Lecture 16 November Application of MoUM to our 2-sided testing problem

Transcription

1 STATS 300A: Theory of Statistics Fall 2015 Lecture 16 November 17 Lecturer: Lester Mackey Scribe: Reginald Long, Colin Wei Warning: These notes may contain factual and/or typographic errors Recap Last time we discussed UMP unbiased tests for the null hypothesis H 0 : θ Ω 0 versus the alternative given by H 1 : θ Ω 1 and introduced the following key concepts: Definition 1. φ is an unbiased test at level α if β φ θ 0 α θ 0 Ω 0 and β φ θ 1 α θ 1 Ω 1. Definition 2. φ is α-similar if β φ θ = α θ ω, where ω = Ω 0 Ω 1. Lemma 1. TSH If β φ θ is continuous on Ω for all φ, and φ 0 is UMP amongst all α-similar level α tests, then φ 0 is UMPU at level α. Last time, we focused on two-sided UMPU tests for one parameter exponential families. Today, we will develop UMPU tests for multiparameter exponential families with nuisance parameters Application of MoUM to our 2-sided testing problem We continue our discussion of MoUM from last lecture, where our goal is to find an UMPU test. In this setting, H 0 : θ = θ 0. We will fix a simple alternative θ = θ θ 0 and hope that our best test has no θ dependence. We would like to maximize power φxp θ xdµx subject to φxp θ0 xdµx = α 16.1 φx d dθ p θ 0 xdµx = For a 1-parameter exponential family, we have p θ x = hxe θt x Aθ and 16.3 d dθ p θx = hxe θt x Aθ T x A θ = p θ x T x E θ [T X

2 By our discussion of MoUM from last lecture, we find that a most powerful test has rejection region defined by p θ x > k 1 p θ0 x + k 2 d dθ p θ 0 x for some values of k 1 and k 2, which is equivalent to e θ θ 0 T x k 1 + k 2T x > const with some rearranging. Now consider the set of values of T x satisfying this constraint. Because the constraint is that an exponential function exceeds a linear function, the set of values of T x satisfying this constraint is either a one-sided interval or all points outside a closed interval The first possibility will not give rise to an unbiased test, because the result would be a one-sided test with monotone power functions. Therefore any optimal φ is of the form 1 if T x > C 1 or T x < C 2 φx = γ i if T x = C i. 0 otherwise A simplification is possible if T x is symmetrically distributed under θ 0. Then the optimal test rejects whenever T x µ > const. Such tests are called equitailed tests

3 Example 1. Suppose X 1,..., X iid n N0, σ 2 with H 0 : σ = σ 0 and H 1 : σ σ 0. The optimal test has an acceptance region of the form i C 1 X2 i C σ0 2 2 The middle expression is a sufficient statistic that is χ 2 n distributed under H 0. How do we choose C 1 and C 2? Let f n be density χ 2 n. The level constraint is: P C 1 T X C 2 = C2 C 1 f n ydy = 1 α, and the derivative constraint 16.2 with substitution 16.4 gives E θ0 [T XφX = E θ0 [T xe θ0 [φx. On the left side φx = 1 on the complement of C 1, C 2, and on the right, the mean of a χ 2 n is n, and the level is α. Thus, C 1,C 2 c yf n y = nα. For χ 2 n distributions, yf n y = nf n+1 y, and the derivative constraint ultimately becomes C2 C 1 f n+2 y = 1 α. We have two integral equations, and we can solve them for the unknown boundaries C 1, C Optimal Unbiased Testing in Multiparameter Exponential Families Let the density of an exponential family with the natural parameters θ, λ 1,, λ k R k+1 be k P θ,λ x = exp θux + λ i T i x Aθ, λ hx We want to frame a test for the null hypothesis given by H 0 : θ θ 0 against the alternative given by H 1 : θ > θ 0 in the presence of nuisance parameters λ. On the boundary ω = {θ, λ : θ = θ 0 }, θ is known which would imply T = T 1, T 2,, T k is sufficient for λ. Hence the conditional distribution of X and subsequently that of UX T X has no λ dependence on ω. This point is worth repeating: conditioning can eliminate the influence of nuisance parameters! What is more, for each t, UX T X = t forms a one parameter exponential family. Here is a proof of this in the discrete case. 16-3

4 Proof. P θ,λ UX = u T X = t = exp θu + k λ it i x:t x=t,ux=u hx x:t x=t exp θux + k λ it i hx = c t θ exp θu gt, u These facts suggest the following strategy: 1. Condition on T X = t. 2. For each value of t, construct a best conditional test φu, t which maximizes conditional power E θ [φu, t T = t θ > θ 0 at conditional level α, i.e., E[φu, t T = t α θ θ Check whether this test is UMPU at level α. Since, for each t, the one-parameter exponential family distribution of UX T X = t has MLR in UX, our best test for a fixed t will be a one-sided test of the form 1 if u > ct φu, t = γt if u = ct. 0 if u < ct By the proof of the MLR theorem TSH Thm 3.4.1, if ct and γt are chosen to satisfy the conditional similarity constraint then it will also satisfy the conditional level constraint and the maximum conditional power property, that E θ0 φu, T T = t = α 16.6 E θ φu, T T = t α θ θ E θ φu, T T = t is maximized θ > θ 0, 16.8 among all tests that satisfy Since 16.7 and 16.8 hold for every t, we may take expectations and deduce that φu, T is level α and UMP amongst level α tests satisfying However, 16.6 is a more stringent requirement than α-similarity, so it remains to show that φu, T is also UMP amongst level α, α-similar tests. This would imply that φ is UMPU unbiased at level α. 16-4

5 16.4 Characterization of Similar Tests We will achieve our goal by developing a new characterization of similar tests based on conditioning on a complete sufficient statistic. We begin by assigning a name to property Definition 3. A test φ has α-neyman structure if T is sufficient for {P γ, γ ω} and φ satisfies EφX T X = α a.s. P γ for all γ ω, ω = Ω 0 Ω 1. Notice that if φ has α-neyman structure, then φ is also α-similar, since E γ φx = E γ EφX T X = α for all γ ω. Moreover, if T is complete and sufficient for {P γ : γ ω}, then every α-similar test has α-neyman structure with respect to T. Proof. Suppose φ is α-similar, and let ΨT = EφX α T, which has no γ dependence as T is sufficient. Then E γ ΨT = E γ φx α = 0, γ ω, which in turn implies that ΨT = 0 a.s. by completeness. Therefore, Neyman structure and similarity are equivalent whenever a complete sufficient statistic T exists. We now focus on exponential families, as they typically yield complete sufficient statistics on the boundary. For exponential family with density fx exp θux + i λ it i x hx. If T is complete on ω, then there exists an UMPU test for θ θ 0 vs. θ > θ 0 with form 1 if U > ct φu, T = γt if U = ct if U < ct with ct and γt chosen such that E θ0 [φu, T T = α. Why is this true? By the previous section, φ is UMP amongst level α tests satisfying 16.2, but 16.2 is equivalent to α-neyman structure for this testing scenario. This implies that φ is UMP amongst tests with α-neyman structure. T is complete and sufficient for P θ0, so every α-similar test has α-neyman structure w.r.t T. Hence, φ is UMP amongst α-similar tests. Finally, since an exponential family has continuous power functions, φ is UMPU at level α. Example 2. Suppose that X Poiν and Y Poiµ for X and Y independent. You might view X and Y as the number of successful recoveries from a disease under two different treatments. Our goal is to test the null hypothesis H 0 : µ ν against the alternative H 1 : µ > ν. Equivalently, we test H 0 : logµ/ν 0 against H 1 : logµ/ν > 0. This rewriting emphasizes our singular interest in the ratio µ/ν. Any additional information in µ, ν can be regarded as nuisance. The joint density of X, Y is given by exp ν µ exp x log ν + y log µ /x!y! which is proportional to: 1 exp y logµ/ν + x + y log ν x!y! The above is a 2-dimensional exponential family with sufficient statistics U = Y and T = X + Y. The natural parameters are θ = logµ/ν and λ = logν where λ acts as the nuisance parameter. 16-5

6 On the boundary, ω = {θ, λ : θ = 0}, U, T is a one-dimensional full-rank exponential family with density hu, t exp tλ. This implies that T is complete sufficient on ω. The conditional distribution of U T on ω is Binp = µ, n = T, so T determines the ν+µ number of coin flips, and the probability of each coin coming up heads is determined by µ and ν. The critical value ct is going to be determined at the boundary point, Binp = 1, n = T since θ = log µ = 0. 2 ν Example 3. X 1,..., X iid n Nµ, σ 2, where both µ and σ 2 are unknown. The joint density of X 1,..., X n is exp 1 x 2 2σ 2 i + µ x σ 2 i i Testing the variance We will consider testing H 0 : σ σ 0 vs. H 1 : σ < σ 0, which is equivalent to testing θ θ 0 vs. θ < θ 0, under the choices θ = 1, λ = µ, U = 2σ 2 σ 2 i x i 2, and T = i x i. The UMPU test rejects for small U, i.e., φx = I n X2 i C α X. To determine the critical value C α X, we notice that α = P σ0 Xi 2 C α X X = P σ0 X i X 2 C α X X = P σ0 X i X 2 C α Basu s theorem n = P σ0 X i X 2 σ0 2 }{{} χ 2 n 1 C α σ0 2 = P σ0 X 2 i n X 2 C α X n X 2 X i n X i X 2 X we can choose C α X constant w.r.t X implying that C α = σ0z 2 n 1,α where z n 1,α is the α quantile of χ 2 n 1. Therefore the UMPU test rejects if n X i X 2 σ0z 2 n 1,α. 16-6