[PPT] - Lecture 20 March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun PowerPoint Presentation

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. .

Biostatistics 602 - Statistical Inference Lecture 20 Uniformly Most Powerful Test

Hyun Min Kang March 28th, 2013

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 1 / 1

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Last Lecture

What are the typical steps for constructing a likelihood ratio test?
Is LRT statistic based on sufficient statistic identical to the LRT

based on the full data?

When multiple parameters need to be estimated, what is the

difference in constructing LRT?

What is unbiased test?

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 2 / 1

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Last Lecture

What are the typical steps for constructing a likelihood ratio test?
Is LRT statistic based on sufficient statistic identical to the LRT

based on the full data?

When multiple parameters need to be estimated, what is the

difference in constructing LRT?

What is unbiased test?

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 2 / 1

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Last Lecture

What are the typical steps for constructing a likelihood ratio test?
Is LRT statistic based on sufficient statistic identical to the LRT

based on the full data?

When multiple parameters need to be estimated, what is the

difference in constructing LRT?

What is unbiased test?

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 2 / 1

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Last Lecture

What are the typical steps for constructing a likelihood ratio test?
Is LRT statistic based on sufficient statistic identical to the LRT

based on the full data?

When multiple parameters need to be estimated, what is the

difference in constructing LRT?

What is unbiased test?

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 2 / 1

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Last Lecture

What are the typical steps for constructing a likelihood ratio test?
Is LRT statistic based on sufficient statistic identical to the LRT

based on the full data?

When multiple parameters need to be estimated, what is the

difference in constructing LRT?

What is unbiased test?

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 2 / 1

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LRT based on sufficient statistics

.

Theorem 8.2.4

. . If T(X) is a sufficient statistic for θ, λ∗(t) is the LRT statistic based on T, and λ(x) is the LRT statistic based on x then T x x for every x in the sample space.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 3 / 1

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LRT based on sufficient statistics

.

Theorem 8.2.4

. . If T(X) is a sufficient statistic for θ, λ∗(t) is the LRT statistic based on T, and λ(x) is the LRT statistic based on x then λ∗[T(x)] = λ(x) for every x in the sample space.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 3 / 1

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LRT based on sufficient statistics

.

Theorem 8.2.4

. . If T(X) is a sufficient statistic for θ, λ∗(t) is the LRT statistic based on T, and λ(x) is the LRT statistic based on x then λ∗[T(x)] = λ(x) for every x in the sample space.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 3 / 1

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Unbiased Test

.

Definition

. . If a test always satisfies Pr(reject H0 when H0 is false ) ≥ Pr(reject H0 when H0 is true ) Then the test is said to be unbiased .

Alternative Definition

. . . . . . . . Recall that Pr reject H . A test is unbiased if for every

c and

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 4 / 1

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Unbiased Test

.

Definition

. . If a test always satisfies Pr(reject H0 when H0 is false ) ≥ Pr(reject H0 when H0 is true ) Then the test is said to be unbiased .

Alternative Definition

. . . . . . . . Recall that Pr reject H . A test is unbiased if for every

c and

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 4 / 1

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Unbiased Test

.

Definition

. . If a test always satisfies Pr(reject H0 when H0 is false ) ≥ Pr(reject H0 when H0 is true ) Then the test is said to be unbiased .

Alternative Definition

. . Recall that β(θ) = Pr(reject H0). A test is unbiased if for every

c and

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 4 / 1

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Unbiased Test

.

Definition

. . If a test always satisfies Pr(reject H0 when H0 is false ) ≥ Pr(reject H0 when H0 is true ) Then the test is said to be unbiased .

Alternative Definition

. . Recall that β(θ) = Pr(reject H0). A test is unbiased if β(θ′) ≥ β(θ) for every

c and

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 4 / 1

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Unbiased Test

.

Definition

. . If a test always satisfies Pr(reject H0 when H0 is false ) ≥ Pr(reject H0 when H0 is true ) Then the test is said to be unbiased .

Alternative Definition

. . Recall that β(θ) = Pr(reject H0). A test is unbiased if β(θ′) ≥ β(θ) for every θ′ ∈ Ωc

0 and θ ∈ Ω0.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 4 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H if x

n

c. Pr X n c Pr X n c Pr X n n c Pr X n c n Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

Pr X n c Pr X n c Pr X n n c Pr X n c n Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

β(θ) = Pr (X − θ0 σ/√n > c ) Pr X n c Pr X n n c Pr X n c n Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

β(θ) = Pr (X − θ0 σ/√n > c ) = Pr (X − θ + θ − θ0 σ/√n > c ) Pr X n n c Pr X n c n Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

β(θ) = Pr (X − θ0 σ/√n > c ) = Pr (X − θ + θ − θ0 σ/√n > c ) = Pr (X − θ σ/√n + θ − θ0 σ/√n > c ) Pr X n c n Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

β(θ) = Pr (X − θ0 σ/√n > c ) = Pr (X − θ + θ − θ0 σ/√n > c ) = Pr (X − θ σ/√n + θ − θ0 σ/√n > c ) = Pr (X − θ σ/√n > c − θ − θ0 σ/√n ) Note that Xi , X n , and

X n

.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example

X1, · · · , Xn

i.i.d.

∼ N(θ, σ2) where σ2 is known, testing H0 : θ ≤ θ0 vs

H1 : θ > θ0. LRT test rejects H0 if x−θ0

σ/√n > c.

β(θ) = Pr (X − θ0 σ/√n > c ) = Pr (X − θ + θ − θ0 σ/√n > c ) = Pr (X − θ σ/√n + θ − θ0 σ/√n > c ) = Pr (X − θ σ/√n > c − θ − θ0 σ/√n ) Note that Xi ∼ N(θ, σ2), X ∼ N(θ, σ2/n), and

X−θ σ/√n ∼ N(0, 1).

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 5 / 1

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Example (cont’d)

Therefore, for Z ∼ N(0, 1) β(θ) = Pr ( Z > c + θ0 − θ σ/√n ) Because the power function is increasing function of , always holds when . Therefore the LRTs are unbiased.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 6 / 1

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Example (cont’d)

Therefore, for Z ∼ N(0, 1) β(θ) = Pr ( Z > c + θ0 − θ σ/√n ) Because the power function is increasing function of θ, β(θ′) ≥ β(θ) always holds when . Therefore the LRTs are unbiased.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 6 / 1

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Example (cont’d)

Therefore, for Z ∼ N(0, 1) β(θ) = Pr ( Z > c + θ0 − θ σ/√n ) Because the power function is increasing function of θ, β(θ′) ≥ β(θ) always holds when θ ≤ θ0 < θ′. Therefore the LRTs are unbiased.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 6 / 1

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Uniformly Most Powerful Test (UMP)

.

Definition

. . Let C be a class of tests between H0 : θ ∈ Ω vs H1 : θ ∈ Ωc

0. A test in C,

with power function β(θ) is uniformly most powerful (UMP) test in class C if β(θ) ≥ β′(θ) for every θ ∈ Ωc

0 and every β′(θ), which is a power

function of another test in C.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 7 / 1

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UMP level α test

. . Consider C be the set of all the level α test. The UMP test in this class is called a UMP level α test. UMP level test has the smallest type II error probability for any

c

in this class.

A UMP test is ”uniform” in the sense that it is most powerful for

every

c.

For simple hypothesis such as H

and H , UMP level test always exists.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 8 / 1

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UMP level α test

. . Consider C be the set of all the level α test. The UMP test in this class is called a UMP level α test. UMP level α test has the smallest type II error probability for any θ ∈ Ωc in this class.

A UMP test is ”uniform” in the sense that it is most powerful for

every

c.

For simple hypothesis such as H

and H , UMP level test always exists.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 8 / 1

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UMP level α test

. . Consider C be the set of all the level α test. The UMP test in this class is called a UMP level α test. UMP level α test has the smallest type II error probability for any θ ∈ Ωc in this class.

A UMP test is ”uniform” in the sense that it is most powerful for

every θ ∈ Ωc

0.

For simple hypothesis such as H

and H , UMP level test always exists.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 8 / 1

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UMP level α test

. . Consider C be the set of all the level α test. The UMP test in this class is called a UMP level α test. UMP level α test has the smallest type II error probability for any θ ∈ Ωc in this class.

A UMP test is ”uniform” in the sense that it is most powerful for

every θ ∈ Ωc

0.

For simple hypothesis such as H0 : θ = θ0 and H1 : θ = θ1, UMP level

α test always exists.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 8 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x R if f x kf x and x Rc if f x kf x For some k and Pr X R , Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k

, then every UMP level test is a size test (satisfies 8.3.2), and every UMP level test satisfies 8.3.1 except perhaps on a set A satisfying Pr X A Pr X A .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x ∈ R if f(x|θ1) > kf(x|θ0) (8.3.1) and x Rc if f x kf x For some k and Pr X R , Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k

, then every UMP level test is a size test (satisfies 8.3.2), and every UMP level test satisfies 8.3.1 except perhaps on a set A satisfying Pr X A Pr X A .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x ∈ R if f(x|θ1) > kf(x|θ0) (8.3.1) and x ∈ Rc if f(x|θ1) < kf(x|θ0) (8.3.2) For some k and Pr X R , Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k

, then every UMP level test is a size test (satisfies 8.3.2), and every UMP level test satisfies 8.3.1 except perhaps on a set A satisfying Pr X A Pr X A .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x ∈ R if f(x|θ1) > kf(x|θ0) (8.3.1) and x ∈ Rc if f(x|θ1) < kf(x|θ0) (8.3.2) For some k ≥ 0 and α = Pr(X ∈ R|θ0), Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k

, then every UMP level test is a size test (satisfies 8.3.2), and every UMP level test satisfies 8.3.1 except perhaps on a set A satisfying Pr X A Pr X A .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x ∈ R if f(x|θ1) > kf(x|θ0) (8.3.1) and x ∈ Rc if f(x|θ1) < kf(x|θ0) (8.3.2) For some k ≥ 0 and α = Pr(X ∈ R|θ0), Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level α

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k

, then every UMP level test is a size test (satisfies 8.3.2), and every UMP level test satisfies 8.3.1 except perhaps on a set A satisfying Pr X A Pr X A .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Neyman-Pearson Lemma

.

Theorem 8.3.12 - Neyman-Pearson Lemma

. . Consider testing H0 : θ = θ0 vs. H1 : θ = θ1 where the pdf or pmf corresponding the θi is f(x|θi), i = 0, 1, using a test with rejection region R that satisfies x ∈ R if f(x|θ1) > kf(x|θ0) (8.3.1) and x ∈ Rc if f(x|θ1) < kf(x|θ0) (8.3.2) For some k ≥ 0 and α = Pr(X ∈ R|θ0), Then,

(Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level α

test

(Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k > 0,

then every UMP level α test is a size α test (satisfies 8.3.2), and every UMP level α test satisfies 8.3.1 except perhaps on a set A satisfying Pr(X ∈ A|θ0) = Pr(X ∈ A|θ1) = 0.

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 9 / 1

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H

vs. H

. Calculating the ratios of the pmfs given, f f f f f f

Suppose that k

, then the rejection region R , and UMP level test always rejects H . Therefore Pr reject H .

Suppose that

k , then R , and UMP level test rejects H if x

r x

. Pr reject Pr x Pr x

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H0 : θ = θ0 = 1/2 vs. H1 : θ = θ1 = 3/4. Calculating the ratios of the pmfs given, f f f f f f

Suppose that k

, then the rejection region R , and UMP level test always rejects H . Therefore Pr reject H .

Suppose that

k , then R , and UMP level test rejects H if x

r x

. Pr reject Pr x Pr x

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H0 : θ = θ0 = 1/2 vs. H1 : θ = θ1 = 3/4. Calculating the ratios of the pmfs given, f(0|θ1) f(0|θ0) = 1 4, f(1|θ1) f(1|θ0) = 3 4, f(2|θ1) f(2|θ0) = 9 4

Suppose that k

, then the rejection region R , and UMP level test always rejects H . Therefore Pr reject H .

Suppose that

k , then R , and UMP level test rejects H if x

r x

. Pr reject Pr x Pr x

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

SLIDE 39

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H0 : θ = θ0 = 1/2 vs. H1 : θ = θ1 = 3/4. Calculating the ratios of the pmfs given, f(0|θ1) f(0|θ0) = 1 4, f(1|θ1) f(1|θ0) = 3 4, f(2|θ1) f(2|θ0) = 9 4

Suppose that k < 1/4, then the rejection region R = {0, 1, 2}, and

UMP level α test always rejects H0. Therefore α = Pr(reject H0|θ = θ0 = 1/2) = 1.

Suppose that

k , then R , and UMP level test rejects H if x

r x

. Pr reject Pr x Pr x

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

SLIDE 40

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H0 : θ = θ0 = 1/2 vs. H1 : θ = θ1 = 3/4. Calculating the ratios of the pmfs given, f(0|θ1) f(0|θ0) = 1 4, f(1|θ1) f(1|θ0) = 3 4, f(2|θ1) f(2|θ0) = 9 4

Suppose that k < 1/4, then the rejection region R = {0, 1, 2}, and

UMP level α test always rejects H0. Therefore α = Pr(reject H0|θ = θ0 = 1/2) = 1.

Suppose that 1/4 < k < 3/4, then R = {1, 2}, and UMP level α test

rejects H0 if x = 1 or x = 2. Pr reject Pr x Pr x

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

SLIDE 41

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Example of Neyman-Pearson Lemma

Let X ∈ Binomial(2, θ), and consider testing H0 : θ = θ0 = 1/2 vs. H1 : θ = θ1 = 3/4. Calculating the ratios of the pmfs given, f(0|θ1) f(0|θ0) = 1 4, f(1|θ1) f(1|θ0) = 3 4, f(2|θ1) f(2|θ0) = 9 4

Suppose that k < 1/4, then the rejection region R = {0, 1, 2}, and

UMP level α test always rejects H0. Therefore α = Pr(reject H0|θ = θ0 = 1/2) = 1.

Suppose that 1/4 < k < 3/4, then R = {1, 2}, and UMP level α test

rejects H0 if x = 1 or x = 2. α = Pr(reject|θ = 1/2) = Pr(x = 1|θ = 1/2) + Pr(x = 2|θ = 1/2) = 3 4

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 10 / 1

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Example of Neyman-Pearson Lemma (cont’d)

Suppose that 3/4 < k < 9/4, then UMP level α test rejects H0 if

x = 2 Pr reject Pr x

If k

the UMP level test always not reject H , and

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 11 / 1

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Example of Neyman-Pearson Lemma (cont’d)

Suppose that 3/4 < k < 9/4, then UMP level α test rejects H0 if

x = 2 α = Pr(reject|θ = 1/2) = Pr(x = 2|θ = 1/2) = 1 4

If k

the UMP level test always not reject H , and

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 11 / 1

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Example of Neyman-Pearson Lemma (cont’d)

Suppose that 3/4 < k < 9/4, then UMP level α test rejects H0 if

x = 2 α = Pr(reject|θ = 1/2) = Pr(x = 2|θ = 1/2) = 1 4

If k > 9/4 the UMP level α test always not reject H0, and α = 0

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 11 / 1

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f x

n i

exp xi f x f x exp

n i

xi

exp

n i

xi

exp

n i

xi

n i

xi exp

n i

xi

n i

xi exp n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f(x|θ) =

n

∏

i=1

[ 1 2πσ2 exp { −(xi − θ)2 2σ2 }] f x f x exp

n i

xi

exp

n i

xi

exp

n i

xi

n i

xi exp

n i

xi

n i

xi exp n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

SLIDE 47

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f(x|θ) =

n

∏

i=1

[ 1 2πσ2 exp { −(xi − θ)2 2σ2 }] f(x|θ1) f(x|θ0) = exp { −

∑n

i=1(xi−θ1)2

2σ2

} exp { −

∑n

i=1(xi−θ0)2

2σ2

} exp

n i

xi

n i

xi exp

n i

xi

n i

xi exp n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

SLIDE 48

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f(x|θ) =

n

∏

i=1

[ 1 2πσ2 exp { −(xi − θ)2 2σ2 }] f(x|θ1) f(x|θ0) = exp { −

∑n

i=1(xi−θ1)2

2σ2

} exp { −

∑n

i=1(xi−θ0)2

2σ2

} = exp [ − ∑n

i=1(xi − θ1)2

2σ2 + ∑n

i=1(xi − θ0)2

2σ2 ] exp

n i

xi

n i

xi exp n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

SLIDE 49

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f(x|θ) =

n

∏

i=1

[ 1 2πσ2 exp { −(xi − θ)2 2σ2 }] f(x|θ1) f(x|θ0) = exp { −

∑n

i=1(xi−θ1)2

2σ2

} exp { −

∑n

i=1(xi−θ0)2

2σ2

} = exp [ − ∑n

i=1(xi − θ1)2

2σ2 + ∑n

i=1(xi − θ0)2

2σ2 ] = exp [∑n

i=1(xi − θ0)2 − ∑n i=1(xi − θ1)2

2σ2 ] exp n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

SLIDE 50

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Example - Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. f(x|θ) =

n

∏

i=1

[ 1 2πσ2 exp { −(xi − θ)2 2σ2 }] f(x|θ1) f(x|θ0) = exp { −

∑n

i=1(xi−θ1)2

2σ2

} exp { −

∑n

i=1(xi−θ0)2

2σ2

} = exp [ − ∑n

i=1(xi − θ1)2

2σ2 + ∑n

i=1(xi − θ0)2

2σ2 ] = exp [∑n

i=1(xi − θ0)2 − ∑n i=1(xi − θ1)2

2σ2 ] = exp [n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 ]

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 12 / 1

SLIDE 51

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Example (cont’d)

UMP level α test rejects if exp [n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 ] > k n

n i

xi log k

n i

xi k Pr

n i

Xi k

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 13 / 1

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Example (cont’d)

UMP level α test rejects if exp [n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 ] > k ⇐ ⇒ n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 > log k

n i

xi k Pr

n i

Xi k

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 13 / 1

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Example (cont’d)

UMP level α test rejects if exp [n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 ] > k ⇐ ⇒ n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 > log k ⇐ ⇒

n

∑

i=1

xi > k∗ Pr

n i

Xi k

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 13 / 1

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Example (cont’d)

UMP level α test rejects if exp [n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 ] > k ⇐ ⇒ n(θ2

0 − θ1)2 + 2 ∑n i=1 xi(θ1 − θ0)

2σ2 > log k ⇐ ⇒

n

∑

i=1

xi > k∗ α = Pr ( n ∑

i=1

Xi > k∗|θ0 )

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 13 / 1

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Example (cont’d)

Under H0, Xi ∼ N(θ0, σ2) X n X n Pr

n i

Xi k Pr Z k n n where Z .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 14 / 1

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Example (cont’d)

Under H0, Xi ∼ N(θ0, σ2) X ∼ N(θ0, σ2/n) X n Pr

n i

Xi k Pr Z k n n where Z .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 14 / 1

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Example (cont’d)

Under H0, Xi ∼ N(θ0, σ2) X ∼ N(θ0, σ2/n) X − θ0 σ/√n ∼ N(0, 1) Pr

n i

Xi k Pr Z k n n where Z .

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 14 / 1

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Example (cont’d)

Under H0, Xi ∼ N(θ0, σ2) X ∼ N(θ0, σ2/n) X − θ0 σ/√n ∼ N(0, 1) α = Pr ( n ∑

i=1

Xi > k∗|θ0 ) = Pr ( Z > k∗/n − θ0 σ/√n ) where Z ∼ N(0, 1).

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 14 / 1

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Example (cont’d)

k∗/n − θ0 σ/√n = zα k n z n Thus, the UMP level test reject if Xi k , or equivalently, reject H if X k n z n

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 15 / 1

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Example (cont’d)

k∗/n − θ0 σ/√n = zα k∗ = n ( θ0 + zα σ √n ) Thus, the UMP level test reject if Xi k , or equivalently, reject H if X k n z n

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 15 / 1

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Example (cont’d)

k∗/n − θ0 σ/√n = zα k∗ = n ( θ0 + zα σ √n ) Thus, the UMP level α test reject if ∑ Xi > k∗, or equivalently, reject H0 if X > k∗/n = θ0 + zασ/√n

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 15 / 1

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Neyman-Pearson Lemma on Sufficient Statistics

.

Corollary 8.3.13

. . Consider H0 : θ = θ0 vs H1 : θ = θ1. Suppose T(X) is a sufficient statistic for θ and g(t|θi) is the pdf or pmf of T. Corresponding θi, i ∈ {0, 1}. Then any test based on T with rejection region S is a UMP level α test if it satisfies t S if g t k g t and t Sc if g t k g t For some k and Pr T S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 16 / 1

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Neyman-Pearson Lemma on Sufficient Statistics

.

Corollary 8.3.13

. . Consider H0 : θ = θ0 vs H1 : θ = θ1. Suppose T(X) is a sufficient statistic for θ and g(t|θi) is the pdf or pmf of T. Corresponding θi, i ∈ {0, 1}. Then any test based on T with rejection region S is a UMP level α test if it satisfies t ∈ S if g(t|θ1) > k · g(t|θ0) and t Sc if g t k g t For some k and Pr T S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 16 / 1

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Neyman-Pearson Lemma on Sufficient Statistics

.

Corollary 8.3.13

. . Consider H0 : θ = θ0 vs H1 : θ = θ1. Suppose T(X) is a sufficient statistic for θ and g(t|θi) is the pdf or pmf of T. Corresponding θi, i ∈ {0, 1}. Then any test based on T with rejection region S is a UMP level α test if it satisfies t ∈ S if g(t|θ1) > k · g(t|θ0) and t ∈ Sc if g(t|θ1) < k · g(t|θ0) For some k and Pr T S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 16 / 1

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Neyman-Pearson Lemma on Sufficient Statistics

.

Corollary 8.3.13

. . Consider H0 : θ = θ0 vs H1 : θ = θ1. Suppose T(X) is a sufficient statistic for θ and g(t|θi) is the pdf or pmf of T. Corresponding θi, i ∈ {0, 1}. Then any test based on T with rejection region S is a UMP level α test if it satisfies t ∈ S if g(t|θ1) > k · g(t|θ0) and t ∈ Sc if g(t|θ1) < k · g(t|θ0) For some k > 0 and α = Pr(T ∈ S|θ0)

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 16 / 1

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} x g T x kg T x By Factorization Theorem: f x

i

h x g T x

i

R x g T x h x kg T x h x x f x kf x By Neyman-Pearson Lemma, this test is the UMP level test, and Pr X R Pr T X S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

SLIDE 67

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} = {x : g(T(x)|θ1) > kg(T(x)|θ0)} By Factorization Theorem: f x

i

h x g T x

i

R x g T x h x kg T x h x x f x kf x By Neyman-Pearson Lemma, this test is the UMP level test, and Pr X R Pr T X S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} = {x : g(T(x)|θ1) > kg(T(x)|θ0)} By Factorization Theorem: f(x|θi) = h(x)g(T(x)|θi) R x g T x h x kg T x h x x f x kf x By Neyman-Pearson Lemma, this test is the UMP level test, and Pr X R Pr T X S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

SLIDE 69

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} = {x : g(T(x)|θ1) > kg(T(x)|θ0)} By Factorization Theorem: f(x|θi) = h(x)g(T(x)|θi) R = {x : g(T(x)|θ1)h(x) > kg(T(x)|θ0)h(x)} x f x kf x By Neyman-Pearson Lemma, this test is the UMP level test, and Pr X R Pr T X S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} = {x : g(T(x)|θ1) > kg(T(x)|θ0)} By Factorization Theorem: f(x|θi) = h(x)g(T(x)|θi) R = {x : g(T(x)|θ1)h(x) > kg(T(x)|θ0)h(x)} = {x : f(x|θ1) > kf(x|θ0)} By Neyman-Pearson Lemma, this test is the UMP level test, and Pr X R Pr T X S

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

SLIDE 71

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Proof

The rejection region in the sample space is R = {x : T(x) = t ∈ S} = {x : g(T(x)|θ1) > kg(T(x)|θ0)} By Factorization Theorem: f(x|θi) = h(x)g(T(x)|θi) R = {x : g(T(x)|θ1)h(x) > kg(T(x)|θ0)h(x)} = {x : f(x|θ1) > kf(x|θ0)} By Neyman-Pearson Lemma, this test is the UMP level α test, and α = Pr(X ∈ R) = Pr(T(X) ∈ S|θ0)

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 17 / 1

SLIDE 72

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Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T X is a sufficient statistic for , where T n . g t

i

n exp t

i

n g t g t exp

t n

exp

t n

exp n t t exp n t

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 73

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Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T = X is a sufficient statistic for θ, where T ∼ N(θ, σ2/n). g t

i

n exp t

i

n g t g t exp

t n

exp

t n

exp n t t exp n t

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 74

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Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T = X is a sufficient statistic for θ, where T ∼ N(θ, σ2/n). g(t|θi) = 1 √ 2πσ2/n exp { −(t − θi)2 2σ2/n } g t g t exp

t n

exp

t n

exp n t t exp n t

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 75

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Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T = X is a sufficient statistic for θ, where T ∼ N(θ, σ2/n). g(t|θi) = 1 √ 2πσ2/n exp { −(t − θi)2 2σ2/n } g(t|θ1) g(t|θ0) = exp { − (t−θ1)2

2σ2/n

} exp { − (t−θ0)2

2σ2/n

} exp n t t exp n t

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 76

. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .

Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T = X is a sufficient statistic for θ, where T ∼ N(θ, σ2/n). g(t|θi) = 1 √ 2πσ2/n exp { −(t − θi)2 2σ2/n } g(t|θ1) g(t|θ0) = exp { − (t−θ1)2

2σ2/n

} exp { − (t−θ0)2

2σ2/n

} = exp { − 1 2σ2/n [ (t − θ1)2 − (t − θ0)2]} exp n t

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 77

. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .

Revisiting the Example of Normal Distribution

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider testing H0 : θ = θ0 vs.

H1 : θ = θ1 where θ1 > θ0. T = X is a sufficient statistic for θ, where T ∼ N(θ, σ2/n). g(t|θi) = 1 √ 2πσ2/n exp { −(t − θi)2 2σ2/n } g(t|θ1) g(t|θ0) = exp { − (t−θ1)2

2σ2/n

} exp { − (t−θ0)2

2σ2/n

} = exp { − 1 2σ2/n [ (t − θ1)2 − (t − θ0)2]} = exp { − 1 2σ2/n [ θ2

1 − θ2 0 − 2t(θ1 − θ0)

]}

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 18 / 1

SLIDE 78

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Revisiting the Example (cont’d)

UMP level α test reject if exp { − 1 2σ2/n [ θ2

1 − θ2 0 − 2t(θ1 − θ0)

]} > k n t log k X t k

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 19 / 1

SLIDE 79

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Revisiting the Example (cont’d)

UMP level α test reject if exp { − 1 2σ2/n [ θ2

1 − θ2 0 − 2t(θ1 − θ0)

]} > k ⇐ ⇒ 1 2σ2/n [ −(θ2

1 − θ2 0) + 2t(θ1 − θ0)

] > log k X t k

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 19 / 1

SLIDE 80

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Revisiting the Example (cont’d)

UMP level α test reject if exp { − 1 2σ2/n [ θ2

1 − θ2 0 − 2t(θ1 − θ0)

]} > k ⇐ ⇒ 1 2σ2/n [ −(θ2

1 − θ2 0) + 2t(θ1 − θ0)

] > log k ⇐ ⇒ X = t > k∗

Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 19 / 1

SLIDE 81

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