2014/2015 Smester II ST5224 Final Exam Solution

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1 014/015 Smester II ST54 Final Exam Solution 1 Suppose that (X 1,, X n ) is a random sample from a distribution with probability density function f(x; θ) = e (x θ) I [θ, ) (x) (i) Show that the family of the joint distribution of (X 1,, X n ) is a monotone likelihood ratio family (ii) Let X (1) = min 1 i n X i and g(θ) = E θ ln(x (1) ), where E θ denotes the expectation Show that g(θ) is an nondecreasing function of θ (iii) Consider testing H 0 : θ θ 0 versus H 1 : θ > θ 0 Give the uniformly most powerful (UMP) test Provide your reason why the test you give is UMP (iv) The following is a sample of size 10 from the distribution: 467, 310, 461, 88, 315, 11, 89, 0, 97, 351 Carry out the UMP test with size α = 005 based on the above sample for H 0 : θ 5 versus H 1 : θ > 5 Solution: (i) The joint pdf of the sample is given by For any θ 1 < θ, f(x; θ) = e n i=1 X i+nθ I [θ, ) (X (1) ) f(x; θ ) f(x; θ 1 ) = I [θ, )(X (1) ) I [θ1, )(X (1) ) = { 0, X(1) θ, 1, X (1) > θ, which is a nondecresing function of X (1) Hence, the distribution has a monotone likelihood ration in X (1)

2 (ii) Since the family has a monotone likilihood ratio in X (1), by the property of monotone likelihood ratio family, for any nondecreasing function ψ(x (1) ), E θ ψ(x (1) ) is nondecreasing in θ Because ln is an increasing function, E θ ln(x (1) ) is nondecreasing in θ (iii) Since the family has a monotone likilihood ratio in X (1), by the property of monotone likelihood ratio family, the UMP test is given by T = 1, X (1) > c, γ X (1) = c, 0, X (1) < c, where c and γ are determined by E θ0 T = α Since X (1) has a continuous distribution, we can take γ = 0 in T (iv) Note that X (1) has pdf ne n(x θ) Hence E θ T = c ne n(x θ) dx = e n(c θ) and c = θ ln α/n When θ = 5, α = 005, n = 10, c = 5 ln(005)/10 = Since X (1) = 0 < c, the UMP test does not reject H 0

3 Let X 1,, X n be a random sample from the log-normal distribution with pdf 1 x 1 e 1 σ (ln x µ), x > 0 πσ Consider testing H 0 : µ = 0 versus H 1 : µ 0 (i) Let Y i = ln X i, Ȳ = 1 Y n i, and S = 1 (Y n 1 i Ȳ ) Show that the likelihood ratio test of size α rejects H 0 if and only if T t n 1,α/, where T = nȳ /S Y and t n 1,α/ is the upper α/ quantile of the t-distribution with degrees of freedom n 1 (ii) Derive the Wald Test statistic (iii) Derive the Score Test statistic Solution: (i) The MLEs under H 0 are ˆµ 0 = 0, ˆσ 0 = 1 [ln X n i ] = 1 n i The global MLEs are ˆµ = Ȳ, ˆσ = 1 (Y n i Ȳ ) The likelihood ratio is then ( ) ˆσ n/ λ = Let c be a generic constant Note that λ c ˆσ c Hence ˆσ = λ c nȳ i i nȳ i c = 1 nȳ nȳ i nȳ c, i nȳ since nȳ n is an increasing function of i=1 Y i nȳ we have λ c n Ȳ /S Y c n i=1 Y i Eventualy,

4 Under H 0, T has a t-distribution with df n 1 Hence c = t n 1,α/ (ii) The hypothesis µ = 0 has the form R(θ) = θ 1 = 0 with θ = (µ, σ ) Hence, C(θ) = ( R(θ)/ θ 1, R(θ)/ θ ) τ = (1, 0) τ Note that I n (θ) = ( n σ 0 0 n σ 4 The MLE (ˆµ, ˆσ ) = (Ȳ, n 1 n S Y ) Hence W n = [R(ˆθ)] τ {[C(ˆθ)] τ [I n (ˆθ)] 1 C(ˆθ)} 1 R(ˆθ) = nˆµ /ˆσ = n Ȳ (n 1)SY ) (iii) The score is given by (Y i µ) s n (θ) = (, n σ σ + (Y i µ) ) τ σ 4 The MLE under H 0 ( θ = (0, 1 n i ) Hence Y i s n (θ) = (, 0) ˆσ 0 and R n = [s n ( θ)] τ [I n ( θ)] 1 s n ( θ) = nȳ n nȳ = nȳ ˆσ 0

5 3 Suppose that for a certain species there are two alleles G and g in the population for a gene Thus the gene has three possible genotypes: GG, Gg and gg A sample of n individuals is drawn from the population The numbers of individuals with genotypes GG, Gg and gg are respectively N 1, N and N 3 Let θ and 1 θ be population allele frequencies of G and g respectively Let p = (p 1, p, p 3 ), where p 1, p and p 3 are the population genotype frequencies of GG, Gg and gg respectively (i) By using a χ -test statistic, give the decision rule for testing H 0 : p = p 0 = ( 1 4, 1, 1 4 ) versus H 1 : p p 0, with asymptotic significance level α (ii) Give the decision rule for testing H 0 : p = p(θ) = (θ, θ(1 θ), (1 θ) ) versus H 1 : p p(θ), with asymptotic significance level α by using the generalized χ -test (iii) Suppose that p = (θ, θ(1 θ), (1 θ) ) Give the decision rule of the likelihood ratio test with asymptotic significance level α for testing H 0 : θ = 1 versus H 1 : θ 1 Solution: (i) The χ statistic is χ = 3 j=1 (N j np j0 ) = (N 1 n/4) + (N n/) np j0 n/4 n/ (N 3 n/4) n/4 Let χ,α be the upper α quantile of the χ -distribution with df The decision rule is: Reject H 0 if and only if χ > χ,α (ii) The MLE of θ is given by ˆθ = N 1+N The generalized χ test n statistic is given by 3 χ [N j np j (ˆθ)] g = np j (ˆθ) j=1

6 The decision rule is: Reject H 0 if and only if χ g > χ 1,α, where χ 1,α is the upper α quantile of the χ -distribution with df 1 (iii) The likelihood function is proportional to θ N 1+N (1 θ) N 3+N The likelihood ratio λ = (1/) nˆθ (N 1 +N ) (1 ˆθ) (N 3+N ) = (1/) nˆθ nˆθ(1 ˆθ) n(1 ˆθ), where ˆθ = N 1+N The decision rule is: reject H n 0 if and only if ln λ > χ 1,α

7 4 Let (X 1,, X n ) be a random sample from a distribution having Lebesgue ( ) density γ γ 1 x I(0,θ) (x), where γ 1 and θ > 0 θ θ (i) Construct a confidence set for (γ, θ) with confidence coefficient 1 α using the cumulative distribution function of the largest order statistic X (n) (ii) Suppose that γ is KNOWN Obtain a confidence interval with confidence coefficient 1 α using a pivotal quantity (iii) Suppose that θ is KNOWN Obtain the uniformly most accurate (UMA) upper bound with confidence coefficient 1 α for γ Solution (i) The cumulative distribution function of X (n) is 0, t 0, F θ,γ (t) = (t/θ) nγ, 0 < t < θ, 1, t θ Since F θ,γ (X (n) ) has the uniform distribution on (0, 1), it is a pivot The confidence set can be determined as C(X (n) ) = {(θ, γ) : α 1 (X (n) /θ) nγ 1 α }, where α 1 + α = α (ii) Note that U(θ) = (X (n) /θ) nγ has the uniform distribution on (0, 1) and, hence, is a pivotal quantity The 1 α confidence interval constructed by solving which yields α 1 (X (n) /θ) nγ 1 α, X (n) (1 α ) 1/nγ θ X (n) α 1/nγ 1

8 (iii) The likelihood function of the sample can be writen as C(γ)e γ n i=1 ln(x i/θ), where C(γ) is a function of γ By the theory of UMP test, for any γ 0, the UMP test reject H 0 : γ γ 0 versus H 1 : γ > γ 0 if and only if n i=1 ln(x i /θ > c, where c is determined by P γ0 ( n i=1 ln(x i /θ) > c) = α Note that Y i = γ ln(x i /θ) has E(0, 1) distribution and hence n i=1 Y i follows a Gamma(n) (ie χ n) distribution Hence we can solve n P ( Y i < γ 0 c) = α, i=1 to obtain c = χ n,1 α/γ 0 UMP test is geven by or Thus the acceptance region of the n ln(x i /θ) χ n,1 α/γ 0 i=1 n γ 0 χ n,1 α/ ln(x i /θ) i=1 By converting the UMP test, we obtain the UMA upper confidence bound for γ as γ = χ n,1 α/ n i=1 ln(x i /θ) END OF PAPER