L(θ X) s 0 (1 θ 0) m s. (s/m) s (1 s/m) m s

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1 Hw 4 (due March ) 83 The LRT statstcs s λ(x) sup θ θ 0 L(θ X) The lkelhood s L(θ) θ P x ( sup Θ L(θ X) θ) m P x ad ad the log-lkelhood s (θ) x log θ +(m x ) log( θ) Let S X Note that the ucostraed MLE s S/m, whle the MLE uder the ull hy- f S/m θ0, pothess s m(s/m, θ 0 ) Thus, λ(s) θ s 0 ( θ 0) m s θ s 0 ( θ 0) m s else, ad we reject H 0 f <c To shows that ths s equvalet to rejectg f s<b, we could show λ(s) s decreasg s so that λ(s) <coccurs for s>b>mθ 0 It s easer to work wth log λ(s), ad we have log λ(s) s log θ 0 +(m s) log( θ 0 ) s log(s/m) (m s) log((m s)/m) ad d log λ(s) log ds θ0 ( m s m ) s/m( θ 0 ) For s/m > θ 0, s/m (m s)/m < θ 0, so the fracto above s less tha, ad the log s less tha 0 Thus, λ s a decreasg fucto s ad λ(s) <cf ad oly f s>b θ Alteratvely, oe could work through the algebra We reject f s 0 ( θ 0) m s <c Whch correspods to θ0 x log +(m s/m x ) log θ 0 s/m log c x log c m log θ 0 s/m log θ 0 log θ 0 s/m s/m b 87(a) L(θ, λ x) e (P x θ)/λ I [θ, )(x ()) λ whch s creasg θ f x () θ (regardless of λ) So, the MLE of θ s ˆθ x () The log L P + x ˆθ 0 ad thus ˆλ x x λ λ λ 2 () Because log L P 2 λ 2 λ 2 x ˆθ λ 3 ( x x () < 0, we have ˆθ x ) 2 () ad ˆλ x x () as the urestrcted MLEs of θ ad λ Uder the restrcto θ 0, the MLE of θ (regardless of λ) s ˆθ 0 f x () > 0, 0 For else x () 9

2 x () > 0, substtutg ˆθ 0 0 ad maxmzg the lkelhood wth respect to λ, asabove, f yelds ˆλ x() 0, 0 x Therefore, λ(x) f x () > 0 where L( x, 0 x) L(ˆλ, ˆθ x) L( x,0 x) L(ˆλ,ˆθ x) (/ x) e x/ x (/ˆλ) e ( x x ())/( x x () ) So reject f λ(x) c s equvalet to rejectg f x () / x c x () x 85 From the Neyma-Pearso lemma, the UMP test rejects H 0 f f(x σ ) f(x σ 0 ) (2πσ2 ) /2 e P x2 /(2σ2 ) (2πσ 2 0) /2 e P σ0 exp x 2 x2 /(2σ2 0 ) (/σ 2 0 /σ 2 σ 2 ) for some k 0 After some algebra, ths s equvalet to rejectg f x 2 > 2 log(k(σ /σ 0 ) ) (/σ 2 0 /σ 2 ) Ths s the UMP test of sze α, where α P σ0 ( X2 >c) Note that X2 /σ 2 0 χ 2 Thus, c α P σ0 ( X 2 /σ 2 0 >c/σ 2 0)P (χ 2 >c/σ 2 0) So we must have c/σ 2 0 χ 2,α ad c σ 2 0χ 2,α 87 a The lkelhood fucto s L(µ, θ x,y)µ ( x ) µ θ m m θ j j y ad the loglkelhood fucto s (µ, θ) log µ +(µ ) log x + m log θ +(θ ) m j log y j Hece, the (ucostraed) MLEs are ˆµ P ad ˆθ m log x Pj Uder H 0, the lkelhood log y j s L(θ x,y)θ +m x j y j θ ad maxmzg wrt θ yelds the restrcted MLE, θ0 ˆ ˆθ ˆθ +m (+m) 0 P log x + P θˆ Hece the LRT statstc s λ(x,y) 0 j log y j ˆµ ˆθ ( m x θ ) ˆ 0 ˆµ j y j b Substtutg the formulas for ˆθ, ˆµ ad ˆθ 0 yelds ( x θ ) ˆ 0 ˆµ ( j y θ j) ˆ 0 ˆθ, ad λ(x,y) +m θˆ 0 ˆµ ˆθ ˆ m θ 0 θ0 ˆ m ˆµ ˆθ m+ m m+ m m m ( T ) m T Ths s a umodal fucto of T So, rejectg H 0 f λ(x,y) c s equvalet to rejectg f T c or T c 2, where c ad c 2 are approprately chose costats 0

3 c Smple trasformatos yeld log X exp(/µ) ad log Y exp(/θ) Therefore, T W/(W + V ) where W ad V are depedet, W Gamma(, /µ) ad V Gamma(m, /θ) Uder H 0, the scale parameters of W ad V are equal The, a smple geeralzato of Exercse 49b yelds T Beta(, m) The costats c ad c 2 are determed by the two equatos P (T c )+P (T c 2 )α ad ( c ) m c ( c 2 ) m c 2 03 (a)(e) a The ull hypothess s H 0 : p p 2, whch we ca wrte as H 0 : p p 2 0 To costruct the Wald test statstc, look at T mle uder H ŝe 0 Uder H 0, the mle of p p 2 s ˆp ˆp 2 S / + S 2 / 2 The varace of ths uder H 0 s se 2 (p p 2 ) p ( p ) + p 2( p 2 ) 2 p( p)(/ +/ 2 ) where we estmate p p p 2 uder H 0 as ˆp S +S Thus, the Wald test statstc s T ˆp ˆp 2 ˆp( ˆp)(/ +/ 2 ) Next wrte T ˆp ˆp 2 se(p) se(p) We kow that ˆp ˆp 2 se(ˆp) se(p) d T N(0, ) ad se(p) se(ˆp) p Usg Slutsky s Theorem, we have that T d T N(0, ) uder H 0 Usg the cotuous mappg theorem (e, trasformato theorem covergece dstrbuto), we the have that T 2 d χ 2 e We have that ˆp 34/40, ˆp 2 9/35, ˆp , ad T 836 Sce χ , , we ca reject H 0 at α 005 The p-value of the test s a Frst calculate the MLEs uder p p 2 p We have L(p x) p x p x 2 p x 3 3 p x 2p 3 P m x p Takg logs ad dfferetatg yeld the followg equatos for the MLEs: logl p logl p x + x 2 2(m x ) p 2p 3 p x x p 2p 3 p 0 where 3,, The solutos are ˆp x +x 2, ˆp 2m x for 3,,, ad m ˆp (m x )/m Except for the frst ad secod cells, we have expected observed, 0

4 sce both are equal to x Note that ths comes from the statemet the problem that the expected cell frequeces are the MLEs of mp, uder the assumpto that p p 2 Thus, the expected frequecy of a cell s mˆp m x x m,for 3,, For the frst two cells, expected mˆp (x + x 2 )/2 Hece, (obs exp) 2 exp, reduces to (x (x + x 2 )/2) 2 (x + x 2 )/2 + (x 2 (x + x 2 )/2) 2 (x + x 2 )/2 (x x 2 ) 2 x + x 2 b Now the hypothess about the codtoal probabltes s gve by H 0 : P(chage-tal p agree) P(chage-tal dsagree) or, terms of the parameters, H 0 : p +p 3 p 2 p 2 +p 4 Ths s the same as testg p p 4 p 2 p 3 whch s ot the same as p p f(x α, β) x α e x/β Thus, L(β) e x /β Γαβ α ad (β) α log β β α x /β The S(β) (β) α/β + x /β 2 (/β)( x/β α) ad (β) α/β 2 2 x /β 3 (/β 2 )(α 2 x/β) E( (β)) α/β 2 a Solvg S(β) (β) 0forβ gves us ˆβ MLE x Note that α (ˆβ MLE ) α 3 / x 2 < 0 b se(ˆβ MLE ) /I(ˆβ MLE ) x 2 /(α 3 ) Thus, a Wald statstc for testg H 0 : β β 0 s x/α β 0 x 2 /(α 3 ) 038 We wll use the outputs from 036: A score statstc for testg H 0 : β β 0 s S(β 0 ) (/β 0 )( x/β 0 α) ( x αβ 0 ) I(β0 ) α/β 2 0 αβ 2 0 Addtoal HW The lkelhood fucto ad log lkelhood fucto for (θ, σ 2 )are L(θ, σ 2 )(2πσ 2 ) /2 exp (X θ) 2 2σ 2 (θ, σ 2 ) 2 log(2πσ2 ) (X θ) 2 2σ 2 a Uder H 0, (θ 0,σ 2 ) σ 2 b I the whole parameter space, (θ,σ2 ) + 2σ 2 (X θ 0 ) 2 0 Thus, σ 2 2(σ 2 ) 2 (X θ) θ (X ˆθ) 2 (X X) 2 Thus, ˆθ X ad ˆσ 2 c The lkelhood rato statstc, λ( X), s λ( X) L(θ 0, σ 2 0) (2π σ2 ) /2 exp L(ˆθ, ˆσ 2 ) (2πˆσ 2 ) /2 exp 0 ad (θ,σ2 ) σ 2 σ 2 ˆσ 2 σ 2 2ˆσ 2 P (X θ 0 ) 2 2 σ 2 P (X ˆθ) 2 P (X θ 0 ) 2 2σ 2 + (X θ) 2 2(σ 2 ) 2 0 /2 (X X) 2 (X θ 0 ) 2 /2 2

5 d We ca smplfy λ( X) the followg way: λ( X) (X X) 2 /2 (X (X X) 2 /2 θ 0 ) 2 (X X) 2 + ( X θ 0 ) 2 + ( X θ 0 ) 2 P (X X) 2 /2 + T 2 where T (X) X θ 0 Thus, λ( X) <cs equvalet to the followgs: S /2 λ(x) <c + T <c T (X) 2 >k 2 ( )(c 2/ ) T (X) > k (X)2 Thus, rejectg H 0 whe λ(x) <c correspods to rejectg H 0 whe T (X) > k e Uder H 0, T (X) T, T dstrbuto wth parameters (I beleve that you ca prove t, as t s related to stat 607) f The level α lkelhood rato test s as follows : rejectg H 0 whe λ(x) < c where α P ( 2 log λ(x) >c ) uder H 0 Here, λ(x) <cs equvalet to T (X) > kad T (X) T uder H 0 where α P ( T (X) > t,α/2 ) Thus, k t,α/2 /2 3