S:194 Statistical Iferece II Homework Assigmets Luke Sprig 003
Assigmet 1 Problem 6.3 Problem 6.6 Due Friday, Jauary 31, 003. Problem 6.9 Problem 6.10 Due Friday, Jauary 31, 003. Problem 6.14 Problem 6.0 Due Friday, Jauary 31, 003. Solutios 6.3 The joit desity for the data is f(x 1,..., x µ, σ) 1 σ exp{ (x i µ)/σ} 1 (µ, ) (x i ) 1 { σ exp σ x + µ } 1 (µ, ) (x (1) ) σ So, by the factorizatio criterio, (X, X (1) ) is sufficiet for (µ, σ). 6.6 The desity of a sigle observatio is 1 f(x α, β) Γ(α)β α xα 1 e x/β 1 {(α Γ(α)β exp 1) log x 1β } x α This is a two-parameter expoetial family, so (T 1, T ) ( log X i, X i ) is sufficiet for (α, β). Sice X i is a oe-to-oe fuctio of log X i, the pair ( X i, X i ) is also sufficiet for (α, β). 6.9 a. Doe i class already. 1
b. Θ R, X R, ad Suppose x (1) y (1). The f(x θ) e x i +θ 1 (θ, ) (x (1) ) f(x θ) exp{ y i x i }f(y θ) for all θ. So k(x, y) exp{ y i x i } works. Suppose x (1) y (1). The for some θ oe of f(x θ), f(y θ) is zero ad the other is ot. So o k(x, y) exists. So T (X) X (1) is miimal sufficiet. c. Θ R, X R. The support does ot deped o θ so we ca work with ratios of desities. The ratio of desities for two samples x ad y is (xi θ) (1 + e (y i θ) ) (1 + e (x i θ) ) f(x θ) f(y θ) e e (y i θ) e xi e y i (1 + e y i e θ ) (1 + e x ieθ ) If the two samples cotai idetical values, i.e. if they have the same order statistics, the this ratio is costat i θ. If the ratio is costat i θ the the ratio of the two product terms is costat. These terms are both polomials of degree i e θ. If two polyomials are equal o a ope subset of the real lie the they are equal o the etire real lie. Hece they have the same roots. The roots are { e x i } ad { e y i } (each of degree ). If those sets are equal the the sets of sample values {x i } ad {y i } are equal, i.e. the two samples must have the same order statistics. So the order statistics (X (1),..., X () ) are miimal sufficiet. d. Same idea: f(x θ) f(y θ) (1 + (yi θ) ) (1 + (xi θ) ) If the two samples have the same order statistics the the ratio is costat. If the ratio is costat for all real θ the two polyomials i θ are equal o the complex plae, ad so the roots must be equal. The roots are the complex umbers θ x j ± i, θ y j ± i with i 1. So agai the order statistics are miimal sufficiet.
e. Θ R, X R. The support does ot deped o θ so we ca work with ratios of desities. The ratio of desities for two samples x ad y is f(x θ) f(y θ) exp{ y i θ x i θ } If the order statistics are the same the the ratio is costat. Suppose the order statistics differ. The there is some ope iterval I cotaiig o x i ad o y i such that #{x i > I} #{y i > I}. The slopes o I of x i θ ad y i θ as fuctios of θ are So y i θ x i θ has slope (#{x i > I}), (#{y i > I}) (#{x i > I} #{y i > I}) 0 ad so the ratio is ot costat o I. So agai the order statistic is miimal sufficiet. 6.10 To thos that the miimal sufficiet statistic is ot complete we eed to fis a fuctio g that is ot idetically zero but has expected value zero for all θ. Now E[X (1) ] θ + 1 + 1 E[X () ] θ + + 1 So g(x (1), X () ) X () X (1) 1 +1 idetically zero for > 1. has expected value zero for all θ but is ot 6.14 X 1,..., X are i.i.d. from f(x θ). This meas Z i X i θ are i.i.d. from f(z). Now So X X Z Z is acillary. 6.0 a. The joit desity of the data is X Z + θ X Z + θ f(x 1,..., x θ) ( xi ) 1 (0,θ) (x () ) 1 θ T X () is sufficiet (ad miimal sufficiet). X () has desity f T (t θ) t 1 1 θ 1 (0,θ)(t) 3
Thus for all θ > 0 meas 0 0 θ 0 θ 0 g(t) θ t 1 dt g(t)t 1 dt for almost all θ > 0, ad this i tur implies g(t)t 1 g(t) 0 for all t > 0. So T X () is complete. b. Expoetial family, T (X) log(1 + X i ), 0 ad hece {w(θ) : θ Θ} (1, ) which is a ope iterval. c. Expoetial family, T (X) X i, {w(θ) : θ Θ} {log θ : θ > 1} (0, ) which is a ope iterval. d. Expoetial family, T (X) e X i, {w(θ) : θ Θ} { e θ : θ R} (, 0) which is a ope iterval. e. Expoetial family, T (X) X i, {w(θ) : θ Θ} {log θ log(1 θ) : 0 θ 1} [, ] which cotais a ope iterval. 4
Assigmet Problem 7.6 Problem 7.11 Due Friday,February 7, 003. Problem 7.13 Problem 7.14 Due Friday, February 7, 003. Solutios 7.6 The joit PDF of the data ca be writte as a. X (1) is sufficiet. f(x θ) θ x i 1 [θ, ) (x (1) ) b. The likelihood icreases up to x (1) ad the is zero. So the MLE is θ X (1). c. The expected value of a sigle observatio is E θ [X] θ xθ 1 x dx θ θ 1 x dx So the (usual) method of momets estimator does ot exist. 7.11 a. The likelihood ad log likelihood are L(θ x) θ ( xi ) θ 1 log L(θ x) log θ + (θ 1) log x i The derivative of the log likelihood ad its uique root are d dθ L(θ x) θ + log x i θ log xi Sice log L(θ x) as θ 0 or θ ad the likelihood is differetiable o the parameter space this root is a global maximum. 5
Now log X i Expoetial(1/θ) Gamma(1, 1/θ). So log X i Gamma(, 1/θ). So [ ] θ E log Xi xγ() x 1 e θx dx ad So 0 Γ( 1) θ Γ() [ ( ) ] Γ( ) E θ log Xi Γ() Var( θ) θ as. 1 ( 1 1 ) 1 b. The mea of a sigle observatio is E[X] 1 0 θx θ dx So X θ θ + 1 is the method of momets equatio, ad θx + X θ θ(x 1) X θ X 1 X 1 θ ( 1)( ) θ θ ( 1) ( ) θ 0 θ θ + 1 We could use the delta method to fid a ormal approximatio to the distributio of θ. The variace of the approximate disrtibtio is larger tha the variace of the MLE. or or 7.13 The likelihood is L(θ x) 1 exp{ x i θ } We kow that the sample media X miimizes x i θ, so θ X. The miimizer is uique for odd. For eve ay value betwee the two middle order statistics is a miimizer. 6
7.14 We eed the joit desity of W, Z: P (W 1, Z [z, z + h)) P (X [z, z + h), Y z + h) + o(h) h 1 λ e z/λ e z/µ + o(h) h 1 λ e z( 1 λ + µ) 1 + o(h) ad, similarly, P (W 0, Z [z, z + h)) h 1 µ e z( 1 λ + 1 µ) + o(h) So ad therefore 1 f(w, z) lim h 0 h P (W w, Z [z, z + h)) 1 λ w µ e z( 1 1 w λ + 1 µ) Sice this factors, f(w 1,..., w, z 1,..., z λ, µ) 1 λ w iµ w i e zi ( 1 λ + 1 µ ) f(w 1,..., w, z 1,..., z λ, µ) 1 λ w i e zi /λ 1 µ w i e zi /µ it is maximized by maximizig each term separately, which produces λ zi wi µ zi w i I words, if the X i represet failure times, the λ total time o test umber of observed failures 7
Assigmet 3 Problem 7. Problem 7.3 Due Friday, February 14, 003. Problem 7.33 Due Friday, February 14, 003. Problem 7.38 Problem 7.39 Due Friday, February 14, 003. Solutios 7. We have a. The joit desity of X, θ is f(x, θ) f(x θ)f(θ) exp X θ N(θ, σ /) θ N(µ, τ ) { σ (x θ) 1 } (θ µ) τ This is a joit ormal distributio. The meas ad variaces are E[θ] µ E[X] E[θ] µ Var(θ) τ Var(X) Var(θ) + σ τ + σ The covariace ad correlatio are Cov(θ, X) E[Xθ] µ E[θ ] µ µ + τ µ τ ρ τ (τ + σ /)σ / b. This meas that the margial distributio of X is N(µ, τ + σ /). 8
c. The posterior distribuito of θ is { f(θ x) exp σ (X θ) 1 } (θ µ) τ {( exp σ x + 1 ) τ µ θ 1 ( σ + 1 ) } θ τ This is a ormal distributio with mea ad variace ( Var(θ X) σ + 1 ) 1 τ σ / τ τ + σ / ( E[θ X] Var(θ X) σ X + 1 ) τ µ τ τ + σ / X + σ / τ + σ / µ 7.3 We have S σ Gamma(( 1)/, σ /( 1)) ad f(σ ) 1 1 ) Γ(α)β α (σ ) α+1 e 1/(βσ The posterior distributio σ S is therefore f(σ s 1 ) (σ ) ( 1)/ e s ( 1)/(σ ) 1 ) (σ ) α+1 e 1/(βσ IG(α + ( 1)/, (1/β + ( 1)s /) 1 ) If Y IG(a, b), the V 1/Y Gamma(a, b). So So the posterior mea of σ is 1 1 E[Y ] E[1/V ] 0 v Γ(a)b a va 1 e v/b dv 1 z a e z dz bγ(a) 0 Γ(a 1) 1 bγ(a) b(a 1) E[σ S ] 1/β + ( 1)S / α + ( 3)/ 9
7.33 From Example 7.3.5 the MSE of p B is ( ) E[( p B p) p(1 p) p + α ] (α + β + ) + α + β + p ( p(1 p) ( /4 + /4 + ) + p + /4 p /4 + /4 + which is costat i p. 7.38 a. The populatio desity is p(1 p) + (p + /4 p( + )) ( + ) p(1 p) + ( /4 p ) ( + ) ( ( ) p(1 p) + (1/ p) + ) ( ( p p + 1/4 + p p) ) + ) /4 ( + ) θx θ 1 θx 1 e θ log x So T (X) 1 log X i is efficiet for τ(θ) E θ [log X 1 ]. τ(θ) b. The populatio desity is 1 log xθx θ 1 dx 0 0 yθe θy dy 1/θ log θ θ 1 θx log θ log θ ex θ 1 So log f(xi θ) (log log θ log(θ 1)) + xi log θ ) 7.39 Doe i class. θ log f(x i θ) ( 1 θ log θ 1 θ So X is efficiet for τ(θ) ( 1 log θ ) θ 1 + x θ θ ) θ 1 + x θ θ 1 θ 1 log θ ( ( θ x θ 1 1 )) log θ 10
Assigmet 4 Problem 7.44 Problem 7.48 Due Friday, February 1, 003. Problem 7.6 Problem 7.63 Problem 7.64 Due Friday, February 1, 003. Solutios 7.44 X 1,..., X are i.i.d. N(θ, 1). W X 1/ has E[W ] θ + 1 1 θ Sice X is sufficiet ad complete, W is the UMVUE of θ. The CRLB is Now (θ) I (θ) (θ) 4θ So E[X ] θ + 1 E[X 4 ] E[(θ + X/ ) 4 ] θ 4 + 4θ 3 1 E[Z] + 6θ 1 E[Z ] + 4θ 1 3/ E[Z3 ] + 1 E[Z4 ] θ 4 + 6θ 1 + 3 Var(W ) Var(X ) θ 4 + 6θ 1 + 3 θ4 1 θ 4 θ + > 4 θ 11
7.48 a. The MLE p 1 Xi has variace Var( p) p(1 p). The iformatio is [ I (p) E ( ) ) ] X i log(1 p) 7.6 a. ( Xi log p + θ )] X i 1 p [ ( Xi E θ p ( p p p ) (1 p) p + 1 p p(1 p) So the CRLB is p(1 p). b. E[X 1 X X 3 X 4 ] p 4. X i is sufficiet ad complete. E[W X i t] P (X 1 X X 3 X 4 1 X i t) { 0 t < 4 P (X 1 X X 3 X 4 1, 5 X it 4) P ( t 4 1 X it) 0 t < 4 p 4 ( 4 t 4)p t 4 (1 p) t t 4 So the UMVUE is (for 4) ( t)p t (1 p) t t(t 1)(t )(t 3) ( 1)( )( 3) p( p 1/)( p /)( p 3/) (1 1/)(1 /)(1 3/) No ubiased estimator eists for < 4. R(θ, ax + b) E θ [(ax + b θ) ] a Var(X) + (aθ + b θ) a σ + (b (1 a)θ) b. For η σ τ +σ, So δ π E[θ X] (1 η)x + ηµ R(θ, δ π ) (1 η) σ + (ηµ ηθ) (1 η) σ + η (µ θ) η(1 η)τ + η (µ θ) 1
c. B(π, δ π ) E[E[(θ δ π ) X]] E[E[(θ E[θ X]) X]] [ ] σ τ E[Var(θ X)] E σ τ σ + τ σ + τ ητ 7.63 From the previous problem, whe the prior mea is zero the risk of the Bayes rule is R(θ, δ π ) τ 4 + θ (1 + τ ) So for τ 1 R(θ, δ π ) 1 4 + 1 4 θ ad for τ 10 R(θ, δ π ) 100 11 + 1 11 θ With a smaller τ the risk is lower ear the prior mea ad higher far from the prior mea. 7.64 For ay a (a 1,..., a ) E[ L(θ i, a i ) X x] E[L(θ i, a i ) X x] The idepedece assumptios imply that (θ 1, X i ) is idepedet of {X j : j i} ad therefore. E[L(θ i, a i ) X x] E[L(θ i, a i ) X i x i ] for each i. Sice δ π i is a Bayes rule for estimatig θ i with loss L(θ i, a i ) we have E[L(θ i, a i ) X i x i ] E[L(θ i, δ π i (X i )) X i x i ] E[L(θ i, δ π i (X i )) X x] with the fial equality agai followig from the idepedece assumptios. So E[L(θi, a i ) X i x i ] E[L(θ i, δ π i (X i )) X x] ad therefore E[ L(θ i, δ π i (X i )) X x] E[ L(θ i, a i ) X x] E[ L(θ i, a i ) X i x i ] for all a, which implies that δ π (X) (δ π 1 (X 1 ),..., δ π (X )) is a Bayes rule for estimatig θ (θ 1,..., θ ) with loss L(θ i, a i ) ad prior π i (θ i ). 13
Assigmet 5 Problem 8.5 Problem 8.6 Due Friday, February 8, 003. Solutios 8.5 a. The likelihood ca be writte as L(θ, ν) θ ν θ 1 x θ+1 [ν, ) (x (1) ) i For fixed θ, this icreases i ν for ν x (1) ad is the zero. So ν x (1), ad L (θ) max L(θ, ν) θ ( ) x θ (1) 1 ν x i xi θ e θt So θ /T. b. The likelihood ratio criterio is Λ(x) L (1) L ( θ) e T ( T ) e cost T e T This is a uimodal fuctio of T ; it icreases from zero to a maximum at T ad the decreases back to zero. Therefore for ay c > 0 R {x : Λ(x) < c} {x : T < c 1 or T > c } c. The coditioal desity of X,..., X, give X 1 x ad X i X 1, is f(x 1 ) f(x ) f(x,..., x x 1, x i x 1 ) f(x 1 )P (X > X 1 X 1 x 1 ) P (X > X 1 X 1 x 1 ) f(x ) f(x ) P (X > x 1 ) P (X > x 1 ) ad P (X i > y) y θν θ νθ dx xθ+1 y θ 14
So f(x,..., x x 1, x i x 1 ) θ 1 Let Y 1 X i /x 1, i,...,. The x θ 1 1 x θ+1 {xi >x 1 } i i f Y (y,..., y x 1, x i > x 1 ) x 1 1 f(y x 1,..., y x 1 x 1, x i > x 1 ) θ 1 y θ+1,..., y θ+1 i.e. Y,..., Y are i.i.d. with desity θ/y θ+1, ad T log Y + + log Y. If Z log Y, the f Z (z) f Y (y) dy dz θ e (θ+1)z ez θe θz ad thus T {X 1 x 1, X i > X 1 } Gamma( 1, 1/θ). By symmetry, this meas that T X (1) Gamma( 1, 1/θ), which is idepetet of X (1), so T has this distributio ucoditioaly as well. For θ 1, T Gamma( 1, 1) T Gamma( 1, ) χ 1 8.6 a. The likelihood ratio criterio is ( ) +m +m Xi + Y i e m Λ ( ) ( ) m Xi e Yi e m ( + m)+m ( Xi ) ( Y i ) m m m ( X i + Y i ) +m The test rejects if this is small. b. The likelihood ratio criterio is of the form Λ cost T (1 T ) m. So the test rejects if T is too small or too large. c. Uder H 0, T Beta(, m). 15
Assigmet 6 Problem 8.14 Problem 8.17 Due Friday, March 7, 003. Problem 8.15 Problem 8.5 Due Friday, March 7, 003. Problem 8.8 Problem 8.33 Due Friday, March 7, 003. Solutios 8.14 Use R {x : x i > c}. α 0.01 meas 0.01 P ( X i > c p 0.49) P ( Z > c 0.49 ) 0.49 0.51 So c 0.49 0.49 0.51.36 β 0.99 implies 0.99 P ( X i > c p 0.51) P ( Z > c 0.51 ) 0.49 0.51 So So c 0.51 0.49 0.51.36 c 0.49.36 0.49 0.51 1 c 0.51.36 0.49 0.51 1 or 0.00.36 0.49 0.51 (100) (.36) 0.49 0.51 1351 16
8.17 For X 1,..., X i.i.d. Beta(µ, 1) L(µ x) µ x µ 1 i µ e µ log x i e log x i So ad Λ(x) µ log xi ( ) L( µ x) exp{ log x i } log xi ( ) +m +m log xi + log y i exp{ m log xi log y i } ( ) ( ) m log xi exp{ log xi } m log yi exp{ m log yi } So for suitable c 1, c. ( + m)+m T (1 T ) m m m {Λ < c} {T < c 1 or T > c } Uder H 0, log X i, log Y i are i.i.d. expoetial, so T Beta(, m). To fid c 1 ad c either cheat ad use equal tail probabilities (the right thig to do by symmetry if m), or solve umerically. 8.15 L(σ 1 x) exp (σ ) / ( ) L(σ1) / L(σ0) σ0 exp σ 1 { 1 } x σ i { 1 ( x 1 i 1 )} σ0 σ1 If σ 1 > σ the this is icreasig i x i. So L(σ1)/L(σ 0) > k for some k if ad oly if Xi > c for some c. Uder H 0 : σ σ 0, Xi /σ0 χ, so c σ 0χ,α 17
8.5 a. For θ > θ 1 { (x θ ) g(x θ ) exp g(x θ 1 ) { } cost exp{x(θ θ 1 )/σ } exp (x θ 1) σ σ } This i icreasig i x sice θ θ 1 > 0. b. For θ > θ 1 g(x θ ) g(x θ 1 ) θx e θ /x! θ x 1e θ 1 /x! cost which is icreasig x sice θ /θ > 1. c. For θ > θ 1 g(x θ ) g(x θ 1 ) ( ) x θ ( ) x θ s (1 θ ) x ( ) θ s 1 (1 θ 1 ) cost θ /(1 θ ) x θ 1 /(1 θ 1 ) ( x This is icreasig i x sice θ/(1 θ) is icreasig i θ. θ 1 ) x 8.8 a. Let f(x θ ) f(x θ 1 ) eθ 1 θ (1 + e x θ1 ) (1 + e x θ ) cost g(y) A + y B + y g (y) B + y A y (B + y) B A (B + y) ( e θ 1 + e x e θ + e x ) The g (y) 0 if B A. So we have MLR i x. b. Sice the ratio is icreasig i x, the most powerful test is of the form R {x > c}. Now 1 F θ (x) 1 1 + e x θ So for H 0 : θ 0 ad α 0. 1/(1 + e c ), so The power is β(1) 1 + e c 5 e c 4 c log 4 1.386 1 1 + e log(4) 1 1 1 + 4/e 0.405 c. Sice we have MLR, the test is UMP. This is true for ay θ 0. This oly works for 1; otherwise there is o oe-dimesioal sufficiet statistic. 18
8.33 a. b. P (Y 1 > k or Y > 1 θ 0) P (Y 1 > k θ 0) (1 k) α So k 1 α 1/. c. β(θ) P (Y > 1 or Y 1 > k θ) P (Y > 1 θ) + P (Y 1 > k ad Y 1) { 1 θ > 1 1 (1 θ) + (1 max{k, θ}) θ 1 f(x θ) 1 (θ, ) (Y 1 )1 (,θ+1) (Y ) Fix θ > 0. Suppose k θ. The β(θ ) 1. Suppose k > θ. Take k 1 i the NP lemma. The f(x θ ) < f(x θ 0 ) f(x θ ) > f(x θ 0 ) 0 < Y 1 < θ < k, so x R 1 < Y < θ + 1, so x R So R is a NP test for ay θ. So R is UMP. d. The power is oe for all if θ > 1. 19
Assigmet 7 Problem 8.31 Problem 8.34 Due Friday, March 14, 003. Problem 8.49 Problem 8.54 Due Friday, March 14, 003. Problem 8.55 Problem 8.56 Due Friday, March 14, 003. Solutios 8.31 a. The joit PMF of the data is For λ > λ 1, f(x λ) λ x i e λ xi! f(x λ ) f(x λ 1 ) ( λ λ 1 ) x i e (λ 1 λ ) is icreasig i x i, so has MLR. So a test which rejects the ull hypothesis if X > c is UMP of its size. b. If λ 1, the X AN(1, 1/), so c 1 + z α /. If λ, the X AN(, /), so P (X > 1 + z α / ( ( ) ) zα λ ) P Z > 1 ( P Z > z ) α For α 0.05, z α 1.645. For β() 0.9, z α z 0.1 1.8 0
so (z α + z 1 β ) (1.645 + 1.8) 11.7 so this suggest usig 1. It might be better to use the variace-stabilizig trasformatio X. Either way, use of the CLT is a bit questioable. 8.34 a. T f(t θ), T 0 f(t). So This is icreasig i θ. P (T > c θ) P (T 0 + θ > c) b. First approach: MLR implies that T > c is UMP of θ 1 agaist θ for size α P (T > c θ 1 ). Sice φ (t) α is size α ad β (t) E[φ (T ) θ] α for all θ, UMP implies that β(θ ) β (θ ) α β(θ 1 ). Secod approach: Let α P (T > c θ 1 ), h(t) g(t θ )/g(t θ 1 ). The P (T > c θ ) α E[φ(T ) α θ ] Ca also go back to the NP proof. E[(φ(T ) α)h(t ) θ 1] E[h(T ) θ 1 ] h(c)e[φ(t ) α θ 1] E[h(T ) θ 1 ] h(c)(α α) E[h(T ) θ 1 ] 0 8.49 a. The p-value is P (X 7 θ 0.5) 1 P (X 6 θ 0.5) 0.171875. This ca be computed i R with > 1 - pbiom(6,10,0.5) [1] 0.171875 b. The p-value is P (X 3 λ 1) 1 P (X λ 1) 0.0803014. This ca be computed i R with > 1 - ppois(, 1) [1] 0.0803014 c. If λ 1 the the sufficiet statistic T X 1 + X + X 3 has a Poisso distributio with mea λ T 3. The observed value of T is t 9, so the p-value is P (T 9 λ T 3) 1 P (T 8 λ T 3) 0.00110488. 1
8.54 a. From problem 7. the posterior distributio of θ x is ormal with mea ad variace So P (θ 0 x) P ( τ E[θ X x] τ + σ / x τ Var(θ X x) 1 + τ /σ ) Z (τ /(τ + σ /))x τ /(1 + τ /σ ) P ( Z ) τ (σ /)(τ + σ /) x b. The p-value is P (X x θ 0) P ( Z 1 ) σ/ x c. For τ σ 1 the Bayes probability is larger tha the p-value for x > 0 sice 1 (σ /)(1 + σ /) < 1 1/ d. As, τ (σ /)(τ + σ /) x 1 (σ /)(1 + σ /(τ )) x 1 σ/ ad therefore P (θ 0 x) coverges to the p-value. 8.55 The risk fuctios for these tests are give by { b(θ 0 θ)(1 β(θ)) for θ < θ 0 R(θ, δ) c(θ θ 0 )β(θ) for θ θ 0 { b(θ 0 θ)p (Z + θ > θ 0 z α ) for θ < θ 0 c(θ θ 0 )P (Z + θ θ 0 z α ) for θ θ 0 { b(θ 0 θ)(1 Φ(θ 0 θ z α ) for θ < θ 0 c(θ θ 0 )Φ(θ 0 θ z α ) for θ θ 0 where Z is a stadard ormal radom variable ad Φ is the stadard ormal CDF. 8.56 The risk fuctio for a test δ ad zero-oe loss is { 1 β(θ) for θ Θ 1 R(θ, δ) β(θ) for θ Θ 0
For the two tests i the problem this produces { 1 P (X 1 p) for p > 1/3 R(p, δ I ) P (X 1 p) for p 1/3 ad R(p, δ I ) A graph of the risk fuctios is { 1 P (Xge41 p) for p > 1/3 P (X 4 p) for p 1/3 r 0.0 0. 0.4 0.6 0.8 Test I Test II 0.0 0. 0.4 0.6 0.8 1.0 p Test II has lower risk for large ad small values of p; Test I has lower risk for p betwee 1/3 ad approximately 5.5. These risk fuctio graphs were produced i R usig b <- pbiom(1, 5, p) b1 <- pbiom(1, 5, p) b <- 1-pbiom(3, 5, p) r1<-ifelse(p < 1/3, b1, 1-b1) r<-ifelse(p < 1/3, b, 1-b) plot(p,r,type"l") lies(p,r1, lty ) leged(0.7,0.7,c("test I", "Test II"), ltyc(1,)) 3
Assigmet 8 Problem 9.1 Problem 9. Due Friday, March 8, 003. Problem 9.4 Due Friday, March 8, 003. Problem 9.1 Problem 9.13 Due Friday, March 8, 003. Solutios 9.1 Sice L U, Also So ad thus {L θ U} c {L > θ} {U < θ} {L > θ} {U < θ} P ({L θ U} c ) P (L > θ) + P (U < θ) α 1 + α P (L θ U) 1 α 1 α 9. This ca be iterpreted coditioally or ucoditioally. Give X 1,..., X, P (X +1 x ± 1.96/ x) P (Z x θ ± 1.96/ ) Equality holds oly if 1 ad x θ. Ucoditioally, P (Z 0 ± 1.96/ ) < 0.95 if > 1 0.95 if 1 P (X +1 X ± 1.96/ ) P (Z Z 0 ± 1.96/ ( ) P Z 1.96 0 ± 1 + 1/ P (Z 0 ± 1.96/ + 1) < 0.95 for 1 4
9.4 I this problem a. with X i are i.i.d. N(0, σ X) Y i / λ 0 are i.i.d. N(0, σ X) Λ ( ) (+m)/ +m X i + Yi /λ 0 ( ) / ( X i CT / (1 T ) m/ T 1 + 1 Y i λ 0 X i So Λ < k if ad oly if F < c 1 or F > c. ) m/ m Y i /λ 0 1 1 + m F b. F/λ 0 F m,. Choose c 1, c so c 1 F m,,1 α1, c F m,,α, α 1 + α α, ad f(c 1 ) f(c ) for ( ) / ( 1 f(t) 1 + mt 1 1 ) m/ 1 + mt c. This is a 1 α level CI. 9.1 All of the followig are possible choices: 1. (X θ)/s t 1. ( 1)S /θ χ 1 3. (X θ)/ θ N(0, 1) A(λ) {X, Y : c 1 F c } { } Y X, Y : c 1 i /m λ Xi / c { } Y C(λ) λ : i /m Y c X i / λ i /m c 1 X i / [ ] Y i /m Y c X i /, i /m c 1 X i / The first two produce the obvious itervals. For the third, look for those θ with z α/ < Q(X, θ) X θ θ/ < z α/ If x 0, the Q(x, ) is decreasig, ad the cofidece set is a iterval. 5
z α/ 0 L U θ -z α/ If x < 0, the Q(X, θ) is egative with a sigle mode. If x is large eough (close eough to zero, the the cofidece set is a iterval, correspodig to the two solutios to Q(x, θ) z α/. If x is too small, the there are o solutios ad the cofidece set is empty. 0 L U 0 -z α/ -z α/ 9.13 a. U log X exp(1/θ), θu exp(1). So P (Y/ < θ < Y ) P (1/ < θu < 1) e 1 e 1/ 0.39 6
b. θu exp(1). P ( log(1 α/) < θu < log(α/)) 1 α ( log(1 α/) P < θ < log(α/) ) 1 α U U [ log(1 α/)y, log(α/)y ] [0.479Y, 0.966Y ] c. The iterval i b. is a little shorter, though it is ot of optimal legth. b a 0.487 0.5 7
Assigmet 9 Problem 9.7 Problem 9.33 Due Friday, April 4, 003. Problem 10.1 Due Friday, April 4, 003. Solutios 9.7 a. The posterior desity is π(λ X) 1 λ e x i /λ 1 λ 1 e [1/b+ λ+a+1 a+1 e 1/(bλ) xi ]/λ IG( + a, [1/b + x i ] 1 ) The iverse gamma desity is uimodal, so the HPD regio is a iterval [c 1, c ] with c 1, c chose to have equal desity values ad P (Y > 1/c 1 ) + P (Y < 1/c ) α, with Y Gamma( + a, [1/b + x i ] 1 )..b The distributio of S is Gamma(( 1)/, σ /( 1)). The resultig posterior desity is therefore π(σ s (s ) ( 1)/ 1 1 ) (σ /( 1)) ( 1)/ e ( 1)s /σ ) (σ ) a+1 e 1/(bσ 1 ]/σ (σ e [1/b+( 1)s )( 1)/+a+1 IG(( 1)/ + a, [1/b + ( 1)s ] 1 ) As i the previous part, the HPD regio is a iterval that ca be determied by solvig a correspodig set of equatios. c. The limitig posterior distributio is IG(( 1)/, [( 1)s ] 1 ). The limitig HPD regio is a iterval [c 1, c ] with c 1 ( 1)s /χ 1,α 1 ad c ( 1)s /χ 1,1 α where α 1 + α α ad c 1, c have equal posterior desity values. 8
9.33 a. Sice 0 C a (x) for all a, x, P µ0 (0 C a (X)) 1 For µ < 0, P µ (µ C a (X)) P µ (mi{0, X a} µ) P µ (X a µ) P µ (X µ a) 1 α if a z α. For µ > 0, P µ (µ C a (X)) P µ (max{0, X a} µ) P µ (X + a µ) P µ (X µ a) 1 α if a z α. b. For π(µ) 1, f(µ x) N(x, 1). P (mi{0, x a} µ max{0, x a} X x) P (x a µ x + a X x) 1 α if a z α ad z α x z α. For a z α ad x > z α, P (mi{0, x a} µ max{0, x a} X x) P ( x Z a) P (Z z) 1 α as x. 10.1 The mea is µ θ/3, so the method of momets estimator is W 3X. By the law of large umbers X P µ θ/3, so W 3X P θ. 9
Assigmet 10 Problem 10.3 Problem 10.9 (but oly for e λ ; do ot do λe λ ) Due Friday, April 11, 003. Problem: Fid the approximate joit distributio of the maximum likelihood estimators i problem 7.14 of the text. Due Friday, April 11, 003. Problem: I the settig of problem 7.14 of the text, suppose 100, W i 71, ad Z i 780. Also assume a smooth, vague prior distributio. Fid the posterior probability that λ > 100. Due Friday, April 11, 003. Solutios 10.3 a. The derivative of the log likelihood is ( ) θ log θ (Xi θ) θ θ + (Xi θ) + (Xi θ) 1θ θ θ + X i θ θ W θ θ θ So the MLE is a root of the quadratic equatio θ + θ W 0. The roots are θ 1, 1 ± 1 4 + W The MLE is the larger root sice that represets a local maximum ad sice the smaller root is egative. b. The Fisher iformatio is So θ AN(θ, θ (θ+1/) ). I (θ) E E[W ] θ 3 [ ( )] θ + θ W θ θ ] θ/ E[W θ θ 3 θ + θ θ/ θ + 1/ θ 3 θ 30
10.9 (but oly for e λ ; do ot do λe λ ) The UMVUE of e λ is V (1 1/) X ad the MLE is e X. Sice (V e X ) O(1/) O(1/ ) both (V e λ ) ad (e X e λ ) have the same ormal limitig distributio ad therefore their ARE is oe. I fiite samples oe ca argue that the UMVUE should be preferred if ubiasedess is deemed importat. The MLE is always larger tha the UMVUE i this case, which might i some cotexts be a argumet for usig the UMVUE. A compariso of mea square errors might be useful. For the data provided, e X 0.0009747 ad V 0.0007653. Problem: Fid the approximate joit distributio of the maximum likelihood estimators i problem 7.14 of the text. Solutio: The log-likelihood is So l(λ, µ) w i log λ ( w i ) log µ z i ( 1 λ + 1 µ wi l(λ, µ) λ λ + λ wi l(λ, µ) + µ µ wi l(λ, µ) λ λ λ 3 wi l(λ, µ) µ µ l(λ, µ) 0 λ µ E[W i ] µ λ + µ E[Z i ] λµ λ + µ zi zi zi µ zi µ 3 ) So ] E [ l(λ, µ) λ ] E [ l(λ, µ) µ λ µ λ + µ λ µ λ + µ λ µ λ + µ µ λ λ + µ 31
ad thus ] [ λ AN µ ( [λ ], µ [ λ (λ+µ) 0 µ 0 µ (λ+µ) λ ]) Problem: I the settig of problem 7.14 of the text, suppose 100, W i 71, ad Z i 780. Also assume a smooth, vague prior distributio. Fid the posterior probability that λ > 100. Solutio: The MLE s are Zi λ 109.89 Wi µ Zi W i The observed iformatio is Î ( λ, µ) [ Wi 0 λ 0 W i µ Thus the margial posterior distributio of λ is approximately N( λ, λ / W i ) N(109.89, 170.07) ] So P (λ > 100 X) P ( Z > 100 109.89 170.07 ) 0.7758 3
Assigmet 11 Problem: Let X 1,..., X be a radom sample from a Pareto(1, β) distributio with desity f(x β) β/x β+1 for x 1. Fid the asymptotic relative efficiecy of the method of momets estimator of β to the MLE of β. Due Friday, April 18, 003. Problem: Let X 1,..., X be i.i.d. Poisso(λ) ad let W e X. Fid the parametric bootstrap variace Var (W ) ad show that Var (W )/Var(W ) P 1 as. Due Friday, April 18, 003. 1. Let X 1,..., X be a radom sample that may come from a Poisso distributio with mea λ. Fid the sadwich estimator of the asymptotic variace of the MLE λ X.. Let g(x) e x for x > 0 be a expoetial desity with mea oe ad let f(x θ) be a N(θ, 1) desity. Fid the value θ correspodig to the desity of the form f(x θ) that is closest to g i Kullback-Liebler divergece. Due Friday, April 18, 003. Solutios Problem: Let X 1,..., X be a radom sample from a Pareto(1, β) distributio with desity f(x β) β/x β+1 for x 1. Fid the asymptotic relative efficiecy of the method of momets estimator of β to the MLE of β. Solutio: The mea is fiite ad E[X] β/(β 1) if β > 1. So the method of momets estimator is β MM X/(X 1) if X > 1 ad udefied otherwise. β The variace is fiite ad Var(X) if β >. So for β > the (β 1) (β ) cetral limit theorem implies that ( ) (X β/(β 1)) D β N 0, (β 1) (β ) Sice β MM g(x) with g(x) x/(x 1) ad g (x) 1/(x 1), we have g (β/(β 1)) (β 1) ad the delta method shows that ( ) ( βmm β) D N 0, g (β/(β 1)) β N(0, β(β 1) /(β )) (β 1) (β ) 33
The MLE is β /( log X i ) ad the Fisher iformatio is I (β) /β. So the asymptotic relative efficiecy of the method of momets estimator to the MLE is ARE( β MM, β) β β(β 1) /(β ) β(β ) (β 1) for β >. For β the method of momets estimator will exist for large but will ot he -cosistet; it makes sese to say the asymptotic relative efficiecy is zero i this case. Problem: Let X 1,..., X be i.i.d. Poisso(λ) ad let W e X. Fid the parametric bootstrap variace Var (W ) ad show that Var (W )/Var(W ) P 1 as. Solutio: Usig the MGF of the Poisso distributio we have The variace of W is therefore E[W λ] M X i ( 1/) exp{λ(e 1/ 1)} E[W λ] M X i ( /) exp{λ(e / 1)} Var(W λ) E[W λ] E[W λ] exp{λ(e / 1)} exp{λ(e 1/ 1)} exp{λ(e 1/ 1)}(exp{λ(e / e 1/ + 1)} 1) exp{λ(e 1/ 1)}(exp{λ(e 1/ 1) } 1) λb g(a λ, b λ) with g(x, y) { e x ey 1 y if y 0 e x if y 0 a (e 1/ 1) The bootstrap variace is b (e 1/ 1) a Var (W ) Var(W λ X) exp{x(e 1/ 1)}(exp{X(e 1/ 1) } 1) Xb g(a X, b X) Now g is cotiuous, a 1 ad b 0. So by the law of large umbers, Slutsky s theorem, ad the cotiuous mappig theorem Var (W ) Var(W λ) Xg(a X, b X) λg(a λ, b λ) 34 P λg(λ, 0) λg(λ, 0) 1
Problem: 1. Let X 1,..., X be a radom sample that may come from a Poisso distributio with mea λ. Fid the sadwich estimator of the asymptotic variace of the MLE λ X.. Let g(x) e x for x > 0 be a expoetial desity with mea oe ad let f(x θ) be a N(θ, 1) desity. Fid the value θ correspodig to the desity of the form f(x θ) that is closest to g i Kullback-Liebler divergece. Solutio: 1. For the Poisso distributio X λ log f(x λ) λ λ λ log f(x λ) X λ So the sadwich estimator of the asymptotic variace of the MLE is Var( ( λ λ) 1 ( 1 ) ( log f(x λ i λ) ) 1 (Xi X) log f(x λ i λ). The Kullback-Liebler divergece from ay distributio with desity g to a N(θ, 1) distributio is KL(g, f) log g(x) f(x θ) g(x)dx cost + 1 (x θ) g(x)dx This is miimized by θ E g [X]; for this particular choice of g this meas θ 1. 35
Assigmet 1 Problem 10.30 (b) Due Friday, April 5, 003. Problem: Cosider the settig of Problem 10.31. Derive a expressio for log Λ, where Λ is the likelihood ratio test statistic, ad fid the approximate distributio of this quatity uder the ull hypothesis. Due Friday, April 5, 003. Problem 10.38 Due Friday, April 5, 003. Solutios 10.30 (b) For the Huber M-estimator ψ( ) k ad ψ( ) k, so η 1/(1+1) 1/ ad the breakdow is 50%. The formula for the breakdow give i this problem is oly applicable to mootoe ψ fuctios. For redescedig ψ fuctios the estimatig equatio eed ot have a uique root. To resolve this oe ca specify that a estimator should be determied usig a local root fidig procedure startig at, say, the sample media. I this case the M-estimator iherits the 50% breakdow of the media. See Huber, pages 53 55, for a more complete discussio. Problem: Cosider the settig of Problem 10.31. Derive a expressio for log Λ, where Λ is the likelihood ratio test statistic, ad fid the approximate distributio of this quatity uder the ull hypothesis. Solutio: The restricted likelihood correspods to 1 + Beroulli trials with S 1 + S successes ad commo success probability p, so the MLE of p is p (S 1 +S )/( 1 + ). The urestricted likelihood cosists of two idepedet sets of Beroulli trials with success probabilities p 1 ad p, ad the correpsodig MLS s are p 1 S 1 / 1 ad p S /. The likelihood ratio statistic is therefore p S 1+S (1 p) F 1+F ( ) S1 ( ) S ( ) F1 ( ) F p p 1 p 1 p Λ ad p 1 S 1 (1 p 1 ) F 1 p S (1 p ) F p 1 p 1 p 1 1 p ( log Λ S 1 log p 1 p + S log p p + F 1 log 1 p 1 1 p + F log 1 p ) 1 p 36
The restricted parameter space uder the ull hypothesis is oe-dimesioal ad the urestricted parameter space is two-dimesioal. Thus uder the ull hypothesis the distributio of log Λ is approximately χ 1. 10.38 The log likelihood for a radom sample from a Gamma(α, β) distributio is ( log L(β) log Γ(α) α log β + (α 1) 1 ) log Xi X/β So the score fuctio is V (β) log L(β) β ad the Fisher iformatio is [ α I (β) E β X ] β 3 So the score statistic is V (β) I (β) X αβ αβ ( αβ + Xβ ) x αβ β αβ β 3 α β α β X αβ αβ 37
Assigmet 13 Problem 10.41 Due Friday, May, 003. Problem: Let x 1,..., x be costats, ad suppose Y i β 1 (1 e β x i ) + ε i with the ε i idepedet N(0.σ ) ramdom variables. a. Fid the ormal equatios for the least squares estimators of β 1 ad β. b. Suppose β is kow. Fid the least squares estimator for β 1 as a fuctio of the data ad β. Due Friday, May, 003. Problem: Let x 1,..., x be costats, ad suppose Let y be a costat ad let let x satisfy Y i β 1 + β x i + ε i y β 0 + β 1 x that is, x is the value of x at which the mea respose is y. a. Fid the maximum likelihood estimator x of x. b. Use the delta method to fid the approximate samplig distributio of x. Due Friday, May, 003. Solutios 10.41 This problem should have stated that r is assumed kow. a. The log likelihood for p is log L(p) cost + (r log p + x log(1 p)) 38
The first ad secod partial derivatives with respoct to p are So the Fisher iformatio is I (p) r p p r p x 1 p p r p x (1 p) x 1 p r p (1 p) r p (1 p) ad the score test statistic is ( ) (1 p)r + px r 1 p b. The mea is µ r(1 p)/p. The score statsitic ca be writte i terms of the mea as ( ) (1 p)r + px µ x r 1 p µ + µ /r A cofidece iterval is give by { } µ x C µ : µ + µ /r z α/ The edpoits are the solutios to a quatriatic, r(8x + zα/ rz ) ± α/ 16rx + 16x + rzα/ U, L 8r zα/ To use the cotiuity corectio, replace x with x + 1 poit ad x 1 for the lower ed poit. Problem: Let x 1,..., x be costats, ad suppose for the upper ed Y i β 1 (1 e β x i ) + ε i with the ε i idepedet N(0.σ ) ramdom variables. a. Fid the ormal equatios for the least squares estimators of β 1 ad β. b. Suppose β is kow. Fid the least squares estimator for β 1 as a fuctio of the data ad β. Solutio: 39
a. The mea respose is µ i (β) β 1 (1 e β x i ). So the partial derivatives are So the ormal equatios are β 1 µ i (β) 1 e β x i β µ i (β) β 1 (1 e β x i )x i β 1 (1 e β x i ) i1 β1(1 e β x i ) x i i1 (1 e β x i )Y i i1 β 1 (1 e β x i )x i Y i b. If β is kow the the least squares estimator for β 1 ca be foud by solvig the first ormal equatio: i1 β 1 i1 (1 e β x i )Y i i1 (1 e β x i) Problem: Let x 1,..., x be costats, ad suppose Y i β 1 + β x i + ε i Let y be a costat ad let let x satisfy y β 0 + β 1 x that is, x is the value of x at which the mea respose is y. a. Fid the maximum likelihood estimator x of x. b. Use the delta method to fid the approximate samplig distributio of x. Solutio: This prolem should have explicitly assumed ormal errors. a. Sice x (y β 1 )/β, the MLE is by MLE ivariace. x y β 1 β b. The partial derivatives of the fuctio g(β 1, β ) (y β 1 )/β are g(β 1, β ) 1 β 1 β g(β 1, β ) y β 1 β β 40
So for β 0 the variace of the approximate samplig distributio is Var( x ) gσ (X T X) 1 g T x i x y β 1 1 β β 3 (xi x) + (y β 1 ) β 4 1 β (xi x) + (y β 1 β x) β 4 (xi x) 1 β + (y β 1 β x) β 4 (xi x) So by the delta method x AN(x, Var( x )). The approximatio is reasoably good if β is far from zero, but the actual mea ad variace of x do ot exist. 41