Ho OG o VS H i OE. sup o_ Nxt. sup. OE ou. Ho 0 00 vs H O Oo. Power function. nejeet Ho. Plo p O. Pole ERR KERR IERR.

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1 o OG o VS i OE Nxt sup E sup OE ou o_ 4 i o vs O Oo Power fuctio ejeet o Plo p O Pole ERR 9 4 KERR IERR Blot Eo 4

2 type I emor to whe reject e 3 true B O whe C o Type I error doot reject o che is true BIO whe C i Level d Test sup f lo C o Ed size test sup OE o fl O Mp uiformly most powerful test A test with power fuctio Glo is a UMP test if Glo z p O for all O E where p co is the power fuctio ofay other test restricted to a class of tests all level i tests

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6 8.3. Most powerful tests Defiitio: Let C be a class of tests for testig : : c, where c = \.AtestiCwith power fuctio ( ) isauiformly most powerful (UMP) class C test if ( ) ( ) for all c, where ( ) is the power fuctio of ay other test i C. The uiformly part i this defiitio refers to the fact that the power fuctio ( ) islargertha(i.e.,atleastaslarge as) the power fuctio of ay other class C test for all c. Importat: I this course, we will restrict attetio to tests That is, we will take C = {all level tests}. (x) thatarelevel tests. This restrictio is aalogous to the restrictio we made i the optimal estimatio problem i Chapter 7. Recall that we restricted attetio to ubiased estimators first; we the wated to fid the oe with the smallest variace (uiformly, for all ). I the same spirit, we make the same type of restrictio here by cosiderig oly those tests that are level tests. This is doe so that we ca avoid havig to cosider silly tests, e.g., (x) = forallx X. The power fuctio for this test is ( ) =,forall. This test caot be beate i terms of power whe is true! Ufortuately, it is ot a very good test whe is true. Recall: Atest (x) withpowerfuctio ( ) isalevel test if sup ( ) apple. That is, P (Type I Error ) cabeo larger tha for all. Startig poit: We start by cosiderig the simple--simple test: : = : =. Both ad specify exactly oe probability distributio. Remark: This type of test is rarely of iterest i practice. owever, it is the buildig block situatio for more iterestig problems. PAGE 84

7 Theorem 8.3. (Neyma-Pearso Lemma). Cosider testig : = : = ad deote by f X (x )adf X (x ) the pdfs (pmfs) of X =(X,X,...,X )correspodig to ad, respectively. Cosider the test fuctio 8 f X (x ) ><, f X (x ) >k (x) = f X (x ) >:, f X (x ) <k, for k, where = P (X R) =E [ (X)]. (8.) Su ciecy: Ay test satisfyig the defiitio of (x) above ad Equatio(8.)is a most powerful (MP) level test. Remarks: The ecessity part of the Neyma-Pearso (NP) Lemma is less importat for our immediate purposes (see CB, pp 388). I a simple--simple test, ay MP level test is obviously also UMP level. Recall that the uiformly part i UMP refers to all c. owever, i a simple,thereisolyoevalueof c. I choose to distiguish MP from UMP i this situatio (whereas the authors of CB do ot). Example 8.3. Suppose that X,X,...,X are iid beta(, ), where >; i.e., the populatio pdf is f X (x ) = x I( <x<). Derive the MP level test for : = : =. whereif O Oi If 3 Solutio. The pdf of X =(X,X,...,X )is,for<x i <, f X (x ) iid = Y x i =! Y x i. PAGE 85

8 Form the ratio f X (x ) f X (x ) = f Q X(x ) f X (x ) = ( x i) ( Q x i) = Y x i. The NP Lemma says that the MP level test uses the rejectio rejectio ( ) R = x X : Y x i >k where the costat k satisfies fh! Y = P = (X R) =P X i >k =. Istead of fidig the costat k that satisfies this equatio, we rewrite the rejectio rule { Q x i >k} i a way that makes our life easier. Note that Y x i >k () Y x i > k X () l x i < l( k)=k, say. We have rewritte the rejectio rule { Q x i >k} as { P l x i <k }.Therefore,!! Y X = P X i >k = = P l X i <k =. We have ow chaged the problem to choosig k to solve this equatio above. Q: Why did we do this? A: Because it is easier to fid the distributio of P l X i whe : =istrue. Recall that O O't I X i U(, ) =) l Xi expoetial() X =) l X i gamma(, ). I Tg.i a Therefore, to satisfy the equatio above, we take k = g,,,the(lower) quatile of a gamma(, ) distributio. This otatio for quatiles is cosistet with how CB have defied them o pp 386. Thus, the MP level test of : = : =hasrejectio regio R = ( x X : X l x i <g,, ) Special case: If =ad =.5, the g,, PAGE 86,. 7 It Xi k Rife 3 IT xi3h EsfxiTixisk Kaptur Game vases VS 3 5, 4 i trump for o O I l

9 Q: What is (), the power of this MP test (whe =ad =.5)? A: We calculate! Recall that () = P X l X i < 5.45 =. Therefore, X i beta(, ) =) l Xi expoetial(/) () = Z 5.45 =) X () gamma(, /) pdf l X i gamma(, /). u9 e u du.643. Proof of NP Lemma. We prove the su ciecy part oly. Defie the test fuctio 8 f X (x ) ><, f X (x ) >k (x) = f X (x ) >:, f X (x ) <k, where k ad = P (X R) =E [ (X)]; i.e., (x) isasize test. We wat to show that (x) ismplevel. Therefore,let (x) be the test fuctio for ay other level test of. Note that Thus, E [ (X)] = E [ (X)] apple. Defie E [ (X) b(x) =[ (x) (X)] = E [ (X)] = E [ (X)] apple (x)][f X (x ) kf X (x )].. We wat to show that b(x), for all x X. Case : Suppose f X (x ) kf X (x ) >. The, by defiitio, (x) =. Because apple (x) apple, we have b(x) =[ (x) (x)] [f X (x ) kf X (x )] >. PAGE 87

10 Case : Suppose f X (x ) kf X (x ) <. The, by defiitio, (x) =. Because apple (x) apple, we have b(x) =[ (x) {z (x)] [f } X (x ) kf X (x )] apple <. Case 3: Suppose f X (x ) kf X (x )=. Itistheobviousthatb(x) =. We have show that b(x) =[ (x) (x)][f X (x ) kf X (x )]. Therefore, [ (x) (x)]f X (x ) k[ (x) (x)]f X (x ) () [ (x) (x)]f X (x ) k[ (x) (x)]f X (x ). Itegratig both sides, we get Z [ (x) (x)]f X (x )dx R that is, E [ (X) (X)] Z k [ (x) R ke [ (X) (X)], show above (x)]f X (x )dx, Therefore, E [ (X) (X)] adhecee [ (X)] E [ (X)]. This shows that (x) is more powerful tha (x). Because (x) isaarbitrarylevel test, we are doe.. Corollary (NP Lemma with a su ciet statistic T ). Cosider testig : = : =, f him ffelo ad suppose that T = T (X) isasu cietstatistic. Deotebyg T (t )adg T (t )thepdfs (pmfs) of T correspodig to ad, respectively. Cosider the test fuctio 8 g T (t ) ><, g T (t ) >k (t) = g T (t ) >:, g T (t ) <k, for k, where, with rejectio regio S T, = P (T S) =E [ (T )]. The test that satisfies these specificatios is a MP level test. Proof. See CB (pp 39). PAGE 88

11 Implicatio: I search of a MP test, we ca immediately restrict attetio to those tests based o a su ciet statistic. Example 8.4. Suppose X,X,...,X are iid N (µ, ), where <µ< ad is kow. Fid the MP level test for : µ = µ : µ = µ, where µ <µ. Solutio. The sample mea T = T (X) =X is a su ciet statistic for the N (µ, )family. Furthermore, T N for t R. Formtheratio µ, g T (t µ ) g T (t µ ) = p / e p / e =) g T (t µ) = (t µ ) = e (t µ ) p / e (t µ), [(t µ ) (t µ ) ]. Corollary says that the MP level test rejects whe e b [(t µ ) (t µ ) ] >k () t< l k (µ µ ) (µ µ ) = k, say. Therefore, the MP level test uses the rejectio regio S = t T : g T (t ) g T (t ) >k = {t T : t<k }, where k satisfies = P µ (T<k ) = P Z< k µ / p =) k µ / p = z =) k = µ z / p. Therefore, the MP level test rejects whe X<µ z / p.thisisthesametestwe would have gotte usig f X (x µ )adf X (x µ ) with the origial versio of the NP Lemma (Theorem 8.3.). PAGE 89 trump test for o A Mo i Mc Mo

Lecture 6 Simple alternatives and the Neyman-Pearson lemma

Lecture 6 Simple alternatives and the Neyman-Pearson lemma STATS 00: Itroductio to Statistical Iferece Autum 06 Lecture 6 Simple alteratives ad the Neyma-Pearso lemma Last lecture, we discussed a umber of ways to costruct test statistics for testig a simple ull