ST5224: Advanced Statistical Theory II


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1 ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known population with probability density f j with respect to a σfinite measure ν, j = 0, 1. Let β(p ) be the power function of a UMP (uniformly most powerful) test of size α (0, 1). Show that α < β(p 1 ) unless P 0 = P 1. Solution: Suppose that α = β(p 1 ). Then the test T 0 α is also a UMP test by definition. By the uniqueness of the UMP test, we must have f 1 (x) = cf 0 (x) a.e. ν, which implies c = 1. Therefore, f 1 (x) = f 0 (x) a.e. ν, i.e., P 0 = P 1. 1
2 2. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known population with probability density f j with respect to a σfinite measure ν, j = 0, 1. For any α > 0, define 1 f 1 (X) > c(α)f 0 (X) T α (X) = γ(α) f 1 (X) = c(α)f 0 (X) 0 f 1 (X) < c(α)f 0 (X), where0 γ(α) 1, c(α) 0, E 0 [T α (X)] = α, and E j denotes the expectation with respect to P j. Show that (i) if α 1 < α 2, then c(α 1 ) c(α 2 ). (ii) if α 1 < α 2, then the type II error probability of T α1 is larger than that of T α2, i.e., E 1 [1 T α1 (X)] > E 1 [1 T α2 (X)]. Solution: (i) Assume α 1 < α 2. Suppose that c(α 1 ) < c(α 2 ). Then f 1 (x) c(α 2 )f 0 (x) implies that f 1 (x) > c(α 1 )f 0 (x) unless f 1 (x) = f 0 (x) = 0. Thus, T α1 (x) T α2 (x) a.e. ν, which implies that E 0 [T α1 (X)] E 0 [T α2 (X)]. Then α 1 α 2. This contradiction proves that c(α 1 ) c(α 2 ). (ii) Assume α 1 < α 2. Since T α1 is of level α 2 and T α2 is UMP, E 1 [T α1 (X)] E 1 [T α2 (X)]. The result follows if we can show that the equality can not hold. If E 1 [T α1 (X)] = E 1 [T α2 (X)], then T α1 is also UMP. From part (i), c(α 1 ) c(α 2 ). In the following, we show that E 1 [T α1 (X)] = E 1 [T α2 (X)] and c(α 1 ) c(α 2 ) are contradicting. Suppose c(α 1 ) > c(α 2 ). Then, T 2 = 0 when f 1 /f 0 < c(α 1 ) and T 1 = 1 when f 1 /f 0 > c(α 2 ) by the uniqueness of UMP test, which impies that γ(α 1 ) = 1 and γ(α 2 ) = 0. Thus which is impossible. E θ0 T 2 = P θ0 (f 1 /f 0 > c(α 1 )) α 1 < α 2, Suppose c(α 1 ) = c(α 2 ). Then, γ(α 1 ) < γ(α 2 ) since α 1 < α 2. This implies that A contradiction. E θ1 T 1 = P θ1 (f 1 /f 0 > c(α 1 )) + γ(α 1 )P θ1 (f 1 /f 0 = c(α 1 )) < E θ1 T 2, 2
3 3. Let X be a sample from a population P and P 0 and P 1 be two known populations. Suppose that T is a UMP test of size α (0, 1) for testing H 0 : P = P 0 versus H 1 : P = P 1 and that β < 1, where β is the power of T when H 1 is true. Show that 1 T is a UMP test of size 1 β for testing H 0 : P = P 1 versus H 1 : P = P 0. Solution: Let f j be a probability density for P j, j = 0, 1. By the uniqueness of the UMP test, { 1, f1 (X) > cf 0 (X), 0, f 1 (X) < cf 0 (X). Since α (0, 1) and β < 1, c must be a positive constant. Note that { 1, f0 (X) > c 1 1 f 1 (X), 0, f 0 (X) < c 1 f 1 (X). For testing H 0 : P = P 1 versus H 1 : P = P 0, clearly 1 T has size 1 β. The fact that it is UMP follows from the NeymanPearson Lemma. 3
4 4. Let X = (X 1,..., X n ) be a random sample from a distribution on R with Lebesgue density f θ, θ Θ = (0, ). Let θ 0 be a positive constant. Find a UMP test of size α for testing H 0 : θ θ 0 versus H 1 : θ > θ 0 when (i) f θ (x) = θ 1 e x/θ I (0, ) (x); (ii) f θ (x) = θx θ 1 I (0,1) (x); (iii) f θ (x) is the density of N(1, θ); (iv) f θ (x) = θ c cx c 1 e (x/θ)c I (0, ) (x), where c > 0 is known. Solution: (i) The family of densities has monotone likelihood ratio in T (X) = n i=1 X i, which has the Gamma distribution with shape parameter n and scale parameter θ. Under H 0, 2T/θ 0 has the chisquare distribution χ 2 2n. Hence, the UMP test is { 1, T (X) > θ0 χ 2 2n,α/2, 0, T (X) θ 0 χ 2 2n,α/2, where χ 2 2n,α is the (1 α)th quantile of the chisquare distribution χ 2 2n. (ii) The family of densities has monotone likelihood ratio in T (X) = n i=1 log X i, which has the gamma distribution with shape parameter n and scale parameter θ 1. Therefore, the UMP test is the same as T in part (i) of the solution but with θ 0 replaced by θ0 1. (iii) The family of densities has monotone likelihood ratio in T (X) = n i=1 (X i 1) 2 and T (X)/θ has the chisquare distribution χ 2 n. Therefore, the UMP test is { 1, T (X) > θ0 χ 2 n,α, 0, T (X) θ 0 χ 2 n,α. (iv) The family of densities has monotone likelihood ratio in T (X) = n i=1 Xc i which has the Gamma distribution with shape parameter n and scale parameter θ c. Therefore, the UMP test is the same as T in part (i) of the solution but with θ 0 replaced by θ c 0. 4
5 5. Let X = (X 1,..., X n ) be a random sample from the discrete uniform distribution on points 1,..., θ, where θ = 1, 2, (i) Consider H 0 : θ θ 0 versus H 1 : θ > θ 0, where θ 0 > 0 is known. Show that { 1 X(n) > θ 0 α X (n) θ 0 is a UMP test of size α. (ii) Consider H 0 : θ = θ 0 versus H : θ θ 0, Show that { 1 X(n) > θ 0 or X (n) θ 0 α 1/n 0 otherwise is a UMP test of size α. Solution: Without loss of generality we may assume that θ 0 is an integer. (i) Let P θ be the probability distribution of the largest order statistic X (n) and E θ be the expectation with respect to P θ. The family {P θ : θ = 1, 2,...} is dominated by the counting measure and has monotone likelihood ratio in X (n). Therefore, a UMP test of size α is T 1 (X) = 1, X (n) > c, γ, X (n) = c, 0, X (n) < c, where c is an integer and γ [0, 1] satisfying ( ) n c E θ0 (T 1 ) = 1 + γ cn (c 1) n θ 0 For any θ > θ 0, the power of T 1 is θ n 0 = α. E θ (T 1 ) = P θ (X (n) > c) + γp θ (X (n) = c) = 1 cn θ + γ cn (c 1) n n θ n = 1 (1 α) θn 0 θ n. On the other hand, for θ θ 0, the power of T is E θ (T ) = P θ (X (n) > θ 0 ) + αp θ (X (n) θ 0 ) = 1 θn 0 θ n + αθn 0 θ n. 5
6 Hence, T has the same power as T 1. Since sup E θ (T ) = sup αp θ (X (n) θ 0 ) = αp θ0 (X (n) θ 0 ) = α, θ θ 0 θ θ 0 T is a UMP test of size α. (ii) Consider H 0 : θ = θ 0 versus H 1 : θ > θ 0. The test T 1 in (i) is UMP. For θ > θ 0, E θ (T ) = P θ (X (n) > θ 0 ) + P θ (X (n) θ 0 α 1/n ) = 1 θn 0 θ n + αθn 0 θ n, which is the same as the power of T 1. Now, consider hypotheses H 0 : θ = θ 0 versus H 1 : θ < θ 0. The UMP test is 1, X (n) < d, T 2 (X) = η, X (n) = d, 0, X (n) > d, with E θ0 (T 2 ) = (d 1)n θ n 0 + η dn (d 1) n θ n 0 = α. For θ θ 0, E θ (T ) = P θ (X (n) > θ 0 ) + P θ (X (n) θ 0 α 1/n ) { } = P θ (X (n) θ 0 α 1/n ) = min 1, αθn 0. θ n On the other hand, the power of T 2 when θ θ 0 is E θ (T 2 ) = P θ (X (n) < d) + ηp θ (X (n) = d) = (d 1)n θ n + η dn (d 1) n θ n = α θn 0 θ n. Thus, we conclude that T has size α and its power is the same as the power of T 1 when θ > θ 0 and is no smaller than the power of T 2 when θ < θ 0. Thus, T is UMP. 6
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