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. . . . .. . . .. . . .. . . .. . . .. . . Lecture 20 March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang March 28th, 2013 Hyun Min Kang Uniformly Most Powerful Test Biostatistics 602 - Statistical Inference ..

  1. , X . . Example . .. . . .. . .. Pr . . .. . . .. . i.i.d. Pr Pr Note that X i March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . n X n , and n .. c n X Pr c n n X . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 5 / 1 .. .. . .. . . . . . . .. . . .. . . .. . ∼ N ( θ, σ 2 ) where σ 2 is known, testing H 0 : θ ≤ θ 0 vs X 1 , · · · , X n H 1 : θ > θ 0 . LRT test rejects H 0 if x − θ 0 σ / √ n > c . ( X − θ 0 ) σ / √ n > c β ( θ ) = ( X − θ + θ − θ 0 ) = σ / √ n > c

  2. , X .. . . .. . . .. . . . .. . . .. . . .. .. i.i.d. Example Note that X i March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . n X n , and n . c n X Pr Pr Pr Pr . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 5 / 1 . . . . .. . . . .. . .. .. . .. . . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known, testing H 0 : θ ≤ θ 0 vs X 1 , · · · , X n H 1 : θ > θ 0 . LRT test rejects H 0 if x − θ 0 σ / √ n > c . ( X − θ 0 ) σ / √ n > c β ( θ ) = ( X − θ + θ − θ 0 ) = σ / √ n > c ( X − θ σ / √ n + θ − θ 0 ) σ / √ n > c =

  3. . .. .. . . .. . .. . . . .. . . .. . . . .. . n , and March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . n X , X . Note that X i Pr Pr Pr Pr i.i.d. Example .. . . .. .. . .. .. . . . . . .. . . .. . . . . .. .. . . .. . . .. . . .. . . . 5 / 1 . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known, testing H 0 : θ ≤ θ 0 vs X 1 , · · · , X n H 1 : θ > θ 0 . LRT test rejects H 0 if x − θ 0 σ / √ n > c . ( X − θ 0 ) σ / √ n > c β ( θ ) = ( X − θ + θ − θ 0 ) = σ / √ n > c ( X − θ σ / √ n + θ − θ 0 ) σ / √ n > c = ( X − θ σ / √ n > c − θ − θ 0 ) = σ / √ n

  4. . .. . . .. . . .. . . .. . . .. . . . . Pr March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr Pr Pr i.i.d. . Example . .. . . .. .. .. . .. . .. . . .. . . .. .. . . .. . . . 5 / 1 . .. . . .. . . . . . .. . . .. . . .. ∼ N ( θ, σ 2 ) where σ 2 is known, testing H 0 : θ ≤ θ 0 vs X 1 , · · · , X n H 1 : θ > θ 0 . LRT test rejects H 0 if x − θ 0 σ / √ n > c . ( X − θ 0 ) σ / √ n > c β ( θ ) = ( X − θ + θ − θ 0 ) = σ / √ n > c ( X − θ σ / √ n + θ − θ 0 ) σ / √ n > c = ( X − θ σ / √ n > c − θ − θ 0 ) = σ / √ n Note that X i ∼ N ( θ, σ 2 ) , X ∼ N ( θ, σ 2 / n ) , and X − θ σ / √ n ∼ N (0 , 1) .

  5. . .. .. . .. . . .. . . .. . . .. . . . . Because the power function is increasing function of March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . Therefore the LRTs are unbiased. always holds when , Pr . Example (cont’d) . .. . . .. .. . . . . .. .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . .. 6 / 1 .. . . .. . . Therefore, for Z ∼ N (0 , 1) ( Z > c + θ 0 − θ ) σ / √ n β ( θ ) =

  6. . . .. .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . Example (cont’d) Pr always holds when . Therefore the LRTs are unbiased. Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . .. .. . . .. . . .. . . .. . . . . . .. . . . .. . . .. . . .. . .. . . .. 6 / 1 . Therefore, for Z ∼ N (0 , 1) ( Z > c + θ 0 − θ ) σ / √ n β ( θ ) = Because the power function is increasing function of θ , β ( θ ′ ) ≥ β ( θ )

  7. . .. . .. .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . Example (cont’d) Pr Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . 6 / 1 .. . .. . .. . . Therefore, for Z ∼ N (0 , 1) ( Z > c + θ 0 − θ ) σ / √ n β ( θ ) = Because the power function is increasing function of θ , β ( θ ′ ) ≥ β ( θ ) always holds when θ ≤ θ 0 < θ ′ . Therefore the LRTs are unbiased.

  8. . .. . . .. . . .. . . .. . . .. . . . . Definition March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang function of another test in C . . . . . Uniformly Most Powerful Test (UMP) . .. . . .. .. .. . . .. . .. . . .. . . .. . . .. . . .. . . . . . .. . .. .. . . .. . . .. . . 7 / 1 Let C be a class of tests between H 0 : θ ∈ Ω vs H 1 : θ ∈ Ω c 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 ∈ Ω c 0 and every β ′ ( θ ) , which is a power

  9. • A UMP test is ”uniform” in the sense that it is most powerful for • For simple hypothesis such as H . . . .. . . .. . . . .. . . .. . . .. .. . c . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang test always exists. , UMP level and H every . in this class. c test has the smallest type II error probability for any UMP level . . . .. .. . .. .. . . .. . . . .. . .. . . .. . . . . .. .. . . .. . . .. . . . . . .. . . .. . 8 / 1 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.

  10. • A UMP test is ”uniform” in the sense that it is most powerful for • For simple hypothesis such as H . . .. . . .. . .. . . . .. . . .. . . .. c . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang test always exists. , UMP level and H every . in this class. . . . .. . .. . .. .. . . . .. . . . . . .. . . .. . . . .. .. . . .. . . .. . . .. . .. . . . .. . 8 / 1 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 ∈ Ω c 0

  11. • For simple hypothesis such as H . . .. . . .. . . .. .. . . .. .. . . . . in this class. March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang test always exists. , UMP level and H . . . . .. . . .. .. . . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . 8 / 1 .. . . .. . . 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 ∈ Ω c 0 • A UMP test is ”uniform” in the sense that it is most powerful for every θ ∈ Ω c 0 .

  12. . . . .. .. . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . in this class. Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . 8 / 1 . . . .. . .. .. 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 ∈ Ω c 0 • A UMP test is ”uniform” in the sense that it is most powerful for every θ ∈ Ω c 0 . • For simple hypothesis such as H 0 : θ = θ 0 and H 1 : θ = θ 1 , UMP level α test always exists.

  13. • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level • (Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k . . and kf x if f x R x R that satisfies . Theorem 8.3.12 - Neyman-Pearson Lemma R c . Neyman-Pearson Lemma . .. . . .. . . .. x kf x if f x every UMP level March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . A Pr X A satisfying Pr X test satisfies 8.3.1 except perhaps on a set A test (satisfies 8.3.2), and .. test is a size then every UMP level , test , Then, R Pr X and For some k . . .. .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . . .. . .. . . .. . . .. . . . . . .. . . .. . . .. 9 / 1 Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region

  14. • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level • (Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k . . x R that satisfies . . Theorem 8.3.12 - Neyman-Pearson Lemma . Neyman-Pearson Lemma .. if f x . . .. . . .. . . .. R c For some k kf x test satisfies 8.3.1 except perhaps on a set A March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . A Pr X A satisfying Pr X every UMP level . test (satisfies 8.3.2), and test is a size then every UMP level , test , Then, R Pr X and . .. .. .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . . .. . .. . . .. . . .. . . . . . .. . . .. 9 / 1 Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region x ∈ R if f ( x | θ 1 ) > kf ( x | θ 0 ) (8 . 3 . 1) and

  15. • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level • (Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k . . . Theorem 8.3.12 - Neyman-Pearson Lemma . Neyman-Pearson Lemma . .. . R that satisfies .. . . .. . . .. .. . . and For some k test satisfies 8.3.1 except perhaps on a set A March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . A Pr X A satisfying Pr X every UMP level . test (satisfies 8.3.2), and test is a size then every UMP level , test , Then, R Pr X .. . . .. .. . . .. . . .. . . . .. . .. . . .. . . .. . . . 9 / 1 . . . .. . . .. . . .. .. .. . . .. . . . . Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region x ∈ R if f ( x | θ 1 ) > kf ( x | θ 0 ) (8 . 3 . 1) and x ∈ R c if f ( x | θ 1 ) < kf ( x | θ 0 ) (8 . 3 . 2)

  16. • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level • (Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k . . Neyman-Pearson Lemma . .. . . .. . Theorem 8.3.12 - Neyman-Pearson Lemma .. . . .. . . .. . . . . satisfying Pr X March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . A Pr X A test satisfies 8.3.1 except perhaps on a set A .. every UMP level test (satisfies 8.3.2), and test is a size then every UMP level , test R that satisfies . .. . . . . .. . . .. . .. . . . .. . . .. . . . .. . .. .. . . .. . . .. . . .. . . . .. . .. . . 9 / 1 Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region x ∈ R if f ( x | θ 1 ) > kf ( x | θ 0 ) (8 . 3 . 1) and x ∈ R c if f ( x | θ 1 ) < kf ( x | θ 0 ) (8 . 3 . 2) For some k ≥ 0 and α = Pr ( X ∈ R | θ 0 ) , Then,

  17. • (Necessity) if there exist a test satisfying 8.3.1 and 8.3.2 with k . . . .. . . .. . .. . . . .. . . .. . . Neyman-Pearson Lemma . Theorem 8.3.12 - Neyman-Pearson Lemma satisfying Pr X March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . A Pr X A test satisfies 8.3.1 except perhaps on a set A . every UMP level test (satisfies 8.3.2), and test is a size then every UMP level , test R that satisfies . .. .. . . .. . .. . . .. . .. . . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . .. .. . . 9 / 1 Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region x ∈ R if f ( x | θ 1 ) > kf ( x | θ 0 ) (8 . 3 . 1) and x ∈ R c if f ( x | θ 1 ) < kf ( x | θ 0 ) (8 . 3 . 2) For some k ≥ 0 and α = Pr ( X ∈ R | θ 0 ) , Then, • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level α

  18. . . . . .. . . .. . . .. .. . .. . . .. . . . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang test R that satisfies . Theorem 8.3.12 - Neyman-Pearson Lemma .. . Neyman-Pearson Lemma . .. . . .. . . . . . .. . .. .. . . .. . . .. . . .. . . .. . . .. . .. . 9 / 1 . . .. . . . .. . Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where the pdf or pmf corresponding the θ i is f ( x | θ i ) , i = 0 , 1 , using a test with rejection region x ∈ R if f ( x | θ 1 ) > kf ( x | θ 0 ) (8 . 3 . 1) and x ∈ R c if f ( x | θ 1 ) < kf ( x | θ 0 ) (8 . 3 . 2) For some k ≥ 0 and α = Pr ( X ∈ R | θ 0 ) , Then, • (Sufficiency) Any test that satisfies 8.3.1 and 8.3.2 is a UMP level α • (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 | θ 0 ) = Pr ( X ∈ A | θ 1 ) = 0 .

  19. • Suppose that k • Suppose that . f f f Calculating the ratios of the pmfs given, . vs. H H Example of Neyman-Pearson Lemma . f . . .. . . .. . . .. .. f f rejects H if x March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr x Pr x Pr reject . or x test .. , and UMP level , then R k . Pr reject H test always rejects H . Therefore UMP level , and , then the rejection region R . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 10 / 1 Let X ∈ Binomial (2 , θ ) , and consider testing

  20. • Suppose that k • Suppose that .. f f f f Calculating the ratios of the pmfs given, Example of Neyman-Pearson Lemma . . f . .. . . .. . . .. . f . .. rejects H if x March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr x Pr x Pr reject . or x test , then the rejection region R , and UMP level , then R k . Pr reject H test always rejects H . Therefore UMP level , and . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. . . . .. . .. . . .. . . 10 / 1 Let X ∈ Binomial (2 , θ ) , and consider testing H 0 : θ = θ 0 = 1/2 vs. H 1 : θ = θ 1 = 3/4 .

  21. • Suppose that k • Suppose that . Example of Neyman-Pearson Lemma . .. . . .. . . .. . . .. . .. .. . Calculating the ratios of the pmfs given, , and , then the rejection region R or x March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr x Pr x Pr reject . rejects H if x .. test , and UMP level , then R k . Pr reject H test always rejects H . Therefore UMP level . . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 10 / 1 . .. . .. . . .. . . .. . .. . . . . .. . Let X ∈ Binomial (2 , θ ) , and consider testing H 0 : θ = θ 0 = 1/2 vs. H 1 : θ = θ 1 = 3/4 . f (0 | θ 1 ) f (1 | θ 1 ) f (2 | θ 1 ) f (0 | θ 0 ) = 1 f (1 | θ 0 ) = 3 f (2 | θ 0 ) = 9 4 , 4 , 4

  22. • Suppose that . .. .. . . .. . . . . . .. .. . .. . . . .. . . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr x Pr x Pr reject or x . rejects H if x test , and UMP level , then R k Calculating the ratios of the pmfs given, Example of Neyman-Pearson Lemma .. . . .. .. .. . . .. . . . . . .. . . .. . . . 10 / 1 .. . . . .. . . . . .. . .. . .. . . .. . Let X ∈ Binomial (2 , θ ) , and consider testing H 0 : θ = θ 0 = 1/2 vs. H 1 : θ = θ 1 = 3/4 . f (0 | θ 1 ) f (1 | θ 1 ) f (2 | θ 1 ) f (0 | θ 0 ) = 1 f (1 | θ 0 ) = 3 f (2 | θ 0 ) = 9 4 , 4 , 4 • Suppose that k < 1/4 , then the rejection region R = { 0 , 1 , 2 } , and UMP level α test always rejects H 0 . Therefore α = Pr ( reject H 0 | θ = θ 0 = 1/2) = 1 .

  23. . . .. . . .. . .. .. . . .. . . .. . .. . Calculating the ratios of the pmfs given, March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr x Pr x Pr reject Example of Neyman-Pearson Lemma . . .. . . .. . . . .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . .. . . 10 / 1 . .. .. . . .. . Let X ∈ Binomial (2 , θ ) , and consider testing H 0 : θ = θ 0 = 1/2 vs. H 1 : θ = θ 1 = 3/4 . f (0 | θ 1 ) f (1 | θ 1 ) f (2 | θ 1 ) f (0 | θ 0 ) = 1 f (1 | θ 0 ) = 3 f (2 | θ 0 ) = 9 4 , 4 , 4 • Suppose that k < 1/4 , then the rejection region R = { 0 , 1 , 2 } , and UMP level α test always rejects H 0 . Therefore α = Pr ( reject H 0 | θ = θ 0 = 1/2) = 1 . • Suppose that 1/4 < k < 3/4 , then R = { 1 , 2 } , and UMP level α test rejects H 0 if x = 1 or x = 2 .

  24. . .. . .. . . .. . .. .. . . .. . . . .. . .. . . .. . . .. . Example of Neyman-Pearson Lemma Calculating the ratios of the pmfs given, Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . . .. . 10 / 1 .. . . .. . .. . Let X ∈ Binomial (2 , θ ) , and consider testing H 0 : θ = θ 0 = 1/2 vs. H 1 : θ = θ 1 = 3/4 . f (0 | θ 1 ) f (1 | θ 1 ) f (2 | θ 1 ) f (0 | θ 0 ) = 1 f (1 | θ 0 ) = 3 f (2 | θ 0 ) = 9 4 , 4 , 4 • Suppose that k < 1/4 , then the rejection region R = { 0 , 1 , 2 } , and UMP level α test always rejects H 0 . Therefore α = Pr ( reject H 0 | θ = θ 0 = 1/2) = 1 . • Suppose that 1/4 < k < 3/4 , then R = { 1 , 2 } , and UMP level α test rejects H 0 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

  25. • If k . . . .. . . .. .. . .. . . .. . . . . .. . .. . . .. . Example of Neyman-Pearson Lemma (cont’d) Pr reject Pr x the UMP level test always not reject H , and Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 .. . . . . . .. . . .. . .. .. . . .. . . .. . . . . . .. . . .. . 11 / 1 .. .. . . .. . . • Suppose that 3/4 < k < 9/4 , then UMP level α test rejects H 0 if x = 2

  26. • If k . . .. . . .. . . .. . . .. . . .. .. . . . . .. . . .. . Example of Neyman-Pearson Lemma (cont’d) the UMP level test always not reject H , and Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 .. . .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . .. . 11 / 1 . .. . . .. . . .. • Suppose that 3/4 < k < 9/4 , then UMP level α test rejects H 0 if x = 2 α = Pr ( reject | θ = 1/2) = Pr ( x = 2 | θ = 1/2) = 1 4

  27. . . .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Example of Neyman-Pearson Lemma (cont’d) Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 .. . . . . . .. . . .. . . .. . . .. .. . .. .. . . .. . . . . . .. . . .. . 11 / 1 • Suppose that 3/4 < k < 9/4 , then UMP level α test rejects H 0 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 H 0 , and α = 0

  28. . n x i i n exp f x f x x i exp i f x n i.i.d. X i Example - Normal Distribution . .. . . .. . . exp i .. n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang x i i n n exp x i i x i x i i n exp x i i n x i i n exp .. . . .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . .. . . . .. . . .. . . .. . 12 / 1 . .. . . .. . . .. . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 .

  29. . X i exp x i i n exp f x f x n i.i.d. Example - Normal Distribution i . .. . . .. . . .. . n x i .. i March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang x i i n n exp x i n exp x i i n exp x i i n x i i n . .. . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . 12 / 1 .. . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . − ( x i − θ ) 2 [ 1 { }] ∏ f ( x | θ ) = 2 πσ 2 exp 2 σ 2 i =1

  30. . . exp n i.i.d. X i Example - Normal Distribution . .. . .. exp . . .. . . .. . . exp n . x i March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang x i i n n exp i i n x i i n exp x i i n x i .. .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 12 / 1 . . . . .. . .. . . .. . .. . . . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . − ( x i − θ ) 2 [ 1 { }] ∏ f ( x | θ ) = 2 πσ 2 exp 2 σ 2 i =1 { ∑ n i =1 ( x i − θ 1 ) 2 } − f ( x | θ 1 ) 2 σ 2 = { } f ( x | θ 0 ) ∑ n i =1 ( x i − θ 0 ) 2 − 2 σ 2

  31. . . Example - Normal Distribution . .. . . .. . .. i.i.d. . . .. . . .. . . .. n . exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang x i i n n x i exp i n x i i n exp exp exp .. X i . . .. . . .. . . .. . .. . . . .. . . .. . . .. 12 / 1 . . . .. . . . .. . . .. . . . .. .. .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . − ( x i − θ ) 2 [ 1 { }] ∏ f ( x | θ ) = 2 πσ 2 exp 2 σ 2 i =1 { ∑ n i =1 ( x i − θ 1 ) 2 } − f ( x | θ 1 ) 2 σ 2 = { } f ( x | θ 0 ) ∑ n i =1 ( x i − θ 0 ) 2 − 2 σ 2 i =1 ( x i − θ 1 ) 2 i =1 ( x i − θ 0 ) 2 [ ∑ n ∑ n ] − = + 2 σ 2 2 σ 2

  32. . .. .. . . .. . . . . . .. . . .. . . . .. . exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang x i i n n exp . exp exp exp n i.i.d. .. Example - Normal Distribution .. X i . .. .. . . .. . . . . . .. . . .. . . . .. .. .. . . .. . . .. . . .. . . . 12 / 1 . . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . − ( x i − θ ) 2 [ 1 { }] ∏ f ( x | θ ) = 2 πσ 2 exp 2 σ 2 i =1 { ∑ n i =1 ( x i − θ 1 ) 2 } − f ( x | θ 1 ) 2 σ 2 = { } f ( x | θ 0 ) ∑ n i =1 ( x i − θ 0 ) 2 − 2 σ 2 i =1 ( x i − θ 1 ) 2 i =1 ( x i − θ 0 ) 2 [ ∑ n ∑ n ] − = + 2 σ 2 2 σ 2 i =1 ( x i − θ 0 ) 2 − ∑ n i =1 ( x i − θ 1 ) 2 [∑ n ] = 2 σ 2

  33. . .. .. . . .. . . . . . .. . . .. . . .. .. exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang exp exp exp exp . n .. X i Example - Normal Distribution . .. . . i.i.d. . . . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . . .. .. . 12 / 1 .. . . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . − ( x i − θ ) 2 [ 1 { }] ∏ f ( x | θ ) = 2 πσ 2 exp 2 σ 2 i =1 { ∑ n i =1 ( x i − θ 1 ) 2 } − f ( x | θ 1 ) 2 σ 2 = { } f ( x | θ 0 ) ∑ n i =1 ( x i − θ 0 ) 2 − 2 σ 2 i =1 ( x i − θ 1 ) 2 i =1 ( x i − θ 0 ) 2 [ ∑ n ∑ n ] − = + 2 σ 2 2 σ 2 i =1 ( x i − θ 0 ) 2 − ∑ n i =1 ( x i − θ 1 ) 2 [∑ n ] = 2 σ 2 0 − θ 1 ) 2 + 2 ∑ n [ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ] = 2 σ 2

  34. . . .. . . .. . . .. . Example (cont’d) .. . . .. . . .. . exp . Pr March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k X i i n k n x i i n log k x i i n . .. .. . . . .. . . .. . .. .. . . .. . . .. . . . 13 / 1 . . .. . . .. . . .. . . .. .. . . .. . . . UMP level α test rejects if 0 − θ 1 ) 2 + 2 ∑ n [ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ] > k 2 σ 2

  35. . . . . .. . . .. . . .. . .. .. . . .. .. . . n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k X i i Pr .. k x i i n exp Example (cont’d) . . . .. .. . . . .. . . . . . .. . . .. . . .. 13 / 1 . . . . .. .. . . .. . . .. . . .. . . .. UMP level α test rejects if 0 − θ 1 ) 2 + 2 ∑ n [ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ] > k 2 σ 2 0 − θ 1 ) 2 + 2 ∑ n ⇒ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ⇐ > log k 2 σ 2

  36. . .. .. . . .. . . .. . . .. . . .. . . .. .. n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k X i i Pr . n exp Example (cont’d) . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 13 / 1 . .. .. .. . . .. . . . . .. . . .. . . UMP level α test rejects if 0 − θ 1 ) 2 + 2 ∑ n [ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ] > k 2 σ 2 0 − θ 1 ) 2 + 2 ∑ n ⇒ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ⇐ > log k 2 σ 2 ∑ x i > k ∗ ⇐ ⇒ i =1

  37. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . Example (cont’d) exp n Pr Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . .. .. . . .. . . .. . .. .. . . . .. . . . 13 / 1 . . . . .. . . . .. .. . . .. . . .. UMP level α test rejects if 0 − θ 1 ) 2 + 2 ∑ n [ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ] > k 2 σ 2 0 − θ 1 ) 2 + 2 ∑ n ⇒ n ( θ 2 i =1 x i ( θ 1 − θ 0 ) ⇐ > log k 2 σ 2 ∑ x i > k ∗ ⇐ ⇒ i =1 ( n ) ∑ X i > k ∗ | θ 0 α = i =1

  38. . .. . .. . . .. . . . X i . .. . . .. . . Example (cont’d) X .. k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . where Z n n Z n Pr k X i i n Pr n X .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 14 / 1 . .. . . .. . . .. . . . . .. . . .. . . .. Under H 0 , N ( θ 0 , σ 2 ) ∼

  39. . . .. . . .. . . .. . Example (cont’d) .. . . .. . . .. . X i . k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . where Z n n Z X Pr k X i i n Pr n X . .. .. . . . .. . . .. . .. . . . .. . . .. . . .. 14 / 1 . . . .. . . .. . . .. . .. . . .. . . .. . Under H 0 , N ( θ 0 , σ 2 ) ∼ N ( θ 0 , σ 2 / n ) ∼

  40. . . . . .. . . .. . .. . . . .. . .. .. . .. Example (cont’d) .. k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . where Z n n Z X i Pr k X i i n Pr X . . . .. . .. . . .. . . . . . .. . . .. . . . 14 / 1 .. . .. . . .. . . .. . . .. . .. . . .. . . Under H 0 , N ( θ 0 , σ 2 ) ∼ N ( θ 0 , σ 2 / n ) ∼ X − θ 0 σ / √ n ∼ N (0 , 1)

  41. . . .. . . .. . . .. . .. .. . . .. . .. . X i March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang Pr Pr X Example (cont’d) . . .. . . .. . . . .. . . .. . . .. . . .. .. . . .. . . . 14 / 1 . . . . .. . . . .. . . .. . . .. .. Under H 0 , N ( θ 0 , σ 2 ) ∼ N ( θ 0 , σ 2 / n ) ∼ X − θ 0 σ / √ n ∼ N (0 , 1) ( n ) ∑ X i > k ∗ | θ 0 = α i =1 Z > k ∗ / n − θ 0 ( ) σ / √ n = where Z ∼ N (0 , 1) .

  42. if X .. . .. . . .. . . . .. . . .. . . .. . . . .. k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang n z n k , or equivalently, reject H Example (cont’d) X i test reject if Thus, the UMP level n z n k .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . .. .. . . .. . . .. . . . . . .. . . .. 15 / 1 k ∗ / n − θ 0 σ / √ n = z α

  43. . . . . .. . . .. . . .. . . .. . . .. .. . . k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang n z n if X .. k , or equivalently, reject H X i test reject if Thus, the UMP level n Example (cont’d) . . .. .. .. . . . .. . . . . . .. . . .. . . .. 15 / 1 . . .. . .. .. . . .. . . . .. . .. . . . k ∗ / n − θ 0 σ / √ n = z α ( ) σ k ∗ √ n = θ 0 + z α

  44. . .. . .. . .. .. . . .. . . .. . . . .. . .. . . .. . . .. . Example (cont’d) n Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . 15 / 1 .. . . .. . .. . k ∗ / n − θ 0 σ / √ n = z α ( ) σ k ∗ √ n = θ 0 + z α Thus, the UMP level α test reject if ∑ X i > k ∗ , or equivalently, reject H 0 if X > k ∗ / n = θ 0 + z α σ / √ n

  45. . . Neyman-Pearson Lemma on Sufficient Statistics . .. . . .. . .. Corollary 8.3.13 . . .. . . .. . . . . . k g t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T and For some k if g t . S c t and k g t if g t S t satisfies .. .. . . . . .. . . .. . .. . . . .. . . .. . . .. .. . . . . .. . . .. . . .. . .. .. . . .. . . 16 / 1 Consider H 0 : θ = θ 0 vs H 1 : θ = θ 1 . Suppose T ( X ) is a sufficient statistic for θ and g ( t | θ i ) is the pdf or pmf of T . Corresponding θ i , i ∈ { 0 , 1 } . Then any test based on T with rejection region S is a UMP level α test if it

  46. . . . .. . . .. . .. .. . . .. . . .. . . . .. k g t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T and For some k if g t Neyman-Pearson Lemma on Sufficient Statistics S c t satisfies . . Corollary 8.3.13 . .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . .. .. . . .. . . .. . . . . . .. . . .. 16 / 1 Consider H 0 : θ = θ 0 vs H 1 : θ = θ 1 . Suppose T ( X ) is a sufficient statistic for θ and g ( t | θ i ) is the pdf or pmf of T . Corresponding θ i , i ∈ { 0 , 1 } . Then any test based on T with rejection region S is a UMP level α test if it t ∈ S if g ( t | θ 1 ) > k · g ( t | θ 0 ) and

  47. . . . .. . . .. . .. .. . . .. . . .. . . . For some k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T and satisfies . . . Corollary 8.3.13 . Neyman-Pearson Lemma on Sufficient Statistics . .. . .. .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . 16 / 1 Consider H 0 : θ = θ 0 vs H 1 : θ = θ 1 . Suppose T ( X ) is a sufficient statistic for θ and g ( t | θ i ) is the pdf or pmf of T . Corresponding θ i , i ∈ { 0 , 1 } . Then any test based on T with rejection region S is a UMP level α test if it t ∈ S if g ( t | θ 1 ) > k · g ( t | θ 0 ) and t ∈ S c if g ( t | θ 1 ) < k · g ( t | θ 0 )

  48. . .. . .. .. . . .. . . .. . . .. . . . . Corollary 8.3.13 March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang satisfies . . . . Neyman-Pearson Lemma on Sufficient Statistics . .. . . .. .. . . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . 16 / 1 . . .. . . .. . . Consider H 0 : θ = θ 0 vs H 1 : θ = θ 1 . Suppose T ( X ) is a sufficient statistic for θ and g ( t | θ i ) is the pdf or pmf of T . Corresponding θ i , i ∈ { 0 , 1 } . Then any test based on T with rejection region S is a UMP level α test if it t ∈ S if g ( t | θ 1 ) > k · g ( t | θ 0 ) and t ∈ S c if g ( t | θ 1 ) < k · g ( t | θ 0 ) For some k > 0 and α = Pr ( T ∈ S | θ 0 )

  49. . . kg T x g T x x R The rejection region in the sample space is Proof . .. . f x .. . . .. . . .. . By Factorization Theorem: i .. By Neyman-Pearson Lemma, this test is the UMP level March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T X R Pr X test, and kf x h x g T x f x x h x kg T x h x g T x x R i . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. . . . .. . .. . . .. . . 17 / 1 = { x : T ( x ) = t ∈ S }

  50. . .. By Factorization Theorem: R The rejection region in the sample space is Proof . .. . . . i . .. . . .. . . .. f x h x g T x . By Neyman-Pearson Lemma, this test is the UMP level March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T X R Pr X test, and kf x i f x x h x kg T x h x g T x x R .. . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. . . .. . . .. 17 / 1 = { x : T ( x ) = t ∈ S } = { x : g ( T ( x ) | θ 1 ) > kg ( T ( x ) | θ 0 ) }

  51. . . Proof . .. . . .. . .. R . . .. . . .. .. . The rejection region in the sample space is By Factorization Theorem: . test, and March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T X R Pr X By Neyman-Pearson Lemma, this test is the UMP level R kf x f x x h x kg T x h x g T x x .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 17 / 1 .. . . . .. . . . . .. . . . .. .. . . . .. = { x : T ( x ) = t ∈ S } = { x : g ( T ( x ) | θ 1 ) > kg ( T ( x ) | θ 0 ) } f ( x | θ i ) = h ( x ) g ( T ( x ) | θ i )

  52. . . . .. . . .. . .. .. . . .. . . .. . . . .. test, and March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T X R Pr X By Neyman-Pearson Lemma, this test is the UMP level Proof kf x f x x R By Factorization Theorem: R The rejection region in the sample space is . .. . .. .. . . . .. . . . . . .. . . .. . . . 17 / 1 .. . .. . . . .. .. . .. . . . . .. . .. . . = { x : T ( x ) = t ∈ S } = { x : g ( T ( x ) | θ 1 ) > kg ( T ( x ) | θ 0 ) } f ( x | θ i ) = h ( x ) g ( T ( x ) | θ i ) = { x : g ( T ( x ) | θ 1 ) h ( x ) > kg ( T ( x ) | θ 0 ) h ( x ) }

  53. . . . . .. . . .. . . .. . . .. . . .. .. . . Pr X March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang S Pr T X R test, and .. By Neyman-Pearson Lemma, this test is the UMP level R By Factorization Theorem: R The rejection region in the sample space is Proof . . .. .. .. .. . . . . . . . . .. . . .. . . . .. .. . . .. . . .. . . .. . . 17 / 1 . .. . . .. = { x : T ( x ) = t ∈ S } = { x : g ( T ( x ) | θ 1 ) > kg ( T ( x ) | θ 0 ) } f ( x | θ i ) = h ( x ) g ( T ( x ) | θ i ) = { x : g ( T ( x ) | θ 1 ) h ( x ) > kg ( T ( x ) | θ 0 ) h ( x ) } = { x : f ( x | θ 1 ) > kf ( x | θ 0 ) }

  54. . . .. . . .. . .. .. . . .. . . .. . .. . The rejection region in the sample space is March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang R By Factorization Theorem: R Proof . . .. . . .. . . . .. .. . . .. . . .. . . . . . .. . . .. . . . .. . . . .. 17 / 1 . . . . .. . .. .. = { x : T ( x ) = t ∈ S } = { x : g ( T ( x ) | θ 1 ) > kg ( T ( x ) | θ 0 ) } f ( x | θ i ) = h ( x ) g ( T ( x ) | θ i ) = { x : g ( T ( x ) | θ 1 ) h ( x ) > kg ( T ( x ) | θ 0 ) h ( x ) } = { x : f ( x | θ 1 ) > kf ( x | θ 0 ) } By Neyman-Pearson Lemma, this test is the UMP level α test, and = Pr ( X ∈ R ) = Pr ( T ( X ) ∈ S | θ 0 ) α

  55. . . g t n . , where T X is a sufficient statistic for T i.i.d. X i Revisiting the Example of Normal Distribution .. n . . .. . . .. . . .. i exp .. exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang t n exp t t n n t t exp n t exp g t g t n i . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 18 / 1 ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 .

  56. . . i g t i.i.d. X i Revisiting the Example of Normal Distribution . .. . .. exp . . .. . . .. . . n t . n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang t n exp t t exp i n t exp n t exp g t g t n .. .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . 18 / 1 ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . T = X is a sufficient statistic for θ , where T ∼ N ( θ, σ 2 / n ) .

  57. . . Revisiting the Example of Normal Distribution . .. . . .. . .. i.i.d. . . .. . .. .. . . X i exp . t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang t n exp t n g t exp n t exp n t exp g t .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 18 / 1 .. . .. . . .. . . .. . . .. . . . .. . . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . T = X is a sufficient statistic for θ , where T ∼ N ( θ, σ 2 / n ) . − ( t − θ i ) 2 1 { } g ( t | θ i ) = 2 σ 2 / n √ 2 πσ 2 / n

  58. . .. .. . . .. . .. . . . .. . . .. . . . .. . t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang t n exp t n . exp exp exp exp i.i.d. X i Revisiting the Example of Normal Distribution .. . . .. .. . .. . .. . . . . . .. . . .. . . . 18 / 1 .. . . . .. . . .. . . .. . . .. . . .. . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . T = X is a sufficient statistic for θ , where T ∼ N ( θ, σ 2 / n ) . − ( t − θ i ) 2 1 { } g ( t | θ i ) = 2 σ 2 / n √ 2 πσ 2 / n { } − ( t − θ 1 ) 2 g ( t | θ 1 ) 2 σ 2 / n = g ( t | θ 0 ) { } − ( t − θ 0 ) 2 2 σ 2 / n

  59. . . . .. . . .. . .. .. . . .. . . .. . . . exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang t n exp exp . exp exp i.i.d. X i Revisiting the Example of Normal Distribution . .. . .. .. .. . . .. . . . . . . .. . . .. . . .. 18 / 1 . . . .. . .. . . .. . . .. .. . . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . T = X is a sufficient statistic for θ , where T ∼ N ( θ, σ 2 / n ) . − ( t − θ i ) 2 1 { } g ( t | θ i ) = 2 σ 2 / n √ 2 πσ 2 / n { } − ( t − θ 1 ) 2 g ( t | θ 1 ) 2 σ 2 / n = g ( t | θ 0 ) { } − ( t − θ 0 ) 2 2 σ 2 / n { 1 ( t − θ 1 ) 2 − ( t − θ 0 ) 2 ]} [ = − 2 σ 2 / n

  60. . .. .. . . .. . . . . . .. . . .. . . .. .. exp March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang exp exp exp exp .. i.i.d. X i Revisiting the Example of Normal Distribution . .. . . . . . . . .. . . .. . . . .. . . .. . . .. 18 / 1 . .. .. . .. . . . . .. . . .. . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider testing H 0 : θ = θ 0 vs. H 1 : θ = θ 1 where θ 1 > θ 0 . T = X is a sufficient statistic for θ , where T ∼ N ( θ, σ 2 / n ) . − ( t − θ i ) 2 1 { } g ( t | θ i ) = 2 σ 2 / n √ 2 πσ 2 / n { } − ( t − θ 1 ) 2 g ( t | θ 1 ) 2 σ 2 / n = g ( t | θ 0 ) { } − ( t − θ 0 ) 2 2 σ 2 / n { 1 ( t − θ 1 ) 2 − ( t − θ 0 ) 2 ]} [ = − 2 σ 2 / n { 1 ]} θ 2 1 − θ 2 [ = − 0 − 2 t ( θ 1 − θ 0 ) 2 σ 2 / n

  61. . .. .. . . .. . . . . . .. .. . .. . . .. .. log k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k t X t . n k exp Revisiting the Example (cont’d) . .. . . . . . . . . . . .. . . .. . . .. . . .. .. . .. .. . . .. . . .. . . .. 19 / 1 . . .. . . UMP level α test reject if { 1 ]} θ 2 1 − θ 2 [ − 0 − 2 t ( θ 1 − θ 0 ) > 2 σ 2 / n

  62. . . . . .. . . .. . . .. . . .. . . .. . . log k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k t X k .. exp Revisiting the Example (cont’d) . .. . . .. .. . . . .. . . .. .. . .. .. . . .. . . . 19 / 1 . . . . .. . . .. . . .. .. . . .. . . UMP level α test reject if { 1 ]} θ 2 1 − θ 2 [ − 0 − 2 t ( θ 1 − θ 0 ) > 2 σ 2 / n 1 − ( θ 2 1 − θ 2 [ ] ⇐ ⇒ 0 ) + 2 t ( θ 1 − θ 0 ) > 2 σ 2 / n

  63. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . Revisiting the Example (cont’d) exp k log k Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . .. .. . . .. . . .. . .. .. . . . .. . . . 19 / 1 . . . . .. . . .. . . . .. . . .. .. UMP level α test reject if { 1 ]} θ 2 1 − θ 2 [ − 0 − 2 t ( θ 1 − θ 0 ) > 2 σ 2 / n 1 − ( θ 2 1 − θ 2 [ ] ⇐ ⇒ 0 ) + 2 t ( θ 1 − θ 0 ) > 2 σ 2 / n k ∗ ⇐ ⇒ X = t >

  64. Pr X z n .. . .. . . .. . . . Pr reject H . .. . . .. . . Revisiting the Example (cont’d) . . n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k z n k k k Z Pr n k n X Pr .. . .. . . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . .. . . .. . . .. 20 / 1 Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies

  65. z n . . .. . . .. . . .. . Revisiting the Example (cont’d) .. . . .. . . .. . Pr X . n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k z n k k k Z Pr n k n X Pr .. . .. . .. . . .. . . .. . .. . . . .. . . .. . . . 20 / 1 . .. . .. . . .. . . . .. . .. . . .. . . Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α

  66. z n . . . .. . . .. . . .. .. . .. . . .. . . Revisiting the Example (cont’d) . n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k z n k k .. Z Pr n k n X Pr . .. . .. . .. . . .. . . . . . .. . . .. . . . 20 / 1 .. . .. . . .. . . .. . . .. . .. . . .. . . Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α Pr ( X > k ∗ | θ 0 ) α =

  67. z n . . . .. . . .. . .. .. . . .. . . .. . . . k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k z n n . k Z Pr Pr Revisiting the Example (cont’d) . .. . .. .. .. . . . .. . . . . . .. . . .. . . .. 20 / 1 . .. . . .. .. . . . . . . .. . .. . . .. Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α Pr ( X > k ∗ | θ 0 ) α = σ / √ n > k ∗ − θ 0 ( X − θ 0 ) σ / √ n =

  68. z n . .. . . .. . . .. . . .. . . .. . .. .. . k March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang k z n Pr .. Pr Revisiting the Example (cont’d) . .. . . . . . .. . . .. . . . . . . .. . . .. . . .. 20 / 1 . . . .. .. . . .. . . .. . . .. . . .. Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α Pr ( X > k ∗ | θ 0 ) α = σ / √ n > k ∗ − θ 0 ( X − θ 0 ) σ / √ n = Z > k ∗ − θ 0 ( ) σ / √ n =

  69. z n . . .. . . .. .. . .. . . .. . . .. . .. . . . .. . . .. . Revisiting the Example (cont’d) Pr Pr k Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . . .. . . .. . . .. . . .. .. . . .. . . . 20 / 1 . .. . .. . . .. . . . . . .. . . .. Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α Pr ( X > k ∗ | θ 0 ) α = σ / √ n > k ∗ − θ 0 ( X − θ 0 ) σ / √ n = Z > k ∗ − θ 0 ( ) σ / √ n = k ∗ − θ 0 σ / √ n = z α

  70. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . Revisiting the Example (cont’d) Pr Pr n Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 . .. .. . . .. . . .. . . .. .. . . .. . . . 20 / 1 . . .. . . .. . . . .. . . .. . . .. Under H 0 , X ∼ N ( θ 0 , σ 2 / n ) . k ∗ satisfies Pr ( reject H 0 | θ 0 ) = α Pr ( X > k ∗ | θ 0 ) α = σ / √ n > k ∗ − θ 0 ( X − θ 0 ) σ / √ n = Z > k ∗ − θ 0 ( ) σ / √ n = k ∗ − θ 0 σ / √ n = z α σ k ∗ = θ 0 + z α

  71. . . . . .. . . .. . . .. . . .. . .. .. . . . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang theorems are stated according to the definition. Note: we may define MLR using decreasing function of t . But all following . Definition .. . Monotone Likelihood Ratio . .. . . .. . . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. 21 / 1 . . .. . . .. . . A family of pdfs or pmfs { g ( t | θ ) : θ ∈ Ω } for a univariate random variable T with real-valued parameter θ have a monotone likelihood ratio if g ( t | θ 2 ) g ( t | θ 1 ) is an increasing (or non-decreasing) function of t for every θ 2 > θ 1 on { t : g ( t | θ 1 ) > 0 or g ( t | θ 2 ) > 0 } .

  72. . . . . .. . . .. . . .. . . .. . .. .. . . . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang theorems are stated according to the definition. Note: we may define MLR using decreasing function of t . But all following . Definition .. . Monotone Likelihood Ratio . .. . . .. . . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. 21 / 1 . . .. . . .. . . A family of pdfs or pmfs { g ( t | θ ) : θ ∈ Ω } for a univariate random variable T with real-valued parameter θ have a monotone likelihood ratio if g ( t | θ 2 ) g ( t | θ 1 ) is an increasing (or non-decreasing) function of t for every θ 2 > θ 1 on { t : g ( t | θ 1 ) > 0 or g ( t | θ 2 ) > 0 } .

  73. • If T is from an exponential family with the pdf or pmf . . .. . . .. . .. . . . .. . . .. . . .. .. t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . is a non-decreasing function of Then T has an MLR if w exp w . h t c g t Example of Monotone Likelihood Ratio . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . .. .. . . .. . . 22 / 1 • Normal, Poisson, Binomial have the MLR Property (Exercise 8.25)

  74. . . .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Example of Monotone Likelihood Ratio Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 .. . . . . . .. . . .. . . .. . . .. . .. .. .. . . .. . . . . . .. . . .. . 22 / 1 • Normal, Poisson, Binomial have the MLR Property (Exercise 8.25) • If T is from an exponential family with the pdf or pmf g ( t | θ ) = h ( t ) c ( θ ) exp [ w ( θ ) · t ] Then T has an MLR if w ( θ ) is a non-decreasing function of θ .

  75. t is an increasing function of t . Therefore, g t .. h t c t exp w h t c g t g t Proof . . t . . .. . . .. . . .. exp w c c g t March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . function of is a non-decreasing non-decreasing function of t , and T has MLR if w is a w . exp w and w , then w is a non-decreasing function of If w t w exp w . .. .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 23 / 1 Suppose that θ 2 > θ 1 .

  76. t is an increasing function of t . Therefore, g t . Proof . .. . . .. . . c .. . . .. . . .. .. c w exp w is a March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . function of is a non-decreasing non-decreasing function of t , and T has MLR if w g t .. w exp w and w , then w is a non-decreasing function of If w t . . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 23 / 1 . .. .. . . .. . . .. . . . . .. . . .. . Suppose that θ 2 > θ 1 . g ( t | θ 2 ) h ( t ) c ( θ 2 ) exp [ w ( θ 2 ) t ] = g ( t | θ 1 ) h ( t ) c ( θ 1 ) exp [ w ( θ 1 ) t ]

  77. t is an increasing function of t . Therefore, g t . . . .. . . .. . .. .. . . .. . .. .. . . Proof . is a March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . function of is a non-decreasing non-decreasing function of t , and T has MLR if w g t .. w exp w and w , then w is a non-decreasing function of If w . . . .. . .. . . .. . . . . . .. . . .. . . . 23 / 1 .. . .. . . .. . . .. . . .. . .. . . .. . . Suppose that θ 2 > θ 1 . g ( t | θ 2 ) h ( t ) c ( θ 2 ) exp [ w ( θ 2 ) t ] = g ( t | θ 1 ) h ( t ) c ( θ 1 ) exp [ w ( θ 1 ) t ] c ( θ 2 ) = c ( θ 1 ) exp [ { w ( θ 2 ) − w ( θ 1 ) } t ]

  78. t is an increasing function of t . Therefore, g t . . . .. . . .. . .. .. . . .. .. . .. . . . non-decreasing function of t , and T has MLR if w March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . function of is a non-decreasing is a . g t w exp w Proof . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. 23 / 1 . . .. . Suppose that θ 2 > θ 1 . g ( t | θ 2 ) h ( t ) c ( θ 2 ) exp [ w ( θ 2 ) t ] = g ( t | θ 1 ) h ( t ) c ( θ 1 ) exp [ w ( θ 1 ) t ] c ( θ 2 ) = c ( θ 1 ) exp [ { w ( θ 2 ) − w ( θ 1 ) } t ] If w ( θ ) is a non-decreasing function of θ , then w ( θ 2 ) − w ( θ 1 ) ≥ 0 and

  79. . . . . .. . .. .. . . .. . . .. . .. . . . .. . . .. . . .. . Proof Hyun Min Kang Biostatistics 602 - Lecture 20 March 28th, 2013 .. . . . . . .. . . .. . .. .. . . .. . . .. .. . . .. . . . 23 / 1 . . .. . . .. . Suppose that θ 2 > θ 1 . g ( t | θ 2 ) h ( t ) c ( θ 2 ) exp [ w ( θ 2 ) t ] = g ( t | θ 1 ) h ( t ) c ( θ 1 ) exp [ w ( θ 1 ) t ] c ( θ 2 ) = c ( θ 1 ) exp [ { w ( θ 2 ) − w ( θ 1 ) } t ] If w ( θ ) is a non-decreasing function of θ , then w ( θ 2 ) − w ( θ 1 ) ≥ 0 and exp [ { w ( θ 2 ) − w ( θ 1 ) } t ] is an increasing function of t . Therefore, g ( t | θ 2 ) g ( t | θ 1 ) is a non-decreasing function of t , and T has MLR if w ( θ ) is a non-decreasing function of θ .

  80. . . . . is an MLR family. Then . . Theorem 8.1.17 . Karlin-Rabin Theorem .. vs H . . .. . . .. . . .. 1 For testing H , the UMP level . test is given March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . t Pr T t where by rejecting H if and only if T , the UMP level test is given vs H 2 For testing H . . . t Pr T t where by rejecting H is and only if T . .. .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 24 / 1 Suppose T ( X ) is a sufficient statistic for θ and the family { g ( t | θ ) : θ ∈ Ω }

  81. . . Karlin-Rabin Theorem . .. . . .. . .. Theorem 8.1.17 . . .. . . .. . . . . . by rejecting H if and only if T March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . t Pr T t where test is given . , the UMP level vs H 2 For testing H . . . . is an MLR family. Then .. .. . . . . .. . . .. . .. . . . .. . . .. . . .. .. . . . . .. . . .. . . .. . .. .. . . .. . . 24 / 1 Suppose T ( X ) is a sufficient statistic for θ and the family { g ( t | θ ) : θ ∈ Ω } 1 For testing H 0 : θ ≤ θ 0 vs H 1 : θ > θ 0 , the UMP level α test is given by rejecting H 0 is and only if T > t 0 where α = Pr ( T > t 0 | θ 0 ) .

  82. . . . .. . . .. . .. .. . . .. . . .. . . . . March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang . . . is an MLR family. Then . . . Theorem 8.1.17 . Karlin-Rabin Theorem . .. .. . .. .. . . . .. . . . . . .. . . .. . . . .. .. .. . .. . . .. . . . . . .. . . .. . 24 / 1 Suppose T ( X ) is a sufficient statistic for θ and the family { g ( t | θ ) : θ ∈ Ω } 1 For testing H 0 : θ ≤ θ 0 vs H 1 : θ > θ 0 , the UMP level α test is given by rejecting H 0 is and only if T > t 0 where α = Pr ( T > t 0 | θ 0 ) . 2 For testing H 0 : θ ≥ θ 0 vs H 1 : θ < θ 0 , the UMP level α test is given by rejecting H 0 if and only if T < t 0 where α = Pr ( T < t 0 | θ 0 ) .

  83. n is an increasing function in . Let X i t exp n g t n . , and T X is a sufficient statistic for T X i.i.d. Example Application of Karlin-Rabin Theorem n . .. . . .. . . .. . n exp .. n March 28th, 2013 Biostatistics 602 - Lecture 20 Hyun Min Kang property. . Therefore T is MLR where w t exp w h t c t t exp n exp n t exp n n t . .. . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . 25 / 1 ∼ N ( θ, σ 2 ) where σ 2 is known, Find the UMP level α test for H 0 : θ ≤ θ 0 vs H 1 : θ > θ 0 .

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