Hypothesis Tests - Mathematical Statistics - Lecture Slides | STAT 710, Exams of Mathematical Statistics

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Stat 710: Mathematical Statistics

Lecture 21

Jun Shao

Department of Statistics University of Wisconsin Madison, WI 53706, USA

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Chapter 6. Hypothesis Tests

Lecture 21: UMP tests and Neyman-Pearson lemma

Theory of testing hypotheses

X : a sample from a population P in P, a family of populations. Based on the observed X , we test a given hypothesis H 0 : P ∈ P 0 vs H 1 : P ∈ P 1 where P 0 and P 1 are two disjoint subsets of P and P 0 ∪ P 1 = P. A test for a hypothesis is a statistic T ( X ) taking values in [ 0 , 1 ]. When X = x is observed, we reject (or accept) H 0 with probability T ( x ) (or 1 − T ( x )). If T ( X ) = 1 or 0 a.s. P, then T ( X ) is a nonrandomized test; otherwise T ( X ) is randomized. For a given test T ( X ), the power function of T ( X ) is defined to be β T ( P ) = E [ T ( X )], P ∈ P, which is the type I error probability of T ( X ) when P ∈ P 0 and one minus the type II error probability of T ( X ) when P ∈ P 1.

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Significance tests

With a sample of a fixed size, we are not able to minimize two error probabilities simultaneously. Our approach involves maximizing the power β T ( P ) over all P ∈ P 1 (i.e., minimizing the type II error probability) and over all tests T satisfying sup P ∈P 0

β T ( P ) ≤ α,

where α ∈ [ 0 , 1 ] is a given level of significance. The left-hand side of the last expression is defined to be the size of T.

Definition 6.

A test T ∗ of size α is a uniformly most powerful (UMP) test if and only if β T ∗ ( P ) ≥ β T ( P ) for all P ∈ P 1 and T of level α.

Using sufficient statistics

If U ( X ) is a sufficient statistic for P ∈ P, then for any test T ( X ), E ( T | U ) has the same power function as T and, therefore, to find a UMP test we may consider tests that are functions of U only.

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Significance tests

With a sample of a fixed size, we are not able to minimize two error probabilities simultaneously. Our approach involves maximizing the power β T ( P ) over all P ∈ P 1 (i.e., minimizing the type II error probability) and over all tests T satisfying sup P ∈P 0

β T ( P ) ≤ α,

where α ∈ [ 0 , 1 ] is a given level of significance. The left-hand side of the last expression is defined to be the size of T.

Definition 6.

A test T ∗ of size α is a uniformly most powerful (UMP) test if and only if β T ∗ ( P ) ≥ β T ( P ) for all P ∈ P 1 and T of level α.

Using sufficient statistics

If U ( X ) is a sufficient statistic for P ∈ P, then for any test T ( X ), E ( T | U ) has the same power function as T and, therefore, to find a UMP test we may consider tests that are functions of U only.

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Remarks

Theorem 6.1 shows that when both H 0 and H 1 are simple (a hypothesis is simple iff the corresponding set of populations contains exactly one element), there exists a UMP test that can be determined by Theorem 6.1 uniquely (a.s. P) except on the set B = { x : f 1 ( x ) = cf 0 ( x )}. If ν( B ) = 0, then we have a unique nonrandomized UMP test; otherwise UMP tests are randomized on the set B and the randomization is necessary for UMP tests to have the given size α We can always choose a UMP test that is constant on B.

Proof of Theorem 6.

The proof for the case of α = 0 or 1 is left as an exercise. Assume now that 0 < α < 1. (i) We first show that there exist γ and c such that E 0 [ T ∗( X )] = α, where Ej is the expectation w.r.t. Pj. Let γ( t ) = P 0 ( f 1 ( X ) > tf 0 ( X )). Then γ( t ) is nonincreasing, γ( 0 ) = 1, and γ(∞) = 0 (why?).

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Remarks

Theorem 6.1 shows that when both H 0 and H 1 are simple (a hypothesis is simple iff the corresponding set of populations contains exactly one element), there exists a UMP test that can be determined by Theorem 6.1 uniquely (a.s. P) except on the set B = { x : f 1 ( x ) = cf 0 ( x )}. If ν( B ) = 0, then we have a unique nonrandomized UMP test; otherwise UMP tests are randomized on the set B and the randomization is necessary for UMP tests to have the given size α We can always choose a UMP test that is constant on B.

Proof of Theorem 6.

The proof for the case of α = 0 or 1 is left as an exercise. Assume now that 0 < α < 1. (i) We first show that there exist γ and c such that E 0 [ T ∗( X )] = α, where Ej is the expectation w.r.t. Pj. Let γ( t ) = P 0 ( f 1 ( X ) > tf 0 ( X )). Then γ( t ) is nonincreasing, γ( 0 ) = 1, and γ(∞) = 0 (why?).

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Proof (continued)

∫ [ T ∗( x ) − T ( x )][ f 1 ( x ) − cf 0 ( x )] d ν ≥ 0 ,

i.e., ∫ [ T ∗( x ) − T ( x )] f 1 ( x ) d ν ≥ c

∫ [ T ∗( x ) − T ( x )] f 0 ( x ) d ν.

The left-hand side is E 1 [ T ∗( X )] − E 1 [ T ( X )] and the right-hand side is

c { E 0 [ T ∗( X )] − E 0 [ T ( X )]} = c { α − E 0 [ T ( X )]} ≥ 0.

This proves the result in (i). (ii) Let T ∗∗( X ) be a UMP test of size α. Define A = { x : T ∗( x ) 6 = T ∗∗( x ), f 1 ( x ) 6 = cf 0 ( x )}.

Then [ T ∗( x ) − T ∗∗( x )][ f 1 ( x ) − cf 0 ( x )] > 0 when x ∈ A and = 0 when x ∈ Ac^ , and ∫ [ T ∗( x ) − T ∗∗( x )][ f 1 ( x ) − cf 0 ( x )] d ν = 0 ,

since both T ∗ and T ∗∗ are UMP tests of size α. By Proposition 1.6(ii), ν( A ) = 0. This proves the result.

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Example 6.

Suppose that X is a sample of size 1, P 0 = { P 0 }, and P 1 = { P 1 }, where P 0 is N ( 0 , 1 ) and P 1 is the double exponential distribution DE ( 0 , 2 ) with the p.d.f. 4−^1 e −| x |/^2. Since P ( f 1 ( X ) = cf 0 ( X )) = 0, there is a unique nonrandomized UMP test. By Theorem 6.1, the UMP test T ∗( x ) = 1 if and only if^ π 8 ex

(^2) −| x |

c^2 for

some c > 0, which is equivalent to | x | > t or | x | < 1 − t for some t > 12.

Suppose that α < 13. To determine t , we use

α = E 0 [ T ∗( X )] = P 0 (| X | > t ) + P 0 (| X | < 1 − t ).

If t ≤ 1, then P 0 (| X | > t ) ≥ P 0 (| X | > 1 ) = 0. 3374 > α. Hence t should be larger than 1 and

α = P 0 (| X | > t ) = Φ(− t ) + 1 − Φ( t ).

Thus, t = Φ−^1 ( 1 − α/ 2 ) and T ∗( X ) = I ( t ,∞)(| X |). Note that it is not necessary to find out what c is.

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Example 6.1 (continued)

Intuitively, the reason why the UMP test in this example rejects H 0 when | X | is large is that the probability of getting a large | X | is much higher under H 1 (i.e., P is the double exponential distribution DE ( 0 , 2 )). The power of T ∗ when P ∈ P 1 is

E 1 [ T ∗( X )] = P 1 (| X | > t ) = 1 −

∫ (^) t

− t

e −| x |/^2 dx = e − t /^2.

Example 6.

Let X 1 , ..., Xn be i.i.d. binary random variables with p = P ( X 1 = 1 ). Suppose that H 0 : p = p 0 and H 1 : p = p 1 , where 0 < p 0 < p 1 < 1. By Theorem 6.1, a UMP test of size α is

T ∗( Y ) =

1 λ ( Y ) > c γ λ ( Y ) = c 0 λ ( Y ) < c ,

where Y = (^) ∑ ni = 1 Xi and

λ ( Y ) =

p 1 p 0

) Y (

1 − p 1 1 − p 0

) n − Y .

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Example 6.2 (continued)

Since λ ( Y ) is increasing in Y , there is an integer m > 0 such that

T ∗( Y ) =

1 Y > m γ Y = m 0 Y < m ,

where m and γ satisfy α = E 0 [ T ∗( Y )] = P 0 ( Y > m ) + γ P 0 ( Y = m ). Since Y has the binomial distribution Bi ( p , n ), we can determine m and γ from

α =

n

j = m + 1

n j

p 0 j ( 1 − p 0 ) n − j^ + γ

n m

p 0 m ( 1 − p 0 ) n − m.

Unless

α =

n

j = m + 1

n j

p 0 j ( 1 − p 0 ) n − j

for some integer m , in which case we can choose γ = 0, the UMP test T ∗ is a randomized test.

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Remark

An interesting phenomenon in Example 6.2 is that the UMP test T ∗ does not depend on p 1. In such a case, T ∗ is in fact a UMP test for testing H 0 : p = p 0 versus H 1 : p > p 0.

Lemma 6.

Suppose that there is a test T ∗ of size α such that for every P 1 ∈ P 1 , T ∗ is UMP for testing H 0 versus the hypothesis P = P 1. Then T ∗ is UMP for testing H 0 versus H 1.

Proof

T ∗ is a test since it does not depend on P 1. For any test T of level α, T is also of level α for testing H 0 versus the hypothesis P = P 1 with any P 1 ∈ P 1. Hence β T ∗ ( P 1 ) ≥ β T ( P 1 ). Since P 1 is arbitrary, this proves that T ∗ is UMP for testing H 0 versus H 1.

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Remark

An interesting phenomenon in Example 6.2 is that the UMP test T ∗ does not depend on p 1. In such a case, T ∗ is in fact a UMP test for testing H 0 : p = p 0 versus H 1 : p > p 0.

Lemma 6.

Suppose that there is a test T ∗ of size α such that for every P 1 ∈ P 1 , T ∗ is UMP for testing H 0 versus the hypothesis P = P 1. Then T ∗ is UMP for testing H 0 versus H 1.

Proof

T ∗ is a test since it does not depend on P 1. For any test T of level α, T is also of level α for testing H 0 versus the hypothesis P = P 1 with any P 1 ∈ P 1. Hence β T ∗ ( P 1 ) ≥ β T ( P 1 ). Since P 1 is arbitrary, this proves that T ∗ is UMP for testing H 0 versus H 1.