THE ROYAL STATISTICAL SOCIETY 2010 EXAMINATIONS SOLUTIONS GRADUATE DIPLOMA MODULE 2 STATISTICAL INFERENCE

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1 THE ROYAL STATISTICAL SOCIETY 00 EXAMINATIONS SOLUTIONS GRADUATE DIPLOMA MODULE STATISTICAL INFERENCE The Socety provdes these solutos to assst caddates preparg for the examatos future years ad for the formato of ay other persos usg the examatos The solutos should NOT be see as "model aswers" Rather, they have bee wrtte out cosderable detal ad are teded as learg ads Users of the solutos should always be aware that may cases there are vald alteratve methods Also, the may cases where dscusso s called for, there may be other vald pots that could be made Whle every care has bee tae wth the preparato of these solutos, the Socety wll ot be resposble for ay errors or omssos The Socety wll ot eter to ay correspodece respect of these solutos Note I accordace wth the coveto used the Socety's examato papers, the otato log deotes logarthm to base e Logarthms to ay other base are explctly detfed, eg log 0 RSS 00

2 Graduate Dploma, Module, 00 Questo ( The jot desty s t / (,, ( = Σt f t t = e = e g ( t ; Σ h(t,, t Sce the jot desty s the product of a factor (smply ot volvg the parameter ad a factor oly depedet o the observatos t,, t through Σt, t follows by the factorsato theorem that Y = Σt s suffcet for t/ t/ t/ t/ ( E( T = t e dt = te + e dt = [ 0] + e = E( ˆ = E( T = ( = (for all = So ˆ s a ubased estmator of ( t/ t/ t/ E T = t e dt = t e te dt t [ 0] e t = + dt = E ( T = / 0 Var(T = E(T {E(T } = = Var ( ˆ = Var ( T (ote that the T are depedet = = ( = ( ˆ s a ubased estmator of (for all ad ts varace teds to zero as So ˆ s a cosstet estmator of Soluto cotued o ext page

3 Part (v We requre / t / ( E T = t e dt 0 Ths ca be evaluated by otg that explctly as follows t e / t / Γ ( 3/ 3 s the pdf of a gamma dstrbuto, or Frst, use the substtuto u = t /, wth whch the tegral becomes u e u / udu (ow use symmetry about 0 0 = u / ue du (ow create the pdf of a Normal dstrbuto = π / u e π / u / du The tegrad s u tmes the pdf of N(0, /; thus the tegral s the secod momet (about zero of that dstrbuto; ad ths s the varace plus the square of the mea π = π, as requred = Therefore, otg that T ad T are depedet, 4 4 π E % = E T E T π = π 4 = So % s a ubased estmator of ( ( ( Soluto cotued o ext page

4 Part (v Aga otg that T ad T are depedet, 6 6 E( % = E ( T E( T = π π 6 6 Var ( % = E( % { E( % } = = = 06 π π So the relatve effcecy of % compared to ˆ s ( ˆ ( % Var = = Var (or 805%

5 Graduate Dploma, Module, 00 Questo [Soluto cotues o ext page] ( X has the trucated Posso dstrbuto wth zero mssg Lettg Y deote the full Posso dstrbuto, wth mea λ, we have (for =,, 3, e λ PY ( = ( ( 0! λ e P X = = P Y = Y > = = = PY e e ( > 0! ( λ ( The lelhood s L ( λ x Σx λ e λ e = = = x! e x! e ( Π ( ( λ x λ λ ( ( x Σ Π log L = log log e log! dlog L Σx e = dλ λ e The maxmum lelhood estmator ˆλ therefore satsfes ˆ λ Σx e = ˆ λ e ˆ 0 ( We have d log L Σx e e Σx e dλ λ e e λ e = + + = + ( ( ΣE( X E = λ λ d log L e ( e Now, E( X = ( e = λ e! λ The summato ca be exteded to clude also = 0, as ths s a zero term; hece the summato s E(Posso(λ = e λ λ

6 E = = e e d log L λ e λ λ λ ( ( ( ( λ e e e Therefore the asymptotc varace of ˆλ s λ ( e ( e e (v We use the Newto-Raphso method Iteratos cotue accordg to the scheme descrbed below utl covergece occurs dlog L ˆ ˆ ˆ dλ λ= λ0 λ= λ0 d log L ˆ dλ λ= λ0 A startg value s requred A reasoable tal estmate of λ s smply x, as ths would be the estmate f the Posso dstrbuto had ot bee trucated I the questo, a tal estmate of 0 s gve Isertg ths, = 30 ad Σ = 50, we have X dlog L 50 30e = 30 = 96956, ˆ dλ λ= λ 0 e ad d log L 50 30e = + = ˆ dλ λ= λ0 4 ( e ˆ λ = 0 = 068(

7 Graduate Dploma, Module, 00 Questo 3 [Soluto cotues o ext page] Part (a ( The lelhood s f ( θ f ( x θ x = = Usg the otato defed the questo, the loss whe θ s estmated by ˆ θ (a fucto of X, X,, X s l (; θˆ θ The the rs s ( θ = ( ˆ ; ( ˆ θ θ θ = θ; θ ( θ R E l l f x dx ℵ Note ℵ dx represets tegrato over the sample space Thus, as π ( θ s the pror desty of θ, the Bayes rs of ˆ θ s ( ˆ π ( ( ( ( ˆ ( ; ( x ( x r θ = E R θ = R θ π θ dθ = l θ θ f θ π θ d dθ π The Bayes estmator of θ s the value ˆ Θ Θ ℵ Note dθ represets tegrato Θ over the parameter space θ that mmses ( ˆ r π θ ( We ote that f ( x θ π ( θ = π ( θ x f ( x where π ( θ x s the posteror desty of θ gve x ad ( f x s the margal desty of x, ( ( θ π ( θ f x = f x dθ Θ Thus, from the defto above, we ca wrte the Bayes rs as ( ˆ = ( ˆ ; ( ( = ( ( ˆ ; ( rπ θ l θθπθx f x dxdθ f x l θθπθx dθdx Θ ℵ ℵ Θ Thus to mmse ( ˆ whch s the posteror expected loss r π θ at a gve x, we mmse l θ ˆ θπ( θ x Θ ( ; dθ

8 ( The posteror expected loss, gve x, s L( ˆ = ˆ ( have to fd the ˆ θ that mmses ths Frst, we ote that ( ˆ dy dy θ θ θ π θ x dθ We + f y > 0 = Usg ths, f y < 0 dl θ ˆ d θ θ ( ˆ θ = π θ d θ π ˆ ˆ ( θ d θ ˆ π ( θ d θ θ dθ x = dθ x x Settg ths equal to zero gves that ˆ θ s the meda of π( θ x Part (b We have that the posteror dstrbuto of s N(00, ad that the loss fucto s π ( m, m ; m m m m = 5 + m m otherwse So the posteror expected loss s ( 5 ( ( ( ( 5 ( L = + m m P < m + m m P m m + + m m P > m { ( ( } = m m + 5 Φ m 00 + Φ m 00 where Φ s the cdf of N(0, Ths s to be mmsed by choce of m ad m We have L m ( m = + 5φ 00 L m ad = 5φ ( m 00 where φ s the pdf of N(0, Tag φ (75 as (approxmately 0, as s gve the questo, t follows that these dervatves are zero at m = 9885 ad m = 075 respectvely To chec that these gve a mmum, we ote that L m > 0, L m > 0 ad L m m = 0 L s therefore mmsed at m = 9885 ad m = 075, so these gve the Bayes rule

9 Graduate Dploma, Module, 00 Questo 4 Part (a Frst defe the sze of a test as follows: Also defe the power of a test as follows: sze = P(reject H 0 H 0 true power = P(reject H 0 H true The a test (betwee two smple hypotheses s the most powerful test at level α f ts sze s equal to α ad o other test wth sze α has greater power Part (b We have the Webull dstrbuto wth probablty desty φ φ f ( y = θφ y exp( θ y for y > 0, where θ (> 0 s uow ad φ (> 0 s ow The hypotheses are H 0 : θ = 05; H : θ = 0 φ φ ( The lelhood s L( θ = f ( y = ( θφ ( Πy exp( θσy = The form of the most powerful test s to reject H 0 f L(05 for some L(0 So we requre φ ( ( φ Πy Σy φ φ ( Πy exp( Σy 05 φ exp φ φ e exp ( y Σ ' for some ' e φ Σy '' for some '' ( '' s chose so that the sze of the test s equal to the specfed sgfcace level Soluto cotued o ext page

10 ( Let W = Y φ For w > 0, the cdf of W s ( ( ( φ / φ = = = ( F w P W w P Y w P Y w / φ / φ w w φ φ φ θ w θφ y exp( θ y dy exp( θ y e 0 0 = = = Ths mght be recogsed as the cdf of a expoetal dstrbuto (wth mea /θ, or ths ca be see perhaps more famlarly by fdg the pdf of W, e d dw θ ( e = e w θ w θ ( The test statstc foud part ( s Σw From the result gve the questo, we have θ Σ W ~ χ Thus, for = 0, uder H 0 we have Σ W ~ χ 40 For sgfcace level 005, '' (see part ( must satsfy P( χ '' = Thus, from χ tables, '' = 6509, ad the most powerful test s to reject H 0 f Σw (=Σy φ 6509 (v Uder H we have Σ W ~ χ 40 Thus the power s P ΣW = = P ΣW = = P ( 6509 θ ( 5308 θ ( χ 5308 From the tables, P ( χ 5805 = 09 ad ( 40 lear terpolato we get ( Power = P χ = 095 Thus by 40 40

11 Graduate Dploma, Module, 00 Questo 5 3 j,, 3 = exp = j= σ ( The lelhood s ( ( L σ σ σ πσ 3 Σ / π σ exp σ x j = j =( x the log lelhood s log L( σ, σ, σ3 = log( π logσ σ xj j Thus for each (=,, 3, dlog L = + σ x dσ σ 3 j j whch o settg equal to zero gves soluto ˆ σ = x j j To vestgate whether these gve a maxmum, we eed to cosder for each d log L 4 the secod dervatves = 3σ xj Evaluated at ˆ σ, ths s dσ σ j x = = < ˆ σ ˆ σ ˆ σ ˆ σ ˆ σ 3 ˆ j j 3 σ ad t follows that ˆ σ as obtaed above deed gves the maxmum lelhood estmator for each Soluto cotued o ext page

12 ( H 0 s σ = σ = σ 3 Let ˆ σ deote the restrcted maxmum lelhood estmator uder H 0 The restrcted lelhood s Σ / Σ σ exp x j σ j ( π ad the restrcted log lelhood s log( π logσ x j σ j Dfferetatg ths wth respect to σ gves + 3 σ σ j x j ad settg ths equal to 0 gves ˆ σ = j x j (we ca aga chec that ths s a maxmum by cosderg the secod dervatve evaluated at ths pot Thus the restrcted maxmum log lelhood s x j j log( π log x j j = log( π log Soluto cotued o ext page

13 From part (, the urestrcted maxmum log lelhood s x j j log( π log j xj j j x = x j j log( π log Deotg these maxmum log lelhoods by log ( ˆ, ˆ, ˆ L σ σ σ (urestrcted 3 ad log L( ˆ σ (restrcted, the geeralsed lelhood rato test s to reject H 0 f ( ˆ σ ˆ σ ˆ σ3 ( ˆ σ log L,, log L for some e f x j j log( π log x j j log( π log e f x j x j j j log log + e f x j x j j j log log, for some Soluto cotued o ext page

14 ( 3 ( We use the asymptotc result that L( ˆ σ ˆ σ ˆ σ L ( ˆ σ log,, log ~χ uder H 0 (ote degrees of freedom because there are two costrats uder H 0 Isertg the values from the questo, ( L( ˆ σ ˆ σ ˆ σ3 L ( ˆ σ log,, log = 50 log log + 40 log log + 50 log log = ( ( (40 ( = 34 The 95% pot of χ s 599 So the result s ot sgfcat at the 5% level there s o real evdece agast H 0

15 Graduate Dploma, Module, 00 Questo 6 Part (a Noparametrc ferece refers to stuatos where o specfc dstrbuto f (x θ s assumed for the populato uderlyg the data Ths beg the case, ferece about ts parameter(s θ s ot possble However, there are some propertes of a dstrbuto, such as ts meda, whch do ot requre explct owledge of f (x θ Iferece respect of these propertes s possble Although there s o explct assumpto about the form of f (x θ, some weaer assumptos may be eeded Advatages compared to parametrc ferece: ca be used wth (o-umerc ordal data requres few assumptos about the dstrbuto uderlyg the data sometmes early as powerful as a correspodg parametrc test Dsadvatages: some loss of power correspodg parametrc tests are ofte qute robust to some falure dstrbutoal assumptos ofte harder to terpret tha f parametrc assumptos had bee made Part (b ( We have H 0 : m = 0, H : m 0, where m s the populato meda Let Y be the umber of observatos that are 0, so that Y ~ B(, ½ uder H 0 Ths dstrbuto s symmetrcal about Y = 6, so the crtcal rego wll be of the form Y y or Y y, for some y = 0,,,, 6 Uder H 0 we have P(Y = 0 = (½ = 00004, P(Y = = (½ = 00093, P(Y = = 006, P(Y = 3 = Thus, uder H 0, P(Y = 0093 ad P(Y 3 = Thus wth y = the sze (sgfcace level of the test s , ad wth y = 3 t s greater tha 005 So the requred crtcal rego s Y or Y 0 Soluto cotued o ext page

16 ( A approxmate 95% cofdece terval for m cossts of those values m' such that, f the hypotheses were H 0 : m = m' ad H : m m', the H 0 could ot be rejected at the 5% level Let X (, X (,, X ( be the order statstcs for the observatos Now let Y be the umber of observatos that are m' Part (b( shows that H 0 s ot rejected f 3 Y 9 (ad s rejected otherwse However, Y 3 meas that there are 3 observatos m' ad so X (3 m' Smlarly, Y 9 meas that there are 9 observatos m' ad so X (0 > m' Thus the approxmate 95% cofdece terval for m s from X (3 to X (0

17 Graduate Dploma, Module, 00 Questo 7 ( We have X σ σ ~N, ad Y ~N α, Y σ ~N, α α ad so Y σ X ~N 0, + α α Y X α = σ + α α X Y σ ( + α ( ~ N 0, Thus α X Y σ ( + α ( ad ( s a pvotal quatty for α, because t s a fucto of α but ot of (the other uow parameter ts dstrbuto, N(0,, does ot volve ay uow parameters ( The lelhood s ( x σ ( α, = ( πσ ( πσ ( y α / / L e e = = ( = + σ ( πσ exp ( x ( y α ( log L, = log x + y σ ( α ( πσ ( ( α σ Soluto cotued o ext page

18 log L = α σ ( y α log L α σ σ ad = ( x + ( y α Settg frst log L = 0 α ˆ ˆ = 0, we get ( y α Usg ths log L = 0 ˆ 0 =, so that ˆ = x σ gves ( x Isertg ths the prevous expresso gves y ˆ xα = 0, e ˆ α = y/ x Thus the maxmum lelhood estmator of α s Y / X, as requred [NB the questo dcates that oly the frst partal dervatves of the log lelhood eed be cosdered] ( We have α X Y σ ( + α ~ N(0, ad so α X Y P 96 < < 96 = 095 σ ( + α P Y 96 σ ( X Y 96 ( 095 α α σ α + < < + + = Therefore a approxmate 95% cofdece terval for α s Y σ ± 96 ( + ˆ α, e X X σ Y Y ± 96 + X X X Soluto cotued o ext page

19 (v Step : draw a radom sample of sze, wth replacemet, from the set {X, X,, X } Call the values X *, X *,, X * Step : draw a depedet radom sample of sze, wth replacemet, from the set {Y, Y,, Y } Call the values Y *, Y *,, Y * ΣY * Step 3: let α* = Σ X * Repeat steps to 3 may tmes, say 0,000 tmes, to obta estmates * * * α, α,, α 0,000 * * * Order these estmates (smallest to largest α, α,, α ( ( (0,000 * * The a bootstrap 95% cofdece terval for α s α (50 to α (9750

20 Graduate Dploma, Module, 00 Questo 8 Part (a Let π ( η 0 be the pror probablty of the ull hypothess ad π ( η = π ( η 0 the pror probablty of the alteratve be The the pror odds of the ull hypothess = ( 0 ( π η π η Now deote the data by d, the the posteror odds of the ull hypothess = where the π fuctos here represet the posteror probabltes gve the data ( 0 d ( π η π η d, The Bayes factor = lelhood uder ull hypothess lelhood uder alteratve hypothess = f f ( d η0 ( d η obvous otato π η = I the usual maer of Bayesa ferece, we have ( d Thus posteror odds = ( d ( d ( d η π ( η0 ( d ( π η f 0 0 = π η f η π η f ( d η π( η h ( d = Bayes factor pror odds Soluto cotued o ext page

21 Part (b ( P( X x θ ( θ x = = (for x = 0,,, the lelhood s ( ( 0 = x 0 0 ( ( Σ x ( f x θ = θ θ = θ θ = θ θ H 0 s θ = 05, H s θ = 05, wth the pror probablty of H 0 beg Bayes factor = ( ( ( 05 ( = = Also, the pror odds of H 0 = = 3 the posteror odds of H 0 = = 0845 ( Now H 0 s θ = 05 ad H s θ 05 Also, uder H the pror dstrbuto of θ s θ ( θ (for 0 < θ < 40 ( x 0 ( ( f H = θ θ θ θ dθ 0 4 ( = θ θ dθ 0 We mapulate ths to create the pdf of the beta dstrbuto wth parameters 3 ad 4 see the ht the questo!4! Γ(65 4 = θ ( θ dθ 64! 0 Γ(3 Γ(4! 4! = 64! Bayes factor = ( ( ! =! 4!! 4! 64!, as requred Soluto cotued o ext page

22 ( Let W ~ B(64, 05 The we have 64! 64 4 ( ( 05 Bayes factor part ( ( 05 PW= = =! 4! 4 The Normal approxmato to the dstrbuto of W s N(3, 6 Usg ths, PW ( = Φ Φ 4 4 =Φ( 375 Φ( 65 = = Bayes factor = 05

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