x = Pr ( X (n) βx ) =

Transcription

1 Exercise 93 / page 45 The desity of a variable X i i 1 is fx α α a For α kow let say equal to α α > fx α α x α Pr X i x < x < Usig a Pivotal Quatity: x α 1 < x < α > x α 1 ad We solve i a similar way as i Example 98 page 48 We get a ituitio that the method of pivotal quatities may work for this exercise by observig that is a scale parameter ad X/ is a variable that its distributio does ot deped o The latter ca be easily see by cosiderig the distributio of the variable Y X/ The x y ad < y < 1 ad f Y y α f X y α d y α dy y α 1 α y α 1 for < y < 1 For a kow α X is a suciet statistic for that ca be foud by usig the Factorizatio Theorem So the pivotal quatity ca be based o the sample oly through X The cumulative distributio of X is Pr X x Pr X 1 x X x idepedet α x < x < Pr X i x idetically distributed [Pr X i x] or alteratively oe ca use Theorem 544 page 9 to derive the distributio of X The cumulative distributio of the pivot quatity QX X Pr QX x Pr X x Pr X x x is α x α < x < 1 1 We eed to d a upper codece limit for ie a iterval of the form X UX The lower boud i this specic case is X because the support of the desity is < x < Observe that QX X is decreasig fuctio of Thus for a iterval with codece coeciet 95 we eed to d a value b such that Pr b < QX 95 where Pr QX > b 1 Pr QX b 1 b α Thus ad the 95% codece iterval is 1 b α 95 b α 5 b 5 1/α C : x < & 5 1/α < Qx : x < < x or x 5 1/α 1 : x < & 5 1/α < x x 5 1/α

2 By Pivotig a cotiuous CDF: We use Theorem 91 ad solve i a similar way as i Example 913 page 433 For a kow α X is a suciet statistic for ad its cumulative distributio is F X x x α < x < as we foud above F X x is a decreasig fuctio of for each x We are iterested i oe-sided iterval that gives a upper codece boud Applyig Theorem 91 for α 5 the upper boud U x is deed as follows: F X x U x 5 x U x Thus the 95% codece iterval for is x By Ivertig a Test Statistic: α 5 U x x 5 1/α x 5 1/α We solve i a similar way as i Example 93 page 43 Sice we eed a 95% upper codece iterval we eed to ivert the acceptace regio of a 5% test of the hypothesis H : H 1 : < where the parameter space of is B X As a test statistic we will use the LRT which is of the form: where λx sup L x α sup B L x α L x α fx i α α α 1 α 1 xi α x α i x < From exercise 71 we kow that the MLE of is X Thus λx sup L x α sup B L x α L x α L x x α α α α x α x i α 1 x i α 1 x α Note that the LRT depeds o the sample oly through x which is a suciet statistic for The rejectio regio is R x : λx c x : x α c

3 For a 5% test we have 5 sup Pr X R Pr X R Pr X Pr c 1/α c [1/α ]α c X α c use eq1 Therefore the rejectio regio of the 5% test is R x : x α 5 the acceptace regio is A x : x α > 5 x : x > 5 1/α ad the 95% codece iterval is Cx : x < < x 5 1/α or x x 5 1/α Note that i this exercise the codece itervals derived by all three methods are the same However this is ot always the case b Based o the data give i exercise ˆα MLE 159 ad x 14 5 Thus the 95% codece iterval is 5 C : 5 < < : 5 < < 543 or / Exercise 94 / page 45 Note that i this exercise λ is used to deote the LRT while λ is the value of the ratio λ σ Y /σ X uder H a Let θ be the vector parameter θ σx σ Y The parameter space of θ is the two-dimesioal space Θ σx σ Y : σ X σ Y > Uder H the parameter space is Θ σx σ Y : σ Y λ σx σ X σ Y > The LRT statistic for the hypothesis is: λx y sup L σx σ Y x y θ Θ L σx σ Y x y sup θ Θ For the deomiator supl σx σ Y x y we have to d the MLE of σ X ad σ Y The likelihood θ Θ fuctio is 3

4 m LσX σy x y fx σxfy σ Y fx i σx fy i σy πσx 1/ e 1 σ X x m i πσy 1/ e 1 σ Y π +m/ σ X y i / σ Y m/ e 1 σ X x i e 1 σ Y y i The log-likelihood is: log Lσ X σ Y x y + m logπ log σ X m log σ Y 1 σ X x i 1 σ Y m yi ad its rst partial derivatives are: σ X log Lσ X σ Y x y σ X 1 + σx x i [ ] σ σx 4 X x i σ Y log Lσ X σ Y x y m σ Y 1 + σy m y i m σ 4 Y [ ] σy y i m Solvig the system σx σy log LσX σ Y x y log LσX σ Y x y i terms of σx ad σ Y we get: ˆσ X 1 x i ad ˆσ Y 1 m m yi Thus the poit ˆσ X ˆσ Y is a critical poit ad usig the secod derivatives we check whether it is maximum The secod partial derivatives are σ X log Lσ X σ Y x y σ Y log Lσ X σ Y x y σ 1 X σx 3 m σ 1 Y σy 3 x i m yi σ 6 X m σ 6 Y [ ] σx x i [ ] σy y i m σ X σ Y l Lσ X σ Y x y The values of log Lσ σx X σ Y x y ad log Lσ σy X σ Y x y at the critical poit are: 4

5 σ X log Lσ X σ Y x y σ X σ Y ˆσ X ˆσ Y σ 6 X < σ Y log Lσ X σ Y x y σ Y ˆσ Y m σ 6 Y < The Hessia matrix of log LσX σ Y x y evaluated at the poit ˆσ X ˆσ Y is H ˆσ X ˆσ Y σx 6 m σy 6 ad its determiat is positive Thus the poit ˆσ X ˆσ Y is the maximum poit of log Lσ X σ Y x y The the deomiator of λx y is supl σx σy x y L ˆσ X ˆσ Y x y π ˆσ +m/ X / m/ ˆσ Y e +m θ Θ Regardig the umerator of λx y uder H σy λ σx with σ X σ Y fuctio is modied as follows: > ad the likelihood L σ X σ Y x y Lσ X λ σ X x y π +m/ λ m/ σ X [ +m/ 1 e σ X x i + 1 ] m λ y i set LσX x y So sup L σx σy x y sup LσX x y θ Θ σx :σ > X The log-likelihood is: log Lσ X x y + m logπ m log λ + m [ log σx 1 x σx i + 1 λ yi ] ad its rst derivative is: d dσ X log Lσ X x y + m σ X + m σ 4 X [ 1 + x σx i + 1 λ σ X [ yi x i + 1 λ y i + m ] ] For d log Lσ dσx X x y we get: [ ˆσ 1 x i m λ yi ] 5

6 The secod derivative is: d d σ log X Lσ X x y + m σ 1 X σx 3 [ x i + 1 λ yi ] + m σ 6 X [ σ X x i + 1 λ y i + m ] ad evaluated at the critical poit it gets the value: d d σ X log Lσ X x y σ X ˆσ + m ˆσ < [ ] So the maximum occurs at ˆσ 1 +m x i + 1 m λ y i ad the umerator of λx y is sup L σx σy x y sup LσX x y Lˆσ x y θ Θ σx :σ > X Hece the LRT ca be writte as π +m/ λ m/ ˆσ +m/ e +m λx y π +m/ λ m/ ˆσ +m/ e +m π +m/ ˆσ X / ˆσ Y m/ e +m devide by m y i let W X Y let k +m/ X i Y i + m+m/ λ m/ / m m/ ˆσ X / ˆσ Y m/ λ m/ ˆσ +m/ 1 / 1 m/ x i m y i [ λ m/ 1 +m x i + ] +m/ 1 m λ y i + m +m/ x i y i / λ m/ / m m/ x i m y i + m +m/ w / +m/ λ m/ / m m/ w + 1 λ w / k +m/ w + 1λ set λw + 1 λ +m/ We get the idea to use the statistic W X Y above by the questio b of the exercise Note that λx y depeds o the sample poits x ad y oly through the statistics X i ad m Y i which are the suciet statistics for σx ad σ Y i the case of N σ X ad N σ Y respectively The rejectio regio is R x y : λx y c x y : λw c 6

7 where c is such that sup θ Θ Pr λw c α To see how the rejectio regio ca be expressed i terms of the values of W we will study w the mootoicity of the fuctio λw k / +m/ or equivaletly that of the fuctio w+ 1 λ log λw log k + +m log w log w + 1λ The rst derivative of the latter is d log λw dw w + m w + 1 w λ + m λ w + 1 λ w + λ w λ mw w λ w + 1 λ λ mw + w λ w + 1 For d dw log λw we get w /λ m > ad we have: w /λ m log λw + - log λw λw d dw Therefore the rejectio regio i terms of the values of W is We choose k 1 ad k such that R x y : W x y k 1 or W x y k with < k 1 < k sup Pr W x y k 1 or W x y k α θ Θ b Uder H X 1 X is a radom sample from N σx ad Y 1 Y m is aother radom sample idepedet to the rst oe from N λ σx The X i χ σx Y i λ χ σ m ad X the two statistics are idepedet to each other That meas that uder H the variable F X Y X i σ X Y i mλ σ X X i mλ m Y i W X Ymλ F m Note the the distributio of F X Y is idepedet of the parameters σ X ad σ Y Thus the rejectio regio ca be modied as follows: R x y : W x y k 1 or W x y k with < k 1 < k x y : F x y mλ k 1 or F x y mλ k with < k 1 < k x y : F x y c 1 or F x y c with < c 1 < c For a α level test we have α sup θ Θ Pr F X Y c 1 or F X Y c Pr F m c 1 or F m c 7

8 c 1 F mα1 ad c F m1 α where α 1 + α α c Based o questio b we get that the acceptace regio for the test is Aλ x y : F mα1 < F x y < F m1 α x y : F mα1 < W x y mλ < F m1 α Thus the 1 α codece iterval for the ratio λ is C x y λ : F mα1 mw x y < λ < mw x y F m1 α Exercise 95 / page 455 i Ivertig the acceptace regio of LRT statistic: To obtai a 1 α two-sided codece iterval for µ we ca ivert the acceptace regio of the LRT of level α of the hypothesis H : µ µ H 1 : µ µ where the parameter space of µ is M Y with Y X 1 > The parameter space uder H is M µ Sice Y X 1 is a suciet statistic for µ we ca use Theorem 84 ad the LRT is of the form: sup Lµ y µ M λy Lµ y where the likelihood is of the form sup µ M Lµ y exp y µ µ y The likelihood is a icreasig fuctio of µ Hece it reaches its maximum at the maximum possible value of µ with µ y ie for µ y Thus suplµ y Ly y exp y y µ M For the umerator sup µ M Lµ y we have sup Lµ y µ M Lµ y if µ y if µ > y Thus λy Lµ y Ly y exp y µ exp y µ if µ y if µ > y 8

9 The rejectio regio is R x : λy c x : exp y µ c Observe that λy exp y µ y µ is a decreasig fuctio of y Thus exp y µ c y c with c µ ad the rejectio regio ca be expressed i terms of the values of Y as follows: Sice the test is of level α it should hold R x : y c c µ α sup Pr X R µ sup Pr Y c µ Pr Y c µ µ M µ M ˆ ˆ fy µ dy exp y µ dy [ exp y µ ] c c c exp c µ log α c + µ c µ log α Note that the solutio c µ log α satises the coditio c µ sice a 1 ad log α > Thus the rejectio regio of level α LRT for the above hypothesis is R x : y µ log α The acceptace regio is Aµ ad the 1 α codece iterval is µ log α Cx µ : y < ii Usig a Pivotal Quatity: x : y < µ log α with µ y µ : y + log α < µ y To build a pivotal quatity a reasoable strategy is to d a suciet statistic for the parameter i questio ispect its distributio to see whether it is a locatio ad/or scale parameter distributio ad the cosider a variable which icorporates the suciet statistic ad the parameter i questio appropriately i a similar fashio as these variables suggested o page 47 of the book It is give that Y X 1 is suciet statistic for µ The distributio of Y is a locatio parameter distributio with µ beig the locatio parameter We cosider the quatity QX µ X 1 µ Y µ The distributio of QX µ is f Q q e q q > 9

10 The above ca be easily see by cosiderig the distributio of the trasformed variable Q Y µ Sice y µ the q ad also y q + µ Thus f Q q f Y q + µ d q + µ dq e q+µ µ 1 e q for q > Thus for a 1 α two-sided codece iterval we should d c 1 ad c such that Pr c 1 QX µ c 1 α with < c 1 < c or Pr QX µ < c 1 α 1 ad Pr QX µ > c α with α 1 + α α It is preferable to have two equatios sice we eed to determie two ukow quatities c 1 ad c It has ot bee set α 1 α α/ i order to give a geeral solutio ad also because the desity f Q q e q is ot symmetric ad α 1 Pr QX µ < c 1 ˆ c1 e c 1 1 α 1 c 1 log1 α 1 α Pr QX µ > c c log α ˆ Therefore the 1 α two-sided codece iterval is µ : log1 α 1 y µ log α µ : y + log α c e q [ e q] c 1 1 e c 1 e q [ e q] c e c µ y + log1 α 1 So the three codece itervals are: ˆ µ : y + log α ˆ < µ y based o ivertig the acceptace regio of LRT based o a pivotal quatity ad µ : y + log α µ y + log1 α 1 ˆ µ : y + 1 log α µ y + 1 log1 α by pivotig the CDF of Y foud i Example 913 Note that the rst codece iterval is a special case of the secod whe α 1 ad α α ad the third codece iterval is a special case of the secod whe α 1 α α/ A criterio of compariso of itervals of the same codece coeciet is the legth of the itervals The oe with the shorter legth is preferable The legths of the three itervals matioed above are: ˆ LRT: y log α y log α 1

11 ˆ pivotal quatity: y + log1 α 1 y + log α 1 log1 α 1 log α ad ˆ pivotig the CDF: y + 1 log1 α y + 1 log α 1 log1 α log α To compare the legths it suces to study the fuctio gα 1 log 1 α 1 log α α 1 α 1 α Observe that g log α ad gα/ log1 α log α d gα dα 1 1 α 1 α α 1 1 α 1 α 1 α α 1 > Thus gα 1 is a icreasig fuctio of α 1 reachig its miimum value for α 1 ad the miimum value is g log α Thus the codece iterval based o ivertig the acceptace regio of LRT is the oe with the shorter legth Exercise 934 / page 457 a σ kow We kow that a 1 α codece iterval for µ is µ : x z a/ σ µ x + z a/ σ see Example 91 page 4 ad the legth of the iterval is z a/ σ For a 95% codece iterval the legth is For a legth o more tha σ/4 we get 196 σ 39 σ 39 σ σ b σ ukow I such a case we kow that a 1 α codece iterval for µ is s s µ : x t a/ 1 µ x + t a/ 1 see Example 91 page 49 ad the legth of the iterval is t a/ 1 S For a 95% codece iterval the legth is t 5 1 S 11

12 fuc Figure 1: Fuctio h 1 64t 5 1 χ 9 1 The requiremet regardig the legth is: S Pr t 5 1 σ 9 4 We eed to express the above probability as a fuctio of so that we ca solve the iequality i terms of S Pr t 5 1 σ 4 Pr t S 5 1 Pr χ 1 σ t 5 1 1S Pr σ 1 64t 5 1 Therefore we get: Pr χ t t 5 1 χ t 5 1 χ 9 1 The above is solved umerically for example by tryig dieret values of For 76 we get t 575 χ > That is the smallest value of for which the iequality holds The graph of fuctio h miimum such that h 1 64t 5 1 χ 9 1 is show i Figure 1 We search for the S Commet 1: If the required probability was dieret Pr t 5 1 σ 9 the the 4 solutio for chages The less the probability the smaller the miimum value of is Figure shows the fuctio h whe the probability is set to ad 4 respectively Commet : We ca also solve the iequality 1 64t 5 1 χ 9 1 approximately usig the approximatios that for large a t 5 1 z ad b χ N Thus 1

13 fuctio Figure : Fuctio h whe the probability is ad 4 χ 1 N 1 1 ad χ z The we solve the iequality which is very close to the exact result 76 Exercise 935 / page 457 a I exercise 934 a we showed that for σ kow the legth of a 1 α codece iterval for σ µ is z a/ σ ad E z a/ σ z a/ b I exercise 934 b we showed that for σ ukow the legth of a 1 α codece iterval for µ is t a 1 S Sice the legth depeds o the variable S it is a variable itself The expected legth is S ES E t a 1 t a 1 where Γ ES cσ where c 1Γ regardig the value of c see for example the iformatio give i exercise 75 page 364 Therefore E t a 1 S t a 1 ES t a 1 cσ Thus the expected dierece i legth is equal to σ S σ cσ E z a/ t a 1 z a/ t a 1 σ za/ t a 1 c 13

14 because c 1 ad t a 1 z a/ whe I other words asymptotically the expected legth is Commet: A approximate result ca be derived by usig the approximatio ES σ ote that sice c 1 whe thus ES σ whe The E z a/ σ t a 1 S z a/ σ t a 1 σ σ za/ t a 1 For small values of ad for α 5 the quatities z a/ t a 1 c ad z 5 t 5 1 take the values preseted i the table below We see that the diereces betwee the values are fairly small ad decrease as the sample size icreases z 5 t 5 1 c z 5 t