Theory of Statistical Tests

Transcription

1 Ch 9. Theory of Statistical Tests 9.1 Certain Best Tests How to construct good testing. For simple hypothesis H 0 : θ = θ, H 1 : θ = θ, Page 1 of 100 where Θ = {θ, θ }

2 1. Define the best test for H 0 H NP Theorem 3. An example. Page 2 of 100

3 Use rejection region to define a test. Given C, if (x 1,, x n ) C then reject H 0, otherwise, accept H 0. The best rejection region C the best test. Page 3 of 100

4 Def µthe best rejection region C for the simple hypothesis test of level αµ For any subset A in the sample space satisfying P ((X 1,, X n A)) = α : Page 4 of 100

5 (a). P {(X 1,, X n ) C; H 0 } = α (b). P {(X 1,, X n C; H 1 )} P ((X 1,, X n ) A; H 1 ). Page 5 of 100

6 NP TheoremµX 1,, X n iid. f(x, θ), n L(θ) = f(x i, θ), i=1 where Θ = {θ, θ }. Let { } C = (x 1,, x n ) : L(θ ) L(θ ) k Page 6 of 100

7 C = { } (x 1,, x n ) : L(θ ) L(θ ) > k is a complement of C. α = P {(X 1,, X n ) C; H 0 }. Then C is a best rejection region of size α for testing the simple hypothesis H 0 H 1. Page 7 of 100

8 Proof: 0? = L(θ ) C C A L(θ A ) L(θ ) A C L(θ 1 L(θ ) 1 L(θ ) k C A k A C ) Page 8 of 100

9 = 1 k [ C L(θ ) A ] L(θ ) = 1 (α α) = 0. k Page 9 of 100

10 Example: X 1,, X n iid. N(θ, 1), and H 0 : θ = 0, H 1 : θ = 1. L(θ ) n L(θ ) = exp{ i=1 X i + n 2 } k 1 Xi 1 n 2 1 ln k =: c n Page 10 of 100

11 Take C = {(x 1,, x n ) : 1 n n x i c}. i=1 If α = 0.05, then c = 1.645/ n. P (C; H 0 ) = α, P (C; H 1 ) = P ( X 1 c 1; H 1 ) = 1 Φ( n(c 1)). Page 11 of 100

12 If n = 25, then P (C; H 1 ) = 1 Φ( ) = Φ(3.355) = Page 12 of 100

13 Generalization: H 0 and H 1 are simple hypothesis and X 1,, X n iid. L 0 = g(x 1,, x n ) is joint pdf. of X 1,, X n, L 1 = h(x 1,, x n ) is joint pdf. of X 1,, X n. Page 13 of 100

14 Then the critical region for H 0 H 1 is C = {(x 1,, x n ) : L 0 L 1 k} for some k > 0 with α = P ((x 1,, x n ) C; H 0 ). Page 14 of 100

15 9.2 Uniformly Most Powerful Tests (UMPT) H 0 H 1 Def.: C is a UMPT of α for the simple hypothesis H 0 composite hypothesis H 1 C is a best of α for test H 0 each simple hypothesis in H 1. Page 15 of 100

16 In general UMPT doesn t exist. If it exists, then NP theorem is used. Page 16 of 100

17 Example 2. Let X 1,, X n iid. N(0, θ) H 0 : θ = θ H 1 : θ > θ where Θ = {θ : θ θ }, the joint pdf. of X 1,, X n is ( )n 1 2 n L(θ) = exp{ i=1 x2 i }. 2πθ 2θ Page 17 of 100

18 For any θ > θ, k > 0, L(θ ) L(θ ) k ( θ )n 2 exp{ θ θ θ 2θ θ x 2 i } k x 2 i 2θ θ [ n ln(θ ) ln k] θ θ 2 θ =: c. Page 18 of 100

19 Therefore C = {(x 1,, x n ) : x 2 i c} is the best for H 0 H 1 : θ = θ, where c is determined by α. Page 19 of 100

20 X 2 P ( i c H 0 ) = α θ θ c = θ χ 2 n(α). Page 20 of 100

21 For H 1 : θ = θ, then C = {(x 1,, x n ) : x 2 i c} is the same and also the best, C is UMPT. Page 21 of 100

22 If n = 5, α = 0.05, θ = 3, H 0 : θ = 3, H 1 : θ > 3, then c = 3 χ 2 15(0.05) = 3 25 = 75. Page 22 of 100

23 Let X 1,, X n iid. f(x, θ) (θ Θ). Suppose that Y = u(x 1,, x n ) is a sufficient, then L(θ) = k 1 [u(x 1,, x n ); θ]k 2 (x 1,, x n ), Page 23 of 100

24 L(θ ) L(θ ) = k 1(u(x 1,, x n ); θ ) k 1 (u(x 1,, x n ); θ ) = k 1(y, θ ) k 1 (y, θ ). A best test or UMPT depends on u(x 1,, x n ) which is sufficient. Page 24 of 100

25 If k 1(y,θ ) is an increasing function of y k 1 (y,θ ) for θ < θ, then L(θ ) L(θ ) is called monotone likelihood ratio in the statistic Y = u(x 1,, x n ). Page 25 of 100

26 Example. X 1,, X n iid. N(θ, 1), if θ < θ, L(θ ) L(θ ) = n i=1 exp{ 1 2 (x i θ ) 2 } n i=1 exp{ 1 2 (x i θ) 2 } = exp{(θ θ ) x i n 2 θ 2 + n 2 θ 2 } Page 26 of 100 in x i.

27 Example. If f(x, θ) = exp{p(θ)k(x) + S(x) + q(θ)}, then L(θ ) L(θ ) = exp{(p(θ ) p(θ )) K(x i ) +n(q(θ ) q(θ ))}. Page 27 of 100

28 If p( ), then in Y = L(θ ) L(θ ) n K(x i ). i=1 Page 28 of 100

29 If we test H 0 : θ = θ, H 1 : θ < θ. For any θ < θ, we have L(θ ) L(θ ) k n K(x i ) c. i=1 This provides a UMPT. Page 29 of 100

30 If we test H 0 : θ = θ, H 1 : θ > θ, Contents then { K(x i ) c} is a UMPT, where c depends on α only. Page 30 of 100