Biostatistics 602 - Statistical Inference Apil 23rd, 2013 - PowerPoint PPT Presentation

f X x . .. . .. . . .. . . . Review . .. . . .. .. . . . . . . . . . P2 P1 x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang x . : E Bayes Estimator is a posterior mean of m x Posterior distribution . d f x Marginal distribution m x Bayesian Framework Wrap-up . P4 P3 .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 3 / 31 .. . . . .. . . . . .. . . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . Prior distribution π ( θ ) Sampling distribution x | θ ∼ f X ( x | θ ) Joint distribution π ( θ ) f ( x | θ )

f X x . . . .. . . .. . . . .. . . .. .. . .. .. Review . . . . . . . . x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang x . : E Bayes Estimator is a posterior mean of m x Posterior distribution . Bayesian Framework Wrap-up . P4 P3 P2 P1 . . .. .. . .. . . .. . . . .. . .. . . .. . . . . . . . .. . . .. . . .. . . 3 / 31 .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . Prior distribution π ( θ ) Sampling distribution x | θ ∼ f X ( x | θ ) Joint distribution π ( θ ) f ( x | θ ) ∫ π ( θ ) f ( x | θ ) d θ Marginal distribution m ( x ) =

. .. .. . . .. . . . . . .. . .. .. . . . .. . Bayesian Framework Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang x . : E Bayes Estimator is a posterior mean of Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . .. .. . .. .. . . . . . .. . . .. . . . . .. . . . .. . . .. . . .. . 3 / 31 .. . .. . . . . . . . . . . . . . . . . . . . . . . . . Prior distribution π ( θ ) Sampling distribution x | θ ∼ f X ( x | θ ) Joint distribution π ( θ ) f ( x | θ ) ∫ π ( θ ) f ( x | θ ) d θ Marginal distribution m ( x ) = Posterior distribution π ( θ | x ) = f X ( x | θ ) π ( θ ) m ( x )

. .. .. . . .. . . . . . .. . . .. . . .. .. P4 Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Bayesian Framework Wrap-up . P3 . P2 P1 Review . . . . . . . . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . .. .. . . . . . .. 3 / 31 . .. . . . . . . . . . . . . . . . . . . . . . Prior distribution π ( θ ) Sampling distribution x | θ ∼ f X ( x | θ ) Joint distribution π ( θ ) f ( x | θ ) ∫ π ( θ ) f ( x | θ ) d θ Marginal distribution m ( x ) = Posterior distribution π ( θ | x ) = f X ( x | θ ) π ( θ ) m ( x ) Bayes Estimator is a posterior mean of θ : E [ θ | x ] .

. . Review . . . . . . . . . .. . . .. . P2 .. . . .. . . .. .. P1 P3 .. : E R Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang expected loss. minimizes posterior Bayes Rule Estimator minimizes Bayes risk . Bayes Risk is the average risk across all P4 , the risk function is MSE For squared error loss L . E L Risk Function is the average loss : R Bayesian Decision Theory Wrap-up . . . . . . .. . . .. . . .. . . .. . . .. . . .. . . . 4 / 31 . . .. . . .. . .. .. .. . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Loss Function L ( θ, ˆ θ ) (e.g. ( θ − ˆ θ ) 2 )

. . .. . . .. . . .. . . . . . . . . . .. . . .. .. . .. . Review . : E R Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang expected loss. minimizes posterior Bayes Rule Estimator minimizes Bayes risk . Bayes Risk is the average risk across all P1 , the risk function is MSE For squared error loss L Bayesian Decision Theory Wrap-up . P4 P3 P2 . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 4 / 31 .. .. . . .. . .. . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . Loss Function L ( θ, ˆ θ ) (e.g. ( θ − ˆ θ ) 2 ) Risk Function is the average loss : R ( θ, ˆ θ ) = E [ L ( θ, ˆ θ ) | θ ] .

. . . . .. . . .. . .. . . . .. .. . .. . .. . . . . . . . . .. : E R Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang expected loss. minimizes posterior Bayes Rule Estimator minimizes Bayes risk . Bayes Risk is the average risk across all Review Bayesian Decision Theory Wrap-up . P4 P3 P2 P1 . . . .. . . . . .. . . . .. . .. . . .. . . . .. . . .. . . .. . . .. . . 4 / 31 .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . Loss Function L ( θ, ˆ θ ) (e.g. ( θ − ˆ θ ) 2 ) Risk Function is the average loss : R ( θ, ˆ θ ) = E [ L ( θ, ˆ θ ) | θ ] . For squared error loss L = ( θ − ˆ θ ) 2 , the risk function is MSE

. .. .. . . .. . . . . . .. .. . .. . . . .. . Bayesian Decision Theory Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang expected loss. minimizes posterior Bayes Rule Estimator minimizes Bayes risk Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . .. .. . . .. .. . . . . . .. . . .. . . . 4 / 31 .. . . . .. . . .. . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Loss Function L ( θ, ˆ θ ) (e.g. ( θ − ˆ θ ) 2 ) Risk Function is the average loss : R ( θ, ˆ θ ) = E [ L ( θ, ˆ θ ) | θ ] . For squared error loss L = ( θ − ˆ θ ) 2 , the risk function is MSE Bayes Risk is the average risk across all θ : E [ R ( θ, ˆ θ ) | π ( θ )] .

. . . .. . . .. . .. .. .. . .. . . .. . . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang expected loss. Bayesian Decision Theory Wrap-up P4 . P3 P2 P1 Review . . . . . . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . . .. . . 4 / 31 . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . Loss Function L ( θ, ˆ θ ) (e.g. ( θ − ˆ θ ) 2 ) Risk Function is the average loss : R ( θ, ˆ θ ) = E [ L ( θ, ˆ θ ) | θ ] . For squared error loss L = ( θ − ˆ θ ) 2 , the risk function is MSE Bayes Risk is the average risk across all θ : E [ R ( θ, ˆ θ ) | π ( θ )] . Bayes Rule Estimator minimizes Bayes risk ⇐ ⇒ minimizes posterior

Asymptotic Relative Efficiency ARE V n W n . . .. . . .. . . .. Review . . .. . . .. . .. . . . . . . . . P2 P1 V . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang asymptotically efficient under regularity condition. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always is 1. Asymptotically Efficient ARE with CR-bound of unbiased estimator of W . Delta Method Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Consistency Using law of large numbers, show variance and bias Asymptotics Wrap-up . P4 P3 .. . . . . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . .. . . .. . . .. 5 / 31 . . . . . . . . . . . . . . . . . . . . . converges to zero, for any continuous mapping function τ

Asymptotic Relative Efficiency ARE V n W n . . .. . . .. . . .. Review . . .. . . .. . .. . . . . . . . . P2 P1 V . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang asymptotically efficient under regularity condition. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always is 1. Asymptotically Efficient ARE with CR-bound of unbiased estimator of W . Delta Method Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Consistency Using law of large numbers, show variance and bias Asymptotics Wrap-up . P4 P3 .. . . . . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . .. . . .. . . .. 5 / 31 . . . . . . . . . . . . . . . . . . . . . converges to zero, for any continuous mapping function τ

. . .. . . .. . . .. . . . . . . . . . .. . . .. . .. .. . Review . V . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang asymptotically efficient under regularity condition. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always is 1. Asymptotically Efficient ARE with CR-bound of unbiased estimator of Delta Method P1 Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Consistency Using law of large numbers, show variance and bias Asymptotics Wrap-up . P4 P3 P2 . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 5 / 31 . . .. .. . . . . .. . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . converges to zero, for any continuous mapping function τ Asymptotic Relative Efficiency ARE ( V n , W n ) = σ 2 W / σ 2

. . .. . . .. . . .. . . . . . . . . . .. . . .. . .. .. . Review . Delta Method Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang asymptotically efficient under regularity condition. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always Asymptotically Efficient ARE with CR-bound of unbiased estimator of V . Asymptotic Normality Using central limit theorem, Slutsky Theorem, and P1 Consistency Using law of large numbers, show variance and bias Asymptotics Wrap-up . P4 P3 P2 . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 5 / 31 . .. . . . .. . . .. . . .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . converges to zero, for any continuous mapping function τ Asymptotic Relative Efficiency ARE ( V n , W n ) = σ 2 W / σ 2 τ ( θ ) is 1.

. . .. . . .. . . .. . . . . . . . . . .. . . .. . .. .. . Review . Delta Method Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang asymptotically efficient under regularity condition. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always Asymptotically Efficient ARE with CR-bound of unbiased estimator of V . Asymptotic Normality Using central limit theorem, Slutsky Theorem, and P1 Consistency Using law of large numbers, show variance and bias Asymptotics Wrap-up . P4 P3 P2 . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 5 / 31 . .. . . . .. . . .. . . .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . converges to zero, for any continuous mapping function τ Asymptotic Relative Efficiency ARE ( V n , W n ) = σ 2 W / σ 2 τ ( θ ) is 1.

. P4 Power function c when R Pr X Type II error Hypothesis Testing Wrap-up . P3 R P2 P1 Review . . . . . . . . . .. . . .. . Pr X represents Type I error under H , and power (=1-Type .. x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x log c rejects H when II error) under H . x L x L x LRT test sup Level test sup Size .. . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . 6 / 31 . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0

. Review Pr X Power function Hypothesis Testing Wrap-up . P4 P3 P2 P1 . . . . . . . . represents Type I error under H , and power (=1-Type . .. . . .. . . .. . R II error) under H . .. c Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x log x Size rejects H when x L x L x LRT test sup Level test sup . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . .. . . .. . . 6 / 31 . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0

. . . . . . . . . Hypothesis Testing Wrap-up . P4 P3 P2 P1 Review . II error) under H . .. . . .. . . .. .. . represents Type I error under H , and power (=1-Type Size . c Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x log x test sup rejects H when x L x L x LRT test sup Level .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 6 / 31 . .. . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ )

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. Size . . .. . . .. .. . .. Hypothesis Testing test sup . log Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x c Level x rejects H when x L x L x LRT test sup . . .. .. . .. . . .. . . .. . . . . . .. . . .. . . .. . . . 6 / 31 .. . . .. . . .. . . .. . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 .

. . P4 P3 P2 P1 Review . . . . . . . . . .. . Wrap-up .. . . .. . .. .. . . Hypothesis Testing .. log Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x c Level x rejects H when x L x L x LRT test sup . . . .. . . . .. . . .. . . . . . .. . . .. . . .. . . .. 6 / 31 . . .. . . .. . . .. . . .. . . .. . .. .. . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 . Size α test sup θ ∈ Ω 0 β ( θ ) = α

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . .. .. . . P4 Wrap-up . log Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c x c Hypothesis Testing x rejects H when x L x L x LRT .. . . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . 6 / 31 . . . .. . . .. . . .. . . .. . .. . . . .. . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 . Size α test sup θ ∈ Ω 0 β ( θ ) = α Level α test sup θ ∈ Ω 0 β ( θ ) ≤ α

. . . . .. . . .. . .. . .. . .. . . .. . .. . . . . . . . . .. x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient c log c log Review Hypothesis Testing Wrap-up . P4 P3 P2 P1 . . . .. .. . . . .. . . . . . .. . . .. . . . 6 / 31 .. .. .. . . . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 . Size α test sup θ ∈ Ω 0 β ( θ ) = α Level α test sup θ ∈ Ω 0 β ( θ ) ≤ α LRT λ ( x ) = L (ˆ θ 0 | x ) rejects H 0 when λ ( x ) ≤ c L (ˆ θ | x )

. . . . .. . . .. .. . .. . . .. . . .. .. . . Wrap-up Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient Hypothesis Testing . .. P4 P3 P2 P1 Review . . . . . . . . . . . .. .. . . .. . . . . . . .. . . .. . . .. 6 / 31 . . .. . .. . . .. . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 . Size α test sup θ ∈ Ω 0 β ( θ ) = α Level α test sup θ ∈ Ω 0 β ( θ ) ≤ α LRT λ ( x ) = L (ˆ θ 0 | x ) rejects H 0 when λ ( x ) ≤ c L (ˆ θ | x ) ⇒ − 2 log λ ( x ) ≥ − 2 log c = c ∗ ⇐

. . . . .. . . .. .. . .. . . .. . . .. .. . . Wrap-up Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang statistics are identical. LRT based on sufficient statistics LRT based on full data and sufficient Hypothesis Testing . .. P4 P3 P2 P1 Review . . . . . . . . . . . .. .. . . .. . . . . . . .. . . .. . . .. 6 / 31 . . .. . .. . . .. . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . Type I error Pr ( X ∈ R | θ ) when θ ∈ Ω 0 Type II error 1 − Pr ( X ∈ R | θ ) when θ ∈ Ω c 0 Power function β ( θ ) = Pr ( X ∈ R | θ ) β ( θ ) represents Type I error under H 0 , and power (=1-Type II error) under H 1 . Size α test sup θ ∈ Ω 0 β ( θ ) = α Level α test sup θ ∈ Ω 0 β ( θ ) ≤ α LRT λ ( x ) = L (ˆ θ 0 | x ) rejects H 0 when λ ( x ) ≤ c L (ˆ θ | x ) ⇒ − 2 log λ ( x ) ≥ − 2 log c = c ∗ ⇐

c and . . Test UMP test in the class of all the level UMP level . with a class of test of every other test for every UMP Test UMP Wrap-up P4 Type II error given the upper bound of Type I error) P3 P2 P1 Review . . . . . . . . . .. . . .. test. (smallest Neyman-Pearson For H . T Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. is an UMP level t T T or R t T vs. H R Karlin-Rabin If T is sufficient and has MLR, then test rejecting . is an increasing function of t for every g t MLR g t test for its size. k is a UMP level f x region f x , a test with rejection . .. .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . . .. . . .. . . .. . . 7 / 31 .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 .

. P2 Type II error given the upper bound of Type I error) test. (smallest Test UMP test in the class of all the level UMP level UMP Wrap-up . P4 P3 P1 vs. H Review . . . . . . . . . .. . . .. .. . .. Neyman-Pearson For H , a test with rejection . t Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. is an UMP level t T T or R T region f x T R Karlin-Rabin If T is sufficient and has MLR, then test rejecting . is an increasing function of t for every g t MLR g t test for its size. k is a UMP level f x . . .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . . .. . . .. . . .. . . .. . . .. . . .. 7 / 31 . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 . UMP Test β ( θ ) ≥ β ′ ( θ ) for every θ ∈ Ω c 0 and β ′ ( θ ) of every other test with a class of test C .

. Review Neyman-Pearson For H Type II error given the upper bound of Type I error) UMP Wrap-up . P4 P3 P2 P1 . . . . . . . . , a test with rejection . .. . . .. . .. .. . vs. H region f x .. t Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. is an UMP level t T T or R T f x T R Karlin-Rabin If T is sufficient and has MLR, then test rejecting . is an increasing function of t for every g t MLR g t test for its size. k is a UMP level . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . 7 / 31 .. . .. . . .. . . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 . UMP Test β ( θ ) ≥ β ′ ( θ ) for every θ ∈ Ω c 0 and β ′ ( θ ) of every other test with a class of test C . UMP level α Test UMP test in the class of all the level α test. (smallest

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . .. .. . . P4 Wrap-up . or R Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. is an UMP level t T T t UMP T T R Karlin-Rabin If T is sufficient and has MLR, then test rejecting . is an increasing function of t for every g t MLR g t Type II error given the upper bound of Type I error) .. . . .. . .. . .. . . .. . . . . . .. . . .. . . .. . . .. 7 / 31 . . . . .. . . .. . . .. . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 . UMP Test β ( θ ) ≥ β ′ ( θ ) for every θ ∈ Ω c 0 and β ′ ( θ ) of every other test with a class of test C . UMP level α Test UMP test in the class of all the level α test. (smallest Neyman-Pearson For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , a test with rejection region f ( x | θ 1 )/ f ( x | θ 0 ) > k is a UMP level α test for its size.

. . Review . . . . . . . . . .. . . .. . P2 .. . . .. . . .. . P1 P3 .. or R Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. is an UMP level t T T t P4 T T R Karlin-Rabin If T is sufficient and has MLR, then test rejecting Type II error given the upper bound of Type I error) UMP Wrap-up . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . . .. 7 / 31 . .. . . . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 . UMP Test β ( θ ) ≥ β ′ ( θ ) for every θ ∈ Ω c 0 and β ′ ( θ ) of every other test with a class of test C . UMP level α Test UMP test in the class of all the level α test. (smallest Neyman-Pearson For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , a test with rejection region f ( x | θ 1 )/ f ( x | θ 0 ) > k is a UMP level α test for its size. MLR g ( t | θ 2 )/ g ( t | θ 1 ) is an increasing function of t for every θ 2 > θ 1 .

. .. .. . . .. . . . . . .. .. . .. . . . .. . UMP Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang test for one-sided composite hypothesis. Karlin-Rabin If T is sufficient and has MLR, then test rejecting Type II error given the upper bound of Type I error) Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . .. .. . . .. .. . . . . . .. . . .. . . . 7 / 31 .. . . . .. . . .. . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Unbiased Test β ( θ 1 ) ≥ β ( θ 0 ) for every θ 1 ∈ Ω c 0 and θ 0 ∈ Ω 0 . UMP Test β ( θ ) ≥ β ′ ( θ ) for every θ ∈ Ω c 0 and β ′ ( θ ) of every other test with a class of test C . UMP level α Test UMP test in the class of all the level α test. (smallest Neyman-Pearson For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , a test with rejection region f ( x | θ 1 )/ f ( x | θ 0 ) > k is a UMP level α test for its size. MLR g ( t | θ 2 )/ g ( t | θ 1 ) is an increasing function of t for every θ 2 > θ 1 . R = { T : T > t 0 } or R = { T : T < t 0 } is an UMP level α

Wald Test If W n is a consistent estimator of , and S n is a consistent estimator of Var W n , then Z n S n follows a • Two-sided test : Z n • One-sided test : Z n . Asymptotic Tests and p-Values z standard normal distribution W n d Wrap-up or Z n . P4 P3 P2 P1 Review . . . . . . . . . .. .. . z z . valid p-value Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. is also a valid S x W x S X Pr W X p x p-Value given sufficient statistics For a sufficient statistic S X , is a p-Value A p-value W x Pr W X sup that H is true, p x Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence . and every for is valid if, Pr p X p x .. . . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . .. . . .. . . .. 8 / 31 . . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition.

• Two-sided test : Z n • One-sided test : Z n . P4 z z standard normal distribution d Asymptotic Tests and p-Values Wrap-up . P3 z P2 P1 Review . . . . . . . . . .. . .. .. . or Z n p-Value A p-value .. valid p-value Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. is also a valid S x W x S X Pr W X p x p-Value given sufficient statistics For a sufficient statistic S X , is a p x W x Pr W X sup that H is true, p x Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence . and every for is valid if, Pr p X . . . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . 8 / 31 . . . .. .. . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition. Wald Test If W n is a consistent estimator of θ , and S 2 n is a consistent estimator of Var ( W n ) , then Z n = ( W n − θ 0 )/ S n follows a

. Review standard normal distribution d Asymptotic Tests and p-Values Wrap-up . P4 P3 P2 P1 . . . . . . . . p x . .. . . .. .. . .. . p-Value A p-value is valid if, Pr p X .. p-Value given sufficient statistics For a sufficient statistic S X , Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. is also a valid S x W x S X Pr W X p x valid p-value for is a W x Pr W X sup that H is true, p x Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence . and every . . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . 8 / 31 . . . .. . . .. . . .. .. . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition. Wald Test If W n is a consistent estimator of θ , and S 2 n is a consistent estimator of Var ( W n ) , then Z n = ( W n − θ 0 )/ S n follows a • Two-sided test : | Z n | > z α /2 • One-sided test : Z n > z α /2 or Z n < − z α /2

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . . .. . . P4 Wrap-up . p x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. is also a valid S x W x S X Pr W X p-Value given sufficient statistics For a sufficient statistic S X , Asymptotic Tests and p-Values valid p-value is a W x Pr W X sup that H is true, p x Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence standard normal distribution d .. .. . .. . .. . .. . . .. . . . . . .. . . .. . . .. . . .. 8 / 31 . . . . .. . . .. . .. . . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition. Wald Test If W n is a consistent estimator of θ , and S 2 n is a consistent estimator of Var ( W n ) , then Z n = ( W n − θ 0 )/ S n follows a • Two-sided test : | Z n | > z α /2 • One-sided test : Z n > z α /2 or Z n < − z α /2 p-Value A p-value 0 ≤ p ( x ) ≤ 1 is valid if, Pr ( p ( X ) ≤ α | θ ) ≤ α for every θ ∈ Ω 0 and 0 ≤ α ≤ 1 .

. . . . . . . . . . . .. . . .. . .. P1 . .. .. . . .. . . Review P2 . Pr W X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. is also a valid S x W x S X p x P3 p-Value given sufficient statistics For a sufficient statistic S X , valid p-value standard normal distribution d Asymptotic Tests and p-Values Wrap-up . P4 .. . . . . . .. . . .. . .. . . . .. . . .. . . .. .. . . . . .. . . .. . . .. . .. 8 / 31 . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition. Wald Test If W n is a consistent estimator of θ , and S 2 n is a consistent estimator of Var ( W n ) , then Z n = ( W n − θ 0 )/ S n follows a • Two-sided test : | Z n | > z α /2 • One-sided test : Z n > z α /2 or Z n < − z α /2 p-Value A p-value 0 ≤ p ( x ) ≤ 1 is valid if, Pr ( p ( X ) ≤ α | θ ) ≤ α for every θ ∈ Ω 0 and 0 ≤ α ≤ 1 . Constructing p-Value Theorem 8.3.27 : If large W ( X ) value gives evidence that H 1 is true, p ( x ) = sup θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ ) is a

. .. .. . . .. . . .. . . .. . . .. . . . .. . Asymptotic Tests and p-Values Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang p-value. valid p-value standard normal distribution d Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . .. .. . . .. . . . . . .. . . .. . . . .. .. .. . . .. . . .. . . . 8 / 31 . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . Asymptotic Distribution of LRT For testing, H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , → χ 2 − 2 log λ ( x ) 1 under regularity condition. Wald Test If W n is a consistent estimator of θ , and S 2 n is a consistent estimator of Var ( W n ) , then Z n = ( W n − θ 0 )/ S n follows a • Two-sided test : | Z n | > z α /2 • One-sided test : Z n > z α /2 or Z n < − z α /2 p-Value A p-value 0 ≤ p ( x ) ≤ 1 is valid if, Pr ( p ( X ) ≤ α | θ ) ≤ α for every θ ∈ Ω 0 and 0 ≤ α ≤ 1 . Constructing p-Value Theorem 8.3.27 : If large W ( X ) value gives evidence that H 1 is true, p ( x ) = sup θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ ) is a p-Value given sufficient statistics For a sufficient statistic S ( X ) , p ( x ) = Pr ( W ( X ) ≥ W ( x ) | S ( X ) = S ( x )) is also a valid

. Review if inf Coverage coefficient is Interval Estimation Wrap-up . P4 P3 P2 P1 . . . . . . . . L X . .. . . .. . . .. . Pr U X .. test, Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang interval). confidence set (or is a A X then C X is the acceptance region of a level Confidence interval L X test If A Inverting a level U X L X Pr inf if is U X .. . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . .. . . .. . . .. 9 / 31 . . . . . . . . . . . . . . . . . . . . . Coverage probability Pr ( θ ∈ [ L ( X ) , U ( X )])

. .. . P4 P3 P2 P1 Review . . . . . . . . . . Interval Estimation . .. . . .. . . .. .. Wrap-up Confidence interval L X .. then C X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang interval). confidence set (or is a A X test, U X is the acceptance region of a level test If A Inverting a level U X L X Pr inf if is . . . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . . .. . . .. . 9 / 31 . . . . . . . . . . . . . . . . . . . . . Coverage probability Pr ( θ ∈ [ L ( X ) , U ( X )]) Coverage coefficient is 1 − α if inf θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )]) = 1 − α

. . . . . . . . . . . .. . . .. . .. P1 . . .. . . .. .. . Review P2 . X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang interval). confidence set (or is a A then C X P3 test, is the acceptance region of a level test If A Inverting a level Interval Estimation Wrap-up . P4 .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 9 / 31 . . . .. . . .. . . .. .. . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . Coverage probability Pr ( θ ∈ [ L ( X ) , U ( X )]) Coverage coefficient is 1 − α if inf θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )]) = 1 − α Confidence interval [ L ( X ) , U ( X )]) is 1 − α if inf θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )]) = 1 − α

. . . .. . . .. . .. .. . . .. .. . .. . . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang interval). Interval Estimation Wrap-up P4 . P3 P2 P1 Review . . . . . . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . .. . 9 / 31 . . .. . .. . . . . . . . . . . . . . . . . . . . . . . Coverage probability Pr ( θ ∈ [ L ( X ) , U ( X )]) Coverage coefficient is 1 − α if inf θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )]) = 1 − α Confidence interval [ L ( X ) , U ( X )]) is 1 − α if inf θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )]) = 1 − α Inverting a level α test If A ( θ 0 ) is the acceptance region of a level α test, then C ( X ) = { θ : X ∈ A ( θ ) } is a 1 − α confidence set (or

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . . .. . . P4 Wrap-up . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . H vs. for testing H (c) Show that the test in part (b) is UMP size versus H Practice Problem 1 (continued from last week) H test of (b) Based on one observation X , find the most powerful size (a) Show that this family has an MLR . . Problem . .. .. . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . 10 / 31 . . . .. . . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Let f ( x | θ ) be the logistic location pdf e ( x − θ ) f ( x | θ ) = − ∞ < x < ∞ , −∞ < θ < ∞ (1 + e ( x − θ ) ) 2

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . . .. . . P4 Wrap-up . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . H vs. for testing H (c) Show that the test in part (b) is UMP size versus H Practice Problem 1 (continued from last week) H test of (b) Based on one observation X , find the most powerful size (a) Show that this family has an MLR . . Problem . .. .. . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . 10 / 31 . . . .. . . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Let f ( x | θ ) be the logistic location pdf e ( x − θ ) f ( x | θ ) = − ∞ < x < ∞ , −∞ < θ < ∞ (1 + e ( x − θ ) ) 2

. . . . . . . . . . . .. . . .. . .. P1 . . .. . .. .. . . Review P2 . (c) Show that the test in part (b) is UMP size Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . H vs. for testing H (a) Show that this family has an MLR P3 . . Problem . Practice Problem 1 (continued from last week) Wrap-up . P4 .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 10 / 31 . . .. .. . . .. . . . . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Let f ( x | θ ) be the logistic location pdf e ( x − θ ) f ( x | θ ) = − ∞ < x < ∞ , −∞ < θ < ∞ (1 + e ( x − θ ) ) 2 (b) Based on one observation X , find the most powerful size α test of H 0 : θ = 0 versus H 1 : θ = 1 .

. . . .. . . .. . .. .. . . .. . . .. . . . .. . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang (a) Show that this family has an MLR . . Problem Practice Problem 1 (continued from last week) . . . . . . . . Wrap-up . P4 P3 P2 P1 Review . .. . .. . .. . . .. . . . .. . .. . . .. . . . . . .. .. . . .. . . .. . . . . . .. . 10 / 31 .. . . . . . . . . . . . . . . . . . . . . . . Let f ( x | θ ) be the logistic location pdf e ( x − θ ) f ( x | θ ) = − ∞ < x < ∞ , −∞ < θ < ∞ (1 + e ( x − θ ) ) 2 (b) Based on one observation X , find the most powerful size α test of H 0 : θ = 0 versus H 1 : θ = 1 . (c) Show that the test in part (b) is UMP size α for testing H 0 : θ ≤ 0 vs. H 1 : θ > 0 .

e x e x e x e x e x e x e x e x e x e x . Review . . . . . . . . . .. . . .. . P2 . .. .. . .. . P1 e P3 r x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. x x e x P4 e x Let r x .. Solution for (a) Wrap-up . . . . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . 11 / 31 . . .. . .. . .. .. . . . .. . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2

e x e x e x e x e x e x e x e x . . . . . . . . . .. . . .. . . .. P1 .. . . .. . . Review Wrap-up P2 r x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. x x e x P3 e x Let r x Solution for (a) . . P4 .. . . .. . . .. .. . . .. . . . . . .. . . .. . . .. . . .. 11 / 31 . . . . . .. . . .. . . .. . . .. . . .. .. . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2 ) 2 ( 1 + e ( x − θ 1 ) e ( θ 1 − θ 2 ) = 1 + e ( x − θ 2 )

e x e x e x e x e x e x e x e x . . .. . . .. . . . . .. . . .. . .. P3 . . . . . . . . r x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. x x Solution for (a) Review Wrap-up . P4 .. P2 P1 . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . . 11 / 31 . . .. . .. . .. . . .. . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2 ) 2 ( 1 + e ( x − θ 1 ) e ( θ 1 − θ 2 ) = 1 + e ( x − θ 2 ) Let r ( x ) = (1 + e x − θ 1 )/(1 + e x − θ 2 )

e x e x e x .. .. . . .. .. . . . . . .. . . .. . . . . . . . . . . .. Solution for (a) Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. x x Wrap-up . . P4 P3 P2 P1 Review .. . . . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. . . 11 / 31 .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2 ) 2 ( 1 + e ( x − θ 1 ) e ( θ 1 − θ 2 ) = 1 + e ( x − θ 2 ) Let r ( x ) = (1 + e x − θ 1 )/(1 + e x − θ 2 ) e ( x − θ 1 ) (1 + e ( x − θ 2 ) ) − (1 + e ( x − θ 1 ) ) e ( x − θ 2 ) r ′ ( x ) = (1 + e ( x − θ 2 ) ) 2

. . . .. .. . .. . .. .. . . .. . . .. . . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. Solution for (a) Wrap-up P4 . P3 P2 P1 Review . . . . . . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . . .. . . 11 / 31 . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2 ) 2 ( 1 + e ( x − θ 1 ) e ( θ 1 − θ 2 ) = 1 + e ( x − θ 2 ) Let r ( x ) = (1 + e x − θ 1 )/(1 + e x − θ 2 ) e ( x − θ 1 ) (1 + e ( x − θ 2 ) ) − (1 + e ( x − θ 1 ) ) e ( x − θ 2 ) r ′ ( x ) = (1 + e ( x − θ 2 ) ) 2 e ( x − θ 1 ) − e ( x − θ 2 ) ( ∵ x − θ 1 > x − θ 2 ) = > 0 (1 + e ( x − θ 2 ) ) 2

. . . .. .. . .. . .. .. . . .. . . .. . . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Therefore, the family of X has an MLR. Solution for (a) Wrap-up P4 . P3 P2 P1 Review . . . . . . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . . .. . . 11 / 31 . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . For θ 1 < θ 2 , e ( x − θ 2) f ( x | θ 2 ) (1+ e ( x − θ 2) ) 2 = f ( x | θ 1 ) e ( x − θ 1) (1+ e ( x − θ 1) ) 2 ) 2 ( 1 + e ( x − θ 1 ) e ( θ 1 − θ 2 ) = 1 + e ( x − θ 2 ) Let r ( x ) = (1 + e x − θ 1 )/(1 + e x − θ 2 ) e ( x − θ 1 ) (1 + e ( x − θ 2 ) ) − (1 + e ( x − θ 1 ) ) e ( x − θ 2 ) r ′ ( x ) = (1 + e ( x − θ 2 ) ) 2 e ( x − θ 1 ) − e ( x − θ 2 ) ( ∵ x − θ 1 > x − θ 2 ) = > 0 (1 + e ( x − θ 2 ) ) 2

e x e x , the rejection region of UMP level . .. P4 P3 P2 P1 Review . . . . . . . . . . . Wrap-up .. . . .. . . .. . . e Solution for (b) e x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang log x e x F x test satisfies Because under H , F x .. x X k e x e e x k e x . .. . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . 12 / 31 . . .. . . .. . . .. . . .. . . .. . .. .. . . . . . . . . . . . . . . . . . . . . . . . The UMP test rejects H 0 if and only if ) 2 f ( x | 1) ( 1 + e x = > k 1 + e ( x − 1) f ( x | 0)

e x , the rejection region of UMP level .. P2 P1 Review . . . . . . . . . .. . . . P4 . . .. . . .. . . P3 Wrap-up . e x Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang log x e x F x test satisfies Because under H , F x . x X k e x e e x e Solution for (b) .. .. . .. . . .. .. . . .. . . . . . .. . . .. . . .. . . .. 12 / 31 . . .. . . .. . . .. . .. . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . The UMP test rejects H 0 if and only if ) 2 f ( x | 1) ( 1 + e x = > k 1 + e ( x − 1) f ( x | 0) 1 + e x k ∗ > 1 + e ( x − 1)

e x , the rejection region of UMP level . . . . . . . . . . .. . . .. . . P1 .. . . .. . . .. . Review P3 P2 test satisfies Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang log x e x F x e x .. Because under H , F x x X e Solution for (b) Wrap-up . P4 . .. . . .. . . .. . . . .. . . .. . . .. . . .. . . . 12 / 31 . . . .. . . .. . . .. . .. . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . The UMP test rejects H 0 if and only if ) 2 f ( x | 1) ( 1 + e x = > k 1 + e ( x − 1) f ( x | 0) 1 + e x k ∗ > 1 + e ( x − 1) 1 + e x k ∗∗ > e + e x

e x , the rejection region of UMP level .. . .. . . .. . .. . Review . . .. . . .. . . . . . . . . . . P2 P1 test satisfies Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang log x e x F x e x . Because under H , F x X e Solution for (b) Wrap-up . P4 P3 .. . . . . . .. . .. . .. . .. . . . .. . . .. . . .. 12 / 31 . . . .. . . . .. . . .. . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . The UMP test rejects H 0 if and only if ) 2 f ( x | 1) ( 1 + e x = > k 1 + e ( x − 1) f ( x | 0) 1 + e x k ∗ > 1 + e ( x − 1) 1 + e x k ∗∗ > e + e x > x 0

. .. .. . . .. . . . . . .. . . .. . . .. .. . Solution for (b) Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang log e x X e Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. . . .. .. . . .. . . .. . . . 12 / 31 .. . . .. . . . . . .. .. . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . The UMP test rejects H 0 if and only if ) 2 f ( x | 1) ( 1 + e x = > k 1 + e ( x − 1) f ( x | 0) 1 + e x k ∗ > 1 + e ( x − 1) 1 + e x k ∗∗ > e + e x > x 0 Because under H 0 , F ( x 0 | θ = 0) = 1+ e x , the rejection region of UMP level α test satisfies 1 1 − F ( x | θ = 0) = 1 + e x 0 = α ( 1 − α ) ∼ x 0 α

. .. .. . . .. . . . . . .. . .. .. . . . .. . Solution for (c) Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang , which is identical to the test defined in (b). log Therefore, x X Wrap-up . . P4 P3 P2 P1 Review . . . . . . . . .. . . .. .. .. . .. . . . . . .. . . .. . . . . .. .. . . .. . . .. . . . 13 / 31 . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . Because the family of X has an MLR, UMP size α for testing H 0 : θ ≤ 0 vs. H 1 : θ > 0 should be a form of > x 0 Pr ( X > x 0 | θ = 0) = α

. . . . .. . . .. . . .. .. . .. . . .. .. . . Wrap-up Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang , which is identical to the test defined in (b). X Solution for (c) . .. P4 P3 P2 P1 Review . . . . . . . . . . . .. .. . . .. . . . . . . .. . . .. . . .. 13 / 31 . . . .. . . .. . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Because the family of X has an MLR, UMP size α for testing H 0 : θ ≤ 0 vs. H 1 : θ > 0 should be a form of > x 0 Pr ( X > x 0 | θ = 0) = α ( 1 − α ) Therefore, x 0 = log α

X i is a consistent estimator for X i is asymptotically normal and derive its asymptotic . P3 . . Problem . Practice Problem 2 Wrap-up . P4 P2 n P1 Review . . . . . . . . . .. . . .. . .. (a) Show that x n (d) Find an asymptotic Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . and Var X You may use the fact that E X above test by inverting the confidence interval for . . vs. H test for H (c) Derive the Wald asymptotic size distribution x n n (b) Show that . .. . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . 14 / 31 .. . .. . . .. . . .. .. . . . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples with pdf f X ( x | θ ) = θ exp ( − θ x ) , where x ≥ 0 , θ > 0

X i is asymptotically normal and derive its asymptotic . Review . Practice Problem 2 Wrap-up . P4 P3 P2 P1 . . . . . . . . . . .. . . .. . . .. . Problem (a) Show that . (d) Find an asymptotic Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . and Var X You may use the fact that E X above test by inverting the confidence interval for . .. vs. H test for H (c) Derive the Wald asymptotic size distribution x n n (b) Show that n . .. . .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 14 / 31 . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples with pdf f X ( x | θ ) = θ exp ( − θ x ) , where x ≥ 0 , θ > 0 x =1 X i is a consistent estimator for θ . ∑ n

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. .. . .. Practice Problem 2 Problem . (d) Find an asymptotic Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . and Var X You may use the fact that E X above test by inverting the confidence interval for . . vs. H test for H (c) Derive the Wald asymptotic size distribution n (b) Show that n (a) Show that . . . .. .. . . . .. . . .. . . . . . .. . . .. . . .. . . . .. .. . . .. . . .. . . .. . . .. . .. . . . .. . 14 / 31 . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples with pdf f X ( x | θ ) = θ exp ( − θ x ) , where x ≥ 0 , θ > 0 x =1 X i is a consistent estimator for θ . ∑ n ∑ n x =1 X i is asymptotically normal and derive its asymptotic

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . .. .. . . P4 Wrap-up . confidence interval for Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . and Var X You may use the fact that E X above test by inverting the (d) Find an asymptotic Practice Problem 2 distribution n (b) Show that n (a) Show that . . Problem . .. . . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. 14 / 31 . .. . . .. . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples with pdf f X ( x | θ ) = θ exp ( − θ x ) , where x ≥ 0 , θ > 0 x =1 X i is a consistent estimator for θ . ∑ n ∑ n x =1 X i is asymptotically normal and derive its asymptotic (c) Derive the Wald asymptotic size α test for H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 .

. . . . . . . . . . . .. . . .. . .. P1 . . .. .. . .. . . Review P2 . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang above test distribution n (b) Show that (a) Show that P3 . . Problem . Practice Problem 2 Wrap-up . P4 .. . . . .. . .. . . .. . .. . . . .. . . .. . . .. . . . . . .. . . .. . . . .. .. .. .. . . . 14 / 31 . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples with pdf f X ( x | θ ) = θ exp ( − θ x ) , where x ≥ 0 , θ > 0 x =1 X i is a consistent estimator for θ . ∑ n ∑ n x =1 X i is asymptotically normal and derive its asymptotic (c) Derive the Wald asymptotic size α test for H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . (d) Find an asymptotic (1 − α ) confidence interval for θ by inverting the You may use the fact that E X = 1/ θ and Var ( X ) = 1/ θ 2 .

. Review . . Solution (a) - Consistency Wrap-up . P4 P3 P2 P1 . . . . . . . . x exp . .. . . .. .. . .. . E X x .. 3 By Theorem of continuous map, n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . P X X i i n . exp . . E X P 2 By LLN (Law of Large Number), X . x exp x dx . . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . 15 / 31 . . . .. . . .. . . .. .. . .. . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . 1 Obtain E X = 1/ θ (Derive yourself if not given) ∫ ∞ ∫ ∞ xf ( x | θ ) dx = θ x exp ( − θ x ) dx = 0 0

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. . . .. Solution (a) - Consistency . . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . P X X i i 3 By Theorem of continuous map, n E X . . . E X P 2 By LLN (Law of Large Number), X . x exp . .. .. .. .. . . . .. . . .. . . . . . .. . . .. . . .. . . . 15 / 31 .. .. .. . . .. . . . . . .. . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . 1 Obtain E X = 1/ θ (Derive yourself if not given) ∫ ∞ ∫ ∞ xf ( x | θ ) dx = θ x exp ( − θ x ) dx = 0 0 ∫ ∞ [ − x exp ( − θ x )] ∞ exp ( − θ x ) dx = 0 + 0

. .. . P4 P3 P2 P1 Review . . . . . . . . .. . Solution (a) - Consistency . .. . . .. . . .. . Wrap-up . .. n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . P X X i i 3 By Theorem of continuous map, n . . . . E X P 2 By LLN (Law of Large Number), X . . E X . . . .. . . . .. . . .. . . . . . .. . . .. . . .. . . .. 15 / 31 . .. .. . .. . . .. .. . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . 1 Obtain E X = 1/ θ (Derive yourself if not given) ∫ ∞ ∫ ∞ xf ( x | θ ) dx = θ x exp ( − θ x ) dx = 0 0 ∫ ∞ [ − x exp ( − θ x )] ∞ exp ( − θ x ) dx = 0 + 0 ] ∞ [ − 1 = 1 θ exp ( − θ x ) = 0 + θ 0

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . . .. . . .. . . P4 Wrap-up . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . P X X i i 3 By Theorem of continuous map, n Solution (a) - Consistency . . P 2 By LLN (Law of Large Number), X . . E X . . .. .. . .. . . .. . .. . .. . . . . . .. . . .. . . .. . . .. 15 / 31 . . . . .. . . .. . . .. . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . 1 Obtain E X = 1/ θ (Derive yourself if not given) ∫ ∞ ∫ ∞ xf ( x | θ ) dx = θ x exp ( − θ x ) dx = 0 0 ∫ ∞ [ − x exp ( − θ x )] ∞ exp ( − θ x ) dx = 0 + 0 ] ∞ [ − 1 = 1 θ exp ( − θ x ) = 0 + θ 0 → E X = 1/ θ .

. . . . . . . . . . . .. .. . .. . .. P1 . . .. . . .. . . Review P2 . 2 By LLN (Law of Large Number), X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang P . . P . P3 . E X . . Solution (a) - Consistency Wrap-up . P4 .. . . . . . .. . . .. . .. .. . . . .. . . .. . . .. 15 / 31 . . .. . . .. . . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . 1 Obtain E X = 1/ θ (Derive yourself if not given) ∫ ∞ ∫ ∞ xf ( x | θ ) dx = θ x exp ( − θ x ) dx = 0 0 ∫ ∞ [ − x exp ( − θ x )] ∞ exp ( − θ x ) dx = 0 + 0 ] ∞ [ − 1 = 1 θ exp ( − θ x ) = 0 + θ 0 → E X = 1/ θ . 3 By Theorem of continuous map, n / ∑ n i =1 X i = 1/ X → θ .

. . . . . . . . . Solution (b) - Asymptotic Distribution Wrap-up . P4 P3 P2 P1 Review . . .. . . .. . . .. . . . . . g X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X n n n g g X 2 Apply CLT(Central Limit Theorem), n X i y . y , then g y 3 Apply Delta method. Let g y . . n X .. .. . .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . .. . . . . .. . . .. . . .. . .. . . . .. . . .. 16 / 31 . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here).

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. .. . .. Solution (b) - Asymptotic Distribution . . g X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X n n n g g X . n X i y . y , then g y 3 Apply Delta method. Let g y . . 2 Apply CLT(Central Limit Theorem), . . . .. .. . .. . . .. . . .. . . . . . .. . . .. . . .. . . . 16 / 31 .. . .. . . . .. .. . . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here). ( 1 θ, 1 ) X ∼ AN θ 2 n

. . P4 P3 P2 P1 Review . . . . . . . . . .. . Wrap-up .. . . .. . .. .. . . Solution (b) - Asymptotic Distribution .. g Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X n n n g g X . X n X i . . 2 Apply CLT(Central Limit Theorem), . . . . . . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . . .. .. .. . . .. . . .. . . .. . . . . .. . . 16 / 31 . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here). ( 1 θ, 1 ) X ∼ AN θ 2 n 3 Apply Delta method. Let g ( y ) = 1/ y , then g ′ ( y ) = − 1/ y 2 .

. . Review . . . . . . . . . .. . . .. . P2 .. .. . .. . . .. . P1 P3 .. . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X n n n . P4 2 Apply CLT(Central Limit Theorem), . . . . Solution (b) - Asymptotic Distribution Wrap-up . . . . . .. . . . . . .. . .. . . . .. . . .. . . . .. 16 / 31 . .. .. . . .. .. . . .. . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here). ( 1 θ, 1 ) X ∼ AN θ 2 n 3 Apply Delta method. Let g ( y ) = 1/ y , then g ′ ( y ) = − 1/ y 2 . g (1/ θ ) , [ g ′ (1/ θ )] 2 ∑ X i ( ) = 1/ X = g ( X ) ∼ AN θ 2 n

. . Review . . . . . . . . . .. . . .. .. P2 .. . . .. . . .. . P1 P3 .. . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X n n n . P4 2 Apply CLT(Central Limit Theorem), . . . . Solution (b) - Asymptotic Distribution Wrap-up . . . . . .. . . .. . . .. . .. . . . .. . . .. . . . . 16 / 31 . .. . . .. . . .. .. . .. . . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here). ( 1 θ, 1 ) X ∼ AN θ 2 n 3 Apply Delta method. Let g ( y ) = 1/ y , then g ′ ( y ) = − 1/ y 2 . g (1/ θ ) , [ g ′ (1/ θ )] 2 ∑ X i ( ) = 1/ X = g ( X ) ∼ AN θ 2 n θ, θ 2 ( ) AN =

. .. . .. .. . .. . . . Review . .. . . .. . . . . . . . . . . P1 . 2 Apply CLT(Central Limit Theorem), Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang n n . . . P2 . . . Solution (b) - Asymptotic Distribution Wrap-up . P4 P3 .. . . . . . .. . . .. . .. . . . .. . . .. . . .. .. 16 / 31 . . . .. . . .. . . .. . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . 1 Obtain Var ( X ) = 1/ θ 2 (Derive if needed, omitted here). ( 1 θ, 1 ) X ∼ AN θ 2 n 3 Apply Delta method. Let g ( y ) = 1/ y , then g ′ ( y ) = − 1/ y 2 . g (1/ θ ) , [ g ′ (1/ θ )] 2 ∑ X i ( ) = 1/ X = g ( X ) ∼ AN θ 2 n θ, θ 2 ( ) AN = ( 1 ⇒ √ n ) 0 , θ 2 ) ( ⇐ X − θ = N

. Wrap-up . . n n X i i n W X . . . n P4 P3 P2 P1 Review . . . . . . . . . .. . .. 2 Obtain a constant estimator of Var W n . S Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Slutsky’s Theorem P X X i i n n Continuous Map Theorem i P X X i i n n CLT Var X P X X i .. . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . 17 / 31 . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ :

. P4 2 Obtain a constant estimator of Var W . . n n . . Wrap-up . P3 n P2 P1 Review . . . . . . . . . .. . . .. . n i .. S Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Slutsky’s Theorem P X X i i n n Continuous Map Theorem X i P X X i i n n CLT Var X P X . .. . .. . .. . . .. . . .. . . .. . . . .. . .. . . .. . . .. . . .. . . . 17 / 31 . . . . . .. . . . .. .. . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ : θ, θ 2 ∑ n ( ) i =1 X i W ( X ) = ∼ AN

. P3 . . n n . . Wrap-up . P4 P2 n P1 Review . . . . . . . . . .. . . .. . . n i . S Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Slutsky’s Theorem P X X i i n n Continuous Map Theorem X i P X X i i n n CLT Var X P X .. .. . .. . .. . . .. . . .. . . .. . . . .. . .. . . .. . . .. . . .. . . . 17 / 31 .. . . .. . . . .. . . . .. . . .. .. . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ : θ, θ 2 ∑ n ( ) i =1 X i W ( X ) = ∼ AN 2 Obtain a constant estimator of Var ( W )

. . . . . . . . . . . Wrap-up . P4 P3 P2 P1 Review .. n .. . . .. . . .. . . n . . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Slutsky’s Theorem P X X i i n S . Continuous Map Theorem P X X i i n n P n .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 17 / 31 .. . . . .. . . . .. . . .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ : θ, θ 2 ∑ n ( ) i =1 X i W ( X ) = ∼ AN 2 Obtain a constant estimator of Var ( W ) 1 Var ( X ) = 1 ∑ ( X i − X ) 2 → ( CLT ) θ 2 n − 1 i =1

. . P3 P2 P1 Review . . . . . . . . .. .. . .. . . . .. . . .. . . P4 Wrap-up . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang Slutsky’s Theorem P X X i i n . S P P n . . n n . .. . . .. . . .. .. . . .. . . . . . .. . . .. . . .. . . .. 17 / 31 . .. . . .. . . . .. .. . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ : θ, θ 2 ∑ n ( ) i =1 X i W ( X ) = ∼ AN 2 Obtain a constant estimator of Var ( W ) 1 Var ( X ) = 1 ∑ ( X i − X ) 2 → ( CLT ) θ 2 n − 1 i =1 n − 1 θ 2 → ( Continuous Map Theorem ) . i =1 ( X i − X ) 2 ∑ n

. . . . . . . . . . .. .. . . .. . .. P1 . . .. . . .. . . Review P2 . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang P n P P . P3 . n n . . Wrap-up . P4 .. . . . . . .. . . .. .. . .. . . . .. . . .. . . .. 17 / 31 . . .. . .. . . .. . . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald asymptotic size α test 1 Obtain a consistent estimator of θ : θ, θ 2 ∑ n ( ) i =1 X i W ( X ) = ∼ AN 2 Obtain a constant estimator of Var ( W ) 1 Var ( X ) = 1 ∑ ( X i − X ) 2 → ( CLT ) θ 2 n − 1 i =1 n − 1 θ 2 → ( Continuous Map Theorem ) . i =1 ( X i − X ) 2 ∑ n S 2 = θ 2 → ( Slutsky’s Theorem ) . i =1 ( X i − X ) 2 ∑ n

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. . . .. . region is . X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang z X X i i n X X i Z X i n X i i n n n S W X . .. .. .. . .. . . .. . . .. . . . . . .. . . .. . . .. . . . .. 18 / 31 .. . .. . . .. . . .. . . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald Asymptotic size α test (cont’d) 3 Construct a two-sided asymptotic size α Wald test, whose rejection

. . P3 P2 P1 Review . . . . . . . . . .. . .. . . .. .. . . .. . . P4 Wrap-up . X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang z X X i i n X . X i i n X i i n n region is . .. . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. .. . . . . .. . . .. . . .. . . .. . . .. 18 / 31 . .. . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald Asymptotic size α test (cont’d) 3 Construct a two-sided asymptotic size α Wald test, whose rejection � W ( X ) − θ 0 � � � | Z ( X ) | S / √ n = � � � �

. .. .. .. . . .. . . . Review . .. . . .. . . . . . . . . . . P1 . n Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang z X X i i X P2 n region is . . Wrap-up . P4 P3 .. . . . . . .. . . .. . .. .. . . . .. . . .. . . .. 18 / 31 . . .. . . . .. .. . . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald Asymptotic size α test (cont’d) 3 Construct a two-sided asymptotic size α Wald test, whose rejection � W ( X ) − θ 0 � � � | Z ( X ) | S / √ n = � � � � � � i =1 X i − θ 0 � � ∑ n � � = � � √ ∑ n 1 � � i =1 ( X i − X ) 2 � �

. .. .. . . .. . . . . . .. . . .. . . . .. . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang n n region is . Wrap-up . . P4 P3 .. P1 Review . . . . . . . . .. P2 . . .. . . .. . . .. . . .. .. . . .. . . . 18 / 31 .. . . . .. . . . . .. .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Wald Asymptotic size α test (cont’d) 3 Construct a two-sided asymptotic size α Wald test, whose rejection � W ( X ) − θ 0 � � � | Z ( X ) | S / √ n = � � � � � � i =1 X i − θ 0 � � ∑ n � � = � � √ ∑ n 1 � � i =1 ( X i − X ) 2 � � � � � � 1 ∑ ( X i − X ) 2 ≥ z α /2 � � � = X − θ 0 � � � � � i =1

. Review n A The acceptance region is Wrap-up . P4 P3 P2 .. . . . . . . . . C X . .. . . .. . . .. . By inverting the acceptance region, the confidence interval is X .. X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X X i i n z X X i n i n z X C X which is equivalent to z X X i i . P1 . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . 19 / 31 . . .. . . .. . . .. . .. .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Solution (d) - Asymptotic 1 − α confidence interval  �  � � � 1   ( x i − x ) 2 ≤ z α /2 � ∑ � � =  x : x − θ 0 � � � � �  i =1

. Review n A The acceptance region is Wrap-up . P4 P3 P2 .. . . . . . . . . C X . .. . . .. . . .. . By inverting the acceptance region, the confidence interval is X .. X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X X i i n z X X i n i n z X C X which is equivalent to z X X i i . P1 . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . 19 / 31 . . .. . . .. . . .. . .. .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . Solution (d) - Asymptotic 1 − α confidence interval  �  � � � 1   ( x i − x ) 2 ≤ z α /2 � ∑ � � =  x : x − θ 0 � � � � �  i =1

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. The acceptance region is . . .. . . .. . . .. .. A . X Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang X X i i n z X n X i i n z X C X which is equivalent to n By inverting the acceptance region, the confidence interval is . 19 / 31 .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . .. . . .. . . .. . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Solution (d) - Asymptotic 1 − α confidence interval  �  � � � 1   ( x i − x ) 2 ≤ z α /2 � ∑ � � =  x : x − θ 0 � � � � �  i =1  �  � � � 1   ( X i − X ) 2 ≤ z α /2 ∑ � � � C ( X ) =  θ : X − θ � � � � �  i =1

. . . . .. . . .. . .. . . . .. . . .. . .. . . . . . . . . .. .. Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang which is equivalent to n By inverting the acceptance region, the confidence interval is n A Review The acceptance region is Wrap-up . P4 P3 P2 P1 . 19 / 31 . . . .. . . .. . .. .. . . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . Solution (d) - Asymptotic 1 − α confidence interval  �  � � � 1   ( x i − x ) 2 ≤ z α /2 � ∑ � � =  x : x − θ 0 � � � � �  i =1  �  � � � 1   ( X i − X ) 2 ≤ z α /2 ∑ � � � C ( X ) =  θ : X − θ � � � � �  i =1      1 , 1  z α /2 z α /2  C ( X ) =  θ ∈ X − X +  √∑ n √∑ n i =1 ( X i − X ) 2 i =1 ( X i − X ) 2 

. . . . . . . . . . Practice Problem 3 Wrap-up . P4 P3 P2 P1 Review . . .. . . .. . . .. . . Problem . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . with confidence coefficient , find the upper confidence limit for is a known constant 3 When . . . vs. H , construct a LRT testing H is a known constant 2 When . . and 1 Find the MLEs of . .. .. . .. .. .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . . . . . .. . . .. . . .. . . .. 20 / 31 . . .. . . .. . . . . . . . . . . . . . . . . . . . . . The independent random variables X 1 , · · · , X n have the following pdf β x β − 1 f ( x | θ, β ) = 0 < x < θ, β > 0 θ β

. . Wrap-up . P4 P3 P2 P1 Review . . . . . . . . .. . . . .. . . .. . . .. Practice Problem 3 Problem . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . with confidence coefficient , find the upper confidence limit for is a known constant 3 When . . . vs. H , construct a LRT testing H is a known constant 2 When . . . . . . .. .. .. . . . .. . . .. . . . . . .. . . .. . . .. . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . 20 / 31 . . . . . . . . . . . . . . . . . . . . . The independent random variables X 1 , · · · , X n have the following pdf β x β − 1 f ( x | θ, β ) = 0 < x < θ, β > 0 θ β 1 Find the MLEs of β and θ

. .. P2 P1 Review . . . . . . . . . .. . . . P4 . .. . .. .. . . .. P3 . . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . with confidence coefficient , find the upper confidence limit for is a known constant 3 When . Wrap-up . . . . . Problem . Practice Problem 3 . . .. . . . . . .. . . .. . .. .. . . .. . . .. . . . .. . . . .. . . .. . . .. . . .. . 20 / 31 .. . .. . . . . . . . . . . . . . . . . . . . . . . . The independent random variables X 1 , · · · , X n have the following pdf β x β − 1 f ( x | θ, β ) = 0 < x < θ, β > 0 θ β 1 Find the MLEs of β and θ 2 When β is a known constant β 0 , construct a LRT testing H 0 : θ ≥ θ 0 vs. H 1 : θ < θ 0 .

. . . . . . . . . . . .. . . .. . .. P1 . . .. . . .. . . Review P2 . . Apil 23rd, 2013 Biostatistics 602 - Lecture 26 Hyun Min Kang . . . . . P3 . . Problem . Practice Problem 3 Wrap-up . P4 .. .. . . . .. .. . . .. . .. . . . .. . . .. . . .. . . . . . .. . . .. . . .. 20 / 31 . . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . The independent random variables X 1 , · · · , X n have the following pdf β x β − 1 f ( x | θ, β ) = 0 < x < θ, β > 0 θ β 1 Find the MLEs of β and θ 2 When β is a known constant β 0 , construct a LRT testing H 0 : θ ≥ θ 0 vs. H 1 : θ < θ 0 . 3 When β is a known constant β 0 , find the upper confidence limit for θ with confidence coefficient 1 − α .