y x : = = x Y x y = x s the least squares regresso le through the org. (b Let we have w = y c v = x c e = ε c 8

9 w = v + e, = 1,,,, v s gve, ad e d ~ N(, σ, W ~N(v, σ, = 1,, Sce W 1, W,, W are depedet to each other, the lkelhood s: L = 1 (w v πσ e σ = (πσ / exp [ (w v σ ] The log lkelhood s: ll = l(πσ (w v σ Take dervatve wth respect to, ad set t to zero, we obta the MLE: = v W = x Y /c v x /c Therefore we ca easly see that the MLE ad the LSE are o loger detcal whe the c s are ot all equal to each other. (c Let θ = x /c x, = 1,, /c The θ Y ~N(θ x, θ c σ, = 1,, Furthermore, they are depedet to each other. We have the momet geeratg fucto for : M (t = E[exp(t ] = E [exp (t Therefore we foud: The pvotal quatty for s: θ Y ] = E[exp(tθ Y ] = exp [tθ x + 1 t θ c σ ] = exp [t θ x = exp [t + 1 σ t x ] /c = For the test of: H : = versus H a :, The test statstc s: ~N (, σ σ x /c x /c ~N(,1 + 1 t θ c σ ] = σ x /c H ~ N(,1 9

10 At the sgfcace level, we reject H favor of H a f: P( c H P( c H P( c H P( c H P( c Reject H c H f *** That s all, folks! *** 1