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

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Biostatistics 602 - Statistical Inference Lecture 26 Final Exam Review & Practice Problems for the Final

Hyun Min Kang Apil 23rd, 2013

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 1 / 31

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Review of the second half

Rao-Blackwell : If W X is an unbiased estimator of , T E W X T is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Review of the second half

Rao-Blackwell : If W(X) is an unbiased estimator of τ(θ), φ(T) = E[W(X)|T] is a better unbiased estimator for a sufficient statistic. Uniqueness of MVUE : Theorem 7.3.19 - Best unbiased estimator is unique MVUE and UE of zeros : Theorem 7.3.20 - Best unbiased estimator is uncorrelated with any unbiased estimators of zero UMVE by complete sufficient statistics : Theorem 7.3.23 - Any function

• f complete sufficient statistic is the best unbiased estimator

for its expected value How to get UMVUE Strategies to obtain best unbiased estimators:

• Condition a simple unbiased estimator on complete

sufficient statistics

• Come up with a function of sufficient statistic whose

expected value is τ(θ).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 2 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x fX x Joint distribution f x Marginal distribution m x f x d Posterior distribution x

fX x m x

Bayes Estimator is a posterior mean of : E x .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x|θ ∼ fX(x|θ) Joint distribution f x Marginal distribution m x f x d Posterior distribution x

fX x m x

Bayes Estimator is a posterior mean of : E x .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x|θ ∼ fX(x|θ) Joint distribution π(θ)f(x|θ) Marginal distribution m x f x d Posterior distribution x

fX x m x

Bayes Estimator is a posterior mean of : E x .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x|θ ∼ fX(x|θ) Joint distribution π(θ)f(x|θ) Marginal distribution m(x) = ∫ π(θ)f(x|θ)dθ Posterior distribution x

fX x m x

Bayes Estimator is a posterior mean of : E x .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x|θ ∼ fX(x|θ) Joint distribution π(θ)f(x|θ) Marginal distribution m(x) = ∫ π(θ)f(x|θ)dθ Posterior distribution π(θ|x) = fX(x|θ)π(θ)

m(x)

Bayes Estimator is a posterior mean of : E x .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Framework

Prior distribution π(θ) Sampling distribution x|θ ∼ fX(x|θ) Joint distribution π(θ)f(x|θ) Marginal distribution m(x) = ∫ π(θ)f(x|θ)dθ Posterior distribution π(θ|x) = fX(x|θ)π(θ)

m(x)

Bayes Estimator is a posterior mean of θ : E[θ|x].

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 3 / 31

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Bayesian Decision Theory

Loss Function L(θ, ˆ θ) (e.g. (θ − ˆ θ)2) Risk Function is the average loss : R E L . For squared error loss L , the risk function is MSE Bayes Risk is the average risk across all : E R . Bayes Rule Estimator minimizes Bayes risk minimizes posterior expected loss.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 4 / 31

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Bayesian Decision Theory

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 , the risk function is MSE Bayes Risk is the average risk across all : E R . Bayes Rule Estimator minimizes Bayes risk minimizes posterior expected loss.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 4 / 31

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Bayesian Decision Theory

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 : E R . Bayes Rule Estimator minimizes Bayes risk minimizes posterior expected loss.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 4 / 31

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Bayesian Decision Theory

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 expected loss.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 4 / 31

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Bayesian Decision Theory

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 expected loss.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 4 / 31

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Asymptotics

Consistency Using law of large numbers, show variance and bias converges to zero, for any continuous mapping function τ Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Delta Method Asymptotic Relative Efficiency ARE Vn Wn

W V.

Asymptotically Efficient ARE with CR-bound of unbiased estimator of is 1. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always asymptotically efficient under regularity condition.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 5 / 31

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Asymptotics

Consistency Using law of large numbers, show variance and bias converges to zero, for any continuous mapping function τ Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Delta Method Asymptotic Relative Efficiency ARE Vn Wn

W V.

Asymptotically Efficient ARE with CR-bound of unbiased estimator of is 1. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always asymptotically efficient under regularity condition.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 5 / 31

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Asymptotics

Consistency Using law of large numbers, show variance and bias converges to zero, for any continuous mapping function τ Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Delta Method Asymptotic Relative Efficiency ARE(Vn, Wn) = σ2

W/σ2 V.

Asymptotically Efficient ARE with CR-bound of unbiased estimator of is 1. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always asymptotically efficient under regularity condition.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 5 / 31

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Asymptotics

Consistency Using law of large numbers, show variance and bias converges to zero, for any continuous mapping function τ Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Delta Method Asymptotic Relative Efficiency ARE(Vn, Wn) = σ2

W/σ2 V.

Asymptotically Efficient ARE with CR-bound of unbiased estimator of τ(θ) is 1. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always asymptotically efficient under regularity condition.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 5 / 31

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Asymptotics

Consistency Using law of large numbers, show variance and bias converges to zero, for any continuous mapping function τ Asymptotic Normality Using central limit theorem, Slutsky Theorem, and Delta Method Asymptotic Relative Efficiency ARE(Vn, Wn) = σ2

W/σ2 V.

Asymptotically Efficient ARE with CR-bound of unbiased estimator of τ(θ) is 1. Asymptotic Efficiency of MLE Theorem 10.1.12 MLE is always asymptotically efficient under regularity condition.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 5 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error Pr X R when

c

Power function Pr X R represents Type I error under H , and power (=1-Type II error) under H . Size test sup Level test sup LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function Pr X R represents Type I error under H , and power (=1-Type II error) under H . Size test sup Level test sup LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) represents Type I error under H , and power (=1-Type II error) under H . Size test sup Level test sup LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size test sup Level test sup LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size α test supθ∈Ω0 β(θ) = α Level test sup LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size α test supθ∈Ω0 β(θ) = α Level α test supθ∈Ω0 β(θ) ≤ α LRT x L x L x rejects H when x c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size α test supθ∈Ω0 β(θ) = α Level α test supθ∈Ω0 β(θ) ≤ α LRT λ(x) = L(ˆ θ0|x) L(ˆ θ|x) rejects H0 when λ(x) ≤ c log x log c c LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size α test supθ∈Ω0 β(θ) = α Level α test supθ∈Ω0 β(θ) ≤ α LRT λ(x) = L(ˆ θ0|x) L(ˆ θ|x) rejects H0 when λ(x) ≤ c ⇐ ⇒ −2 log λ(x) ≥ −2 log c = c∗ LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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Hypothesis Testing

Type I error Pr(X ∈ R|θ) when θ ∈ Ω0 Type II error 1 − Pr(X ∈ R|θ) when θ ∈ Ωc Power function β(θ) = Pr(X ∈ R|θ) β(θ) represents Type I error under H0, and power (=1-Type II error) under H1. Size α test supθ∈Ω0 β(θ) = α Level α test supθ∈Ω0 β(θ) ≤ α LRT λ(x) = L(ˆ θ0|x) L(ˆ θ|x) rejects H0 when λ(x) ≤ c ⇐ ⇒ −2 log λ(x) ≥ −2 log c = c∗ LRT based on sufficient statistics LRT based on full data and sufficient statistics are identical.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 6 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test for every

c and

• f every other test

with a class of test . UMP level Test UMP test in the class of all the level

• test. (smallest

Type II error given the upper bound of Type I error) Neyman-Pearson For H

• vs. H

, a test with rejection region f x f x k is a UMP level test for its size. MLR g t g t is an increasing function of t for every . Karlin-Rabin If T is sufficient and has MLR, then test rejecting R T T t

• r R

T T t is an UMP level test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test β(θ) ≥ β′(θ) for every θ ∈ Ωc

0 and β′(θ) of every other test

with a class of test C. UMP level Test UMP test in the class of all the level

• test. (smallest

Type II error given the upper bound of Type I error) Neyman-Pearson For H

• vs. H

, a test with rejection region f x f x k is a UMP level test for its size. MLR g t g t is an increasing function of t for every . Karlin-Rabin If T is sufficient and has MLR, then test rejecting R T T t

• r R

T T t is an UMP level test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test β(θ) ≥ β′(θ) for every θ ∈ Ωc

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 Type II error given the upper bound of Type I error) Neyman-Pearson For H

• vs. H

, a test with rejection region f x f x k is a UMP level test for its size. MLR g t g t is an increasing function of t for every . Karlin-Rabin If T is sufficient and has MLR, then test rejecting R T T t

• r R

T T t is an UMP level test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test β(θ) ≥ β′(θ) for every θ ∈ Ωc

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 Type II error given the upper bound of Type I error) Neyman-Pearson For H0 : θ = θ0 vs. H1 : θ = θ1, a test with rejection region f(x|θ1)/f(x|θ0) > k is a UMP level α test for its size. MLR g t g t is an increasing function of t for every . Karlin-Rabin If T is sufficient and has MLR, then test rejecting R T T t

• r R

T T t is an UMP level test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test β(θ) ≥ β′(θ) for every θ ∈ Ωc

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 Type II error given the upper bound of Type I error) Neyman-Pearson For H0 : θ = θ0 vs. H1 : θ = θ1, a test with rejection region f(x|θ1)/f(x|θ0) > k is a UMP level α test for its size. MLR g(t|θ2)/g(t|θ1) is an increasing function of t for every θ2 > θ1. Karlin-Rabin If T is sufficient and has MLR, then test rejecting R T T t

• r R

T T t is an UMP level test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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UMP

Unbiased Test β(θ1) ≥ β(θ0) for every θ1 ∈ Ωc

0 and θ0 ∈ Ω0.

UMP Test β(θ) ≥ β′(θ) for every θ ∈ Ωc

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 Type II error given the upper bound of Type I error) Neyman-Pearson For H0 : θ = θ0 vs. H1 : θ = θ1, a test with rejection region f(x|θ1)/f(x|θ0) > k is a UMP level α test for its size. MLR g(t|θ2)/g(t|θ1) is an increasing function of t for every θ2 > θ1. Karlin-Rabin If T is sufficient and has MLR, then test rejecting R = {T : T > t0} or R = {T : T < t0} is an UMP level α test for one-sided composite hypothesis.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 7 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of , and Sn is a consistent estimator of Var Wn , then Zn Wn Sn follows a standard normal distribution

• Two-sided test : Zn

z

• One-sided test : Zn

z

• r Zn

z p-Value A p-value p x is valid if, Pr p X for every and . Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence that H is true, p x sup Pr W X W x is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of θ, and S2

n is a consistent

estimator of Var(Wn), then Zn = (Wn − θ0)/Sn follows a standard normal distribution

• Two-sided test : Zn

z

• One-sided test : Zn

z

• r Zn

z p-Value A p-value p x is valid if, Pr p X for every and . Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence that H is true, p x sup Pr W X W x is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of θ, and S2

n is a consistent

estimator of Var(Wn), then Zn = (Wn − θ0)/Sn follows a standard normal distribution

• Two-sided test : |Zn| > zα/2
• One-sided test : Zn > zα/2 or Zn < −zα/2

p-Value A p-value p x is valid if, Pr p X for every and . Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence that H is true, p x sup Pr W X W x is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of θ, and S2

n is a consistent

estimator of Var(Wn), then Zn = (Wn − θ0)/Sn follows a standard normal distribution

• Two-sided test : |Zn| > zα/2
• One-sided test : Zn > zα/2 or Zn < −zα/2

p-Value A p-value 0 ≤ p(x) ≤ 1 is valid if, Pr(p(X) ≤ α|θ) ≤ α for every θ ∈ Ω0 and 0 ≤ α ≤ 1. Constructing p-Value Theorem 8.3.27 : If large W X value gives evidence that H is true, p x sup Pr W X W x is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of θ, and S2

n is a consistent

estimator of Var(Wn), then Zn = (Wn − θ0)/Sn follows a standard normal distribution

• Two-sided test : |Zn| > zα/2
• One-sided test : Zn > zα/2 or Zn < −zα/2

p-Value A p-value 0 ≤ p(x) ≤ 1 is valid if, Pr(p(X) ≤ α|θ) ≤ α for every θ ∈ Ω0 and 0 ≤ α ≤ 1. Constructing p-Value Theorem 8.3.27 : If large W(X) value gives evidence that H1 is true, p(x) = supθ∈Ω0 Pr(W(X) ≥ W(x)|θ) is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Asymptotic Tests and p-Values

Asymptotic Distribution of LRT For testing, H0 : θ = θ0 vs. H1 : θ = θ1, −2 log λ(x)

d

→ χ2

1 under regularity condition.

Wald Test If Wn is a consistent estimator of θ, and S2

n is a consistent

estimator of Var(Wn), then Zn = (Wn − θ0)/Sn follows a standard normal distribution

• Two-sided test : |Zn| > zα/2
• One-sided test : Zn > zα/2 or Zn < −zα/2

p-Value A p-value 0 ≤ p(x) ≤ 1 is valid if, Pr(p(X) ≤ α|θ) ≤ α for every θ ∈ Ω0 and 0 ≤ α ≤ 1. Constructing p-Value Theorem 8.3.27 : If large W(X) value gives evidence that H1 is true, p(x) = supθ∈Ω0 Pr(W(X) ≥ W(x)|θ) is a valid p-value 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 p-value.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 8 / 31

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Interval Estimation

Coverage probability Pr(θ ∈ [L(X), U(X)]) Coverage coefficient is if inf Pr L X U X Confidence interval L X U X is if inf Pr L X U X Inverting a level test If A is the acceptance region of a level test, then C X X A is a confidence set (or interval).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 9 / 31

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Interval Estimation

Coverage probability Pr(θ ∈ [L(X), U(X)]) Coverage coefficient is 1 − α if infθ∈Ω Pr(θ ∈ [L(X), U(X)]) = 1 − α Confidence interval L X U X is if inf Pr L X U X Inverting a level test If A is the acceptance region of a level test, then C X X A is a confidence set (or interval).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 9 / 31

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Interval Estimation

Coverage probability Pr(θ ∈ [L(X), U(X)]) Coverage coefficient is 1 − α if infθ∈Ω Pr(θ ∈ [L(X), U(X)]) = 1 − α Confidence interval [L(X), U(X)]) is 1 − α if infθ∈Ω Pr(θ ∈ [L(X), U(X)]) = 1 − α Inverting a level test If A is the acceptance region of a level test, then C X X A is a confidence set (or interval).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 9 / 31

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Interval Estimation

Coverage probability Pr(θ ∈ [L(X), U(X)]) Coverage coefficient is 1 − α if infθ∈Ω Pr(θ ∈ [L(X), U(X)]) = 1 − α Confidence interval [L(X), U(X)]) is 1 − α if infθ∈Ω Pr(θ ∈ [L(X), U(X)]) = 1 − α Inverting a level α test If A(θ0) is the acceptance region of a level α test, then C(X) = {θ : X ∈ A(θ)} is a 1 − α confidence set (or interval).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 9 / 31

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Practice Problem 1 (continued from last week)

.

Problem

. . Let f(x|θ) be the logistic location pdf f(x|θ) = e(x−θ) (1 + e(x−θ))2 − ∞ < x < ∞, −∞ < θ < ∞ (a) Show that this family has an MLR (b) Based on one observation X, find the most powerful size test of H versus H . (c) Show that the test in part (b) is UMP size for testing H vs. H .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 10 / 31

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Practice Problem 1 (continued from last week)

.

Problem

. . Let f(x|θ) be the logistic location pdf f(x|θ) = e(x−θ) (1 + e(x−θ))2 − ∞ < x < ∞, −∞ < θ < ∞ (a) Show that this family has an MLR (b) Based on one observation X, find the most powerful size test of H versus H . (c) Show that the test in part (b) is UMP size for testing H vs. H .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 10 / 31

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Practice Problem 1 (continued from last week)

.

Problem

. . Let f(x|θ) be the logistic location pdf f(x|θ) = e(x−θ) (1 + e(x−θ))2 − ∞ < x < ∞, −∞ < θ < ∞ (a) Show that this family has an MLR (b) Based on one observation X, find the most powerful size α test of H0 : θ = 0 versus H1 : θ = 1. (c) Show that the test in part (b) is UMP size for testing H vs. H .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 10 / 31

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Practice Problem 1 (continued from last week)

.

Problem

. . Let f(x|θ) be the logistic location pdf f(x|θ) = e(x−θ) (1 + e(x−θ))2 − ∞ < x < ∞, −∞ < θ < ∞ (a) Show that this family has an MLR (b) Based on one observation X, find the most powerful size α test of H0 : θ = 0 versus H1 : θ = 1. (c) Show that the test in part (b) is UMP size α for testing H0 : θ ≤ 0 vs. H1 : θ > 0.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 10 / 31

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

e e x e x Let r x ex ex r x e x e x e x e x e x e x e x e x x x Therefore, the family of X has an MLR.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 11 / 31

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

= e(θ1−θ2) ( 1 + e(x−θ1) 1 + e(x−θ2) )2 Let r x ex ex r x e x e x e x e x e x e x e x e x x x Therefore, the family of X has an MLR.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 11 / 31

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

= e(θ1−θ2) ( 1 + e(x−θ1) 1 + e(x−θ2) )2 Let r(x) = (1 + ex−θ1)/(1 + ex−θ2) r x e x e x e x e x e x e x e x e x x x Therefore, the family of X has an MLR.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 11 / 31

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

= e(θ1−θ2) ( 1 + e(x−θ1) 1 + e(x−θ2) )2 Let r(x) = (1 + ex−θ1)/(1 + ex−θ2) r′(x) = e(x−θ1)(1 + e(x−θ2)) − (1 + e(x−θ1))e(x−θ2) (1 + e(x−θ2))2 e x e x e x x x Therefore, the family of X has an MLR.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 11 / 31

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

= e(θ1−θ2) ( 1 + e(x−θ1) 1 + e(x−θ2) )2 Let r(x) = (1 + ex−θ1)/(1 + ex−θ2) r′(x) = e(x−θ1)(1 + e(x−θ2)) − (1 + e(x−θ1))e(x−θ2) (1 + e(x−θ2))2 = e(x−θ1) − e(x−θ2) (1 + e(x−θ2))2 > 0 (∵ x − θ1 > x − θ2) Therefore, the family of X has an MLR.

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Solution for (a)

For θ1 < θ2, f(x|θ2) f(x|θ1) =

e(x−θ2) (1+e(x−θ2))2 e(x−θ1) (1+e(x−θ1))2

= e(θ1−θ2) ( 1 + e(x−θ1) 1 + e(x−θ2) )2 Let r(x) = (1 + ex−θ1)/(1 + ex−θ2) r′(x) = e(x−θ1)(1 + e(x−θ2)) − (1 + e(x−θ1))e(x−θ2) (1 + e(x−θ2))2 = e(x−θ1) − e(x−θ2) (1 + e(x−θ2))2 > 0 (∵ x − θ1 > x − θ2) Therefore, the family of X has an MLR.

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Solution for (b)

The UMP test rejects H0 if and only if f(x|1) f(x|0) = e ( 1 + ex 1 + e(x−1) )2 > k ex e x k ex e ex k X x Because under H , F x

ex ex , the rejection region of UMP level

test satisfies F x ex x log

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Solution for (b)

The UMP test rejects H0 if and only if f(x|1) f(x|0) = e ( 1 + ex 1 + e(x−1) )2 > k 1 + ex 1 + e(x−1) > k∗ ex e ex k X x Because under H , F x

ex ex , the rejection region of UMP level

test satisfies F x ex x log

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Solution for (b)

The UMP test rejects H0 if and only if f(x|1) f(x|0) = e ( 1 + ex 1 + e(x−1) )2 > k 1 + ex 1 + e(x−1) > k∗ 1 + ex e + ex > k∗∗ X x Because under H , F x

ex ex , the rejection region of UMP level

test satisfies F x ex x log

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Solution for (b)

The UMP test rejects H0 if and only if f(x|1) f(x|0) = e ( 1 + ex 1 + e(x−1) )2 > k 1 + ex 1 + e(x−1) > k∗ 1 + ex e + ex > k∗∗ X > x0 Because under H , F x

ex ex , the rejection region of UMP level

test satisfies F x ex x log

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Solution for (b)

The UMP test rejects H0 if and only if f(x|1) f(x|0) = e ( 1 + ex 1 + e(x−1) )2 > k 1 + ex 1 + e(x−1) > k∗ 1 + ex e + ex > k∗∗ X > x0 Because under H0, F(x0|θ = 0) =

ex 1+ex , the rejection region of UMP level

α test satisfies 1 − F(x|θ = 0) = 1 1 + ex0 = α x0 ∼ log (1 − α α )

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Solution for (c)

Because the family of X has an MLR, UMP size α for testing H0 : θ ≤ 0

• vs. H1 : θ > 0 should be a form of

X > x0 Pr(X > x0|θ = 0) = α Therefore, x log , which is identical to the test defined in (b).

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Solution for (c)

Because the family of X has an MLR, UMP size α for testing H0 : θ ≤ 0

• vs. H1 : θ > 0 should be a form of

X > x0 Pr(X > x0|θ = 0) = α Therefore, x0 = log ( 1−α

α

) , which is identical to the test defined in (b).

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Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples with pdf fX(x|θ) = θ exp(−θx), where x ≥ 0, θ > 0 (a) Show that

n

n x

Xi is a consistent estimator for

. (b) Show that

n

n x

Xi is asymptotically normal and derive its asymptotic

distribution (c) Derive the Wald asymptotic size test for H

• vs. H

. (d) Find an asymptotic confidence interval for by inverting the above test You may use the fact that EX and Var X .

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Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples with pdf fX(x|θ) = θ exp(−θx), where x ≥ 0, θ > 0 (a) Show that

n ∑n

x=1 Xi is a consistent estimator for θ.

(b) Show that

n

n x

Xi is asymptotically normal and derive its asymptotic

distribution (c) Derive the Wald asymptotic size test for H

• vs. H

. (d) Find an asymptotic confidence interval for by inverting the above test You may use the fact that EX and Var X .

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Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples with pdf fX(x|θ) = θ exp(−θx), where x ≥ 0, θ > 0 (a) Show that

n ∑n

x=1 Xi is a consistent estimator for θ.

(b) Show that

n ∑n

x=1 Xi is asymptotically normal and derive its asymptotic

distribution (c) Derive the Wald asymptotic size test for H

• vs. H

. (d) Find an asymptotic confidence interval for by inverting the above test You may use the fact that EX and Var X .

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Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples with pdf fX(x|θ) = θ exp(−θx), where x ≥ 0, θ > 0 (a) Show that

n ∑n

x=1 Xi is a consistent estimator for θ.

(b) Show that

n ∑n

x=1 Xi is asymptotically normal and derive its asymptotic

distribution (c) Derive the Wald asymptotic size α test for H0 : θ = θ0 vs. H1 : θ ̸= θ0. (d) Find an asymptotic confidence interval for by inverting the above test You may use the fact that EX and Var X .

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Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples with pdf fX(x|θ) = θ exp(−θx), where x ≥ 0, θ > 0 (a) Show that

n ∑n

x=1 Xi is a consistent estimator for θ.

(b) Show that

n ∑n

x=1 Xi is asymptotically normal and derive its asymptotic

distribution (c) Derive the Wald asymptotic size α test for H0 : θ = θ0 vs. H1 : θ ̸= θ0. (d) Find an asymptotic (1 − α) confidence interval for θ by inverting the above test You may use the fact that EX = 1/θ and Var(X) = 1/θ2.

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Solution (a) - Consistency

. . 1 Obtain EX = 1/θ (Derive yourself if not given)

EX = ∫ ∞ xf(x|θ)dx = ∫ ∞ θx exp(−θx)dx x exp x exp x dx exp x

. 2 By LLN (Law of Large Number), X P

EX .

. . 3 By Theorem of continuous map, n n i

Xi X

P

.

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Solution (a) - Consistency

. . 1 Obtain EX = 1/θ (Derive yourself if not given)

EX = ∫ ∞ xf(x|θ)dx = ∫ ∞ θx exp(−θx)dx = [−x exp(−θx)]∞

0 +

∫ ∞ exp(−θx)dx exp x

. 2 By LLN (Law of Large Number), X P

EX .

. . 3 By Theorem of continuous map, n n i

Xi X

P

.

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Solution (a) - Consistency

. . 1 Obtain EX = 1/θ (Derive yourself if not given)

EX = ∫ ∞ xf(x|θ)dx = ∫ ∞ θx exp(−θx)dx = [−x exp(−θx)]∞

0 +

∫ ∞ exp(−θx)dx = 0 + [ −1 θ exp(−θx) ]∞ = 1 θ

. . 2 By LLN (Law of Large Number), X P

EX .

. . 3 By Theorem of continuous map, n n i

Xi X

P

.

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Solution (a) - Consistency

. . 1 Obtain EX = 1/θ (Derive yourself if not given)

EX = ∫ ∞ xf(x|θ)dx = ∫ ∞ θx exp(−θx)dx = [−x exp(−θx)]∞

0 +

∫ ∞ exp(−θx)dx = 0 + [ −1 θ exp(−θx) ]∞ = 1 θ

. . 2 By LLN (Law of Large Number), X P

→ EX = 1/θ.

. . 3 By Theorem of continuous map, n n i

Xi X

P

.

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Solution (a) - Consistency

. . 1 Obtain EX = 1/θ (Derive yourself if not given)

EX = ∫ ∞ xf(x|θ)dx = ∫ ∞ θx exp(−θx)dx = [−x exp(−θx)]∞

0 +

∫ ∞ exp(−θx)dx = 0 + [ −1 θ exp(−θx) ]∞ = 1 θ

. . 2 By LLN (Law of Large Number), X P

→ EX = 1/θ.

. . 3 By Theorem of continuous map, n/ ∑n i=1 Xi = 1/X P

→ θ.

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . 2 Apply CLT(Central Limit Theorem),

X n

. . 3 Apply Delta method. Let g y

y, then g y y . Xi n X g X g g n n n X

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . . 2 Apply CLT(Central Limit Theorem),

X ∼ AN (1 θ, 1 θ2n )

. . 3 Apply Delta method. Let g y

y, then g y y . Xi n X g X g g n n n X

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . . 2 Apply CLT(Central Limit Theorem),

X ∼ AN (1 θ, 1 θ2n )

. . 3 Apply Delta method. Let g(y) = 1/y, then g′(y) = −1/y2.

Xi n X g X g g n n n X

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . . 2 Apply CLT(Central Limit Theorem),

X ∼ AN (1 θ, 1 θ2n )

. . 3 Apply Delta method. Let g(y) = 1/y, then g′(y) = −1/y2.

∑ Xi n = 1/X = g(X) ∼ AN ( g(1/θ), [g′(1/θ)]2 θ2n ) n n X

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . . 2 Apply CLT(Central Limit Theorem),

X ∼ AN (1 θ, 1 θ2n )

. . 3 Apply Delta method. Let g(y) = 1/y, then g′(y) = −1/y2.

∑ Xi n = 1/X = g(X) ∼ AN ( g(1/θ), [g′(1/θ)]2 θ2n ) = AN ( θ, θ2 n ) n X

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Solution (b) - Asymptotic Distribution

. . 1 Obtain Var(X) = 1/θ2 (Derive if needed, omitted here). . . 2 Apply CLT(Central Limit Theorem),

X ∼ AN (1 θ, 1 θ2n )

. . 3 Apply Delta method. Let g(y) = 1/y, then g′(y) = −1/y2.

∑ Xi n = 1/X = g(X) ∼ AN ( g(1/θ), [g′(1/θ)]2 θ2n ) = AN ( θ, θ2 n ) ⇐ ⇒ √n ( 1 X − θ ) = N ( 0, θ2)

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W X

n i

Xi n n

. . 2 Obtain a constant estimator of Var W

n

n i

Xi X

P

Var X CLT n

n i

Xi X

P

Continuous Map Theorem S n

n i

Xi X

P

Slutsky’s Theorem

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W(X) = ∑n

i=1 Xi

n ∼ AN ( θ, θ2 n )

. . 2 Obtain a constant estimator of Var W

n

n i

Xi X

P

Var X CLT n

n i

Xi X

P

Continuous Map Theorem S n

n i

Xi X

P

Slutsky’s Theorem

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W(X) = ∑n

i=1 Xi

n ∼ AN ( θ, θ2 n )

. . 2 Obtain a constant estimator of Var(W)

n

n i

Xi X

P

Var X CLT n

n i

Xi X

P

Continuous Map Theorem S n

n i

Xi X

P

Slutsky’s Theorem

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W(X) = ∑n

i=1 Xi

n ∼ AN ( θ, θ2 n )

. . 2 Obtain a constant estimator of Var(W)

1 n − 1

n

∑

i=1

(Xi − X)2

P

→

Var(X) = 1 θ2 (CLT) n

n i

Xi X

P

Continuous Map Theorem S n

n i

Xi X

P

Slutsky’s Theorem

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W(X) = ∑n

i=1 Xi

n ∼ AN ( θ, θ2 n )

. . 2 Obtain a constant estimator of Var(W)

1 n − 1

n

∑

i=1

(Xi − X)2

P

→

Var(X) = 1 θ2 (CLT) n − 1 ∑n

i=1(Xi − X)2 P

→

θ2 (Continuous Map Theorem). S n

n i

Xi X

P

Slutsky’s Theorem

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Solution (c) - Wald asymptotic size α test

. . 1 Obtain a consistent estimator of θ :

W(X) = ∑n

i=1 Xi

n ∼ AN ( θ, θ2 n )

. . 2 Obtain a constant estimator of Var(W)

1 n − 1

n

∑

i=1

(Xi − X)2

P

→

Var(X) = 1 θ2 (CLT) n − 1 ∑n

i=1(Xi − X)2 P

→

θ2 (Continuous Map Theorem). S2 = n ∑n

i=1(Xi − X)2 P

→

θ2 (Slutsky’s Theorem).

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Solution (c) - Wald Asymptotic size α test (cont’d)

. . 3 Construct a two-sided asymptotic size α Wald test, whose rejection

region is Z X W X S n

n

n i

Xi

n i

Xi X

X

n i

Xi X z

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Solution (c) - Wald Asymptotic size α test (cont’d)

. . 3 Construct a two-sided asymptotic size α Wald test, whose rejection

region is |Z(X)| =

• W(X) − θ0

S/√n

• n

n i

Xi

n i

Xi X

X

n i

Xi X z

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Solution (c) - Wald Asymptotic size α test (cont’d)

. . 3 Construct a two-sided asymptotic size α Wald test, whose rejection

region is |Z(X)| =

• W(X) − θ0

S/√n

• =
• n

∑n

i=1 Xi − θ0

1

√∑n

i=1(Xi−X)2

• X

n i

Xi X z

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Solution (c) - Wald Asymptotic size α test (cont’d)

. . 3 Construct a two-sided asymptotic size α Wald test, whose rejection

region is |Z(X)| =

• W(X) − θ0

S/√n

• =
• n

∑n

i=1 Xi − θ0

1

√∑n

i=1(Xi−X)2

• =
• 1

X − θ0

• n

∑

i=1

(Xi − X)2 ≥ zα/2

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Solution (d) - Asymptotic 1 − α confidence interval

The acceptance region is A =   x :

• 1

x − θ0

• n

∑

i=1

(xi − x)2 ≤ zα/2    By inverting the acceptance region, the confidence interval is C X X

n i

Xi X z which is equivalent to C X X z

n i

Xi X X z

n i

Xi X

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Solution (d) - Asymptotic 1 − α confidence interval

The acceptance region is A =   x :

• 1

x − θ0

• n

∑

i=1

(xi − x)2 ≤ zα/2    By inverting the acceptance region, the confidence interval is C X X

n i

Xi X z which is equivalent to C X X z

n i

Xi X X z

n i

Xi X

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Solution (d) - Asymptotic 1 − α confidence interval

The acceptance region is A =   x :

• 1

x − θ0

• n

∑

i=1

(xi − x)2 ≤ zα/2    By inverting the acceptance region, the confidence interval is C(X) =   θ :

• 1

X − θ

• n

∑

i=1

(Xi − X)2 ≤ zα/2    which is equivalent to C X X z

n i

Xi X X z

n i

Xi X

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Solution (d) - Asymptotic 1 − α confidence interval

The acceptance region is A =   x :

• 1

x − θ0

• n

∑

i=1

(xi − x)2 ≤ zα/2    By inverting the acceptance region, the confidence interval is C(X) =   θ :

• 1

X − θ

• n

∑

i=1

(Xi − X)2 ≤ zα/2    which is equivalent to C(X) =   θ ∈   1 X − zα/2 √∑n

i=1(Xi − X)2

, 1 X + zα/2 √∑n

i=1(Xi − X)2

    

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Practice Problem 3

.

Problem

. . The independent random variables X1, · · · , Xn have the following pdf f(x|θ, β) = βxβ−1 θβ 0 < x < θ, β > 0

. . 1 Find the MLEs of

and

. . 2 When

is a known constant , construct a LRT testing H

• vs. H

.

. . 3 When

is a known constant , find the upper confidence limit for with confidence coefficient .

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Practice Problem 3

.

Problem

. . The independent random variables X1, · · · , Xn have the following pdf f(x|θ, β) = βxβ−1 θβ 0 < x < θ, β > 0

. . 1 Find the MLEs of β and θ . . 2 When

is a known constant , construct a LRT testing H

• vs. H

.

. . 3 When

is a known constant , find the upper confidence limit for with confidence coefficient .

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Practice Problem 3

.

Problem

. . The independent random variables X1, · · · , Xn have the following pdf f(x|θ, β) = βxβ−1 θβ 0 < x < θ, β > 0

. . 1 Find the MLEs of β and θ . . 2 When β is a known constant β0, construct a LRT testing H0 : θ ≥ θ0

• vs. H1 : θ < θ0.

. 3 When

is a known constant , find the upper confidence limit for with confidence coefficient .

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Practice Problem 3

.

Problem

. . The independent random variables X1, · · · , Xn have the following pdf f(x|θ, β) = βxβ−1 θβ 0 < x < θ, β > 0

. . 1 Find the MLEs of β and θ . . 2 When β is a known constant β0, construct a LRT testing H0 : θ ≥ θ0

• vs. H1 : θ < θ0.

. . 3 When β is a known constant β0, find the upper confidence limit for θ

with confidence coefficient 1 − α.

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of and positive only when x n x n l x n log log xi n log l n log xi n log n n log log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) x n l x n log log xi n log l n log xi n log n n log log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) ˆ θ = x(n) l x n log log xi n log l n log xi n log n n log log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) ˆ θ = x(n) l(θ, β|x) = n log β + (β − 1) ∑ log xi − nβ log θ l n log xi n log n n log log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) ˆ θ = x(n) l(θ, β|x) = n log β + (β − 1) ∑ log xi − nβ log θ ∂l ∂β = n β + ∑ log xi − n log θ = 0 n n log log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) ˆ θ = x(n) l(θ, β|x) = n log β + (β − 1) ∑ log xi − nβ log θ ∂l ∂β = n β + ∑ log xi − n log θ = 0 ˆ β = n n log ˆ θ − ∑ log xi n nx n log xi

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(a) - MLE

L(θ, β|x) = βn (∏n

i=1 xi)β−1

θnβ I(x(n) ≤ θ) Because L is a decreasing function of θ and positive only when θ ≥ x(n) ˆ θ = x(n) l(θ, β|x) = n log β + (β − 1) ∑ log xi − nβ log θ ∂l ∂β = n β + ∑ log xi − n log θ = 0 ˆ β = n n log ˆ θ − ∑ log xi = n nx(n) − ∑ log xi

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(b) - LRT

λ(x) = supθ∈Ω0 L(ˆ θ|x) supθ∈Ω L(ˆ θ|x) x n

L x L x n x

x n x n

x n

n n

x n c x n c

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(b) - LRT

λ(x) = supθ∈Ω0 L(ˆ θ|x) supθ∈Ω L(ˆ θ|x) = { 1 θ0 < x(n)

L(θ0|x) L(x(n)|x)

θ0 ≥ x(n) x n

x n

n n

x n c x n c

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(b) - LRT

λ(x) = supθ∈Ω0 L(ˆ θ|x) supθ∈Ω L(ˆ θ|x) = { 1 θ0 < x(n)

L(θ0|x) L(x(n)|x)

θ0 ≥ x(n) =    1 θ0 < x(n) (x(n))

nβ0

θnβ0

θ0 ≥ x(n) ≤ c x n c

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(b) - LRT

λ(x) = supθ∈Ω0 L(ˆ θ|x) supθ∈Ω L(ˆ θ|x) = { 1 θ0 < x(n)

L(θ0|x) L(x(n)|x)

θ0 ≥ x(n) =    1 θ0 < x(n) (x(n))

nβ0

θnβ0

θ0 ≥ x(n) ≤ c x(n) θ0 ≤ c∗

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(b) - size α LRT

α = Pr (x(n) θ0 ≤ c∗ ) c

n

c

n

Therefore, the rejection region for size LRT is is R x x n

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(b) - size α LRT

α = Pr (x(n) θ0 ≤ c∗ ) = (c∗)nβ0 c

n

Therefore, the rejection region for size LRT is is R x x n

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(b) - size α LRT

α = Pr (x(n) θ0 ≤ c∗ ) = (c∗)nβ0 c∗ = α

1 nβ0

Therefore, the rejection region for size LRT is is R x x n

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(b) - size α LRT

α = Pr (x(n) θ0 ≤ c∗ ) = (c∗)nβ0 c∗ = α

1 nβ0

Therefore, the rejection region for size α LRT is is R x x n

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(b) - size α LRT

α = Pr (x(n) θ0 ≤ c∗ ) = (c∗)nβ0 c∗ = α

1 nβ0

Therefore, the rejection region for size α LRT is is R = { x : x(n) ≤ θ0α

1 nβ0

}

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A x x n

n

By inserting the acceptance region, the confidence interval becomes C X X n

n

X n

n

Therefore, the upper confidence limit is X n

n

.

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A(θ0) = { x : x(n) > θ0α

1 nβ0

} By inserting the acceptance region, the confidence interval becomes C X X n

n

X n

n

Therefore, the upper confidence limit is X n

n

.

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A(θ0) = { x : x(n) > θ0α

1 nβ0

} By inserting the acceptance region, the 1 − α confidence interval becomes C X X n

n

X n

n

Therefore, the upper confidence limit is X n

n

.

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A(θ0) = { x : x(n) > θ0α

1 nβ0

} By inserting the acceptance region, the 1 − α confidence interval becomes C(X) = { θ : X(n) > θα

1 nβ0

} X n

n

Therefore, the upper confidence limit is X n

n

.

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A(θ0) = { x : x(n) > θ0α

1 nβ0

} By inserting the acceptance region, the 1 − α confidence interval becomes C(X) = { θ : X(n) > θα

1 nβ0

} = { θ : θ < X(n)α−

1 nβ0

} Therefore, the upper confidence limit is X n

n

.

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(c) - Upper 1 − α confidence limit

The acceptance region of size α LRT is A(θ0) = { x : x(n) > θ0α

1 nβ0

} By inserting the acceptance region, the 1 − α confidence interval becomes C(X) = { θ : X(n) > θα

1 nβ0

} = { θ : θ < X(n)α−

1 nβ0

} Therefore, the upper 1 − α confidence limit is X(n)α−

1 nβ0 . Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 24 / 31

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Practice Problem 4

.

Problem

. . A random sample X1, · · · , Xn is drawn from a population N(θ, θ) where θ > 0. (a) Find the , the MLE of (b) Find the asymptotic distribution of . (c) Compute ARE X . Determine whether is asymptotically more efficient than X or not. You may use the following fact: Var X .

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Practice Problem 4

.

Problem

. . A random sample X1, · · · , Xn is drawn from a population N(θ, θ) where θ > 0. (a) Find the ˆ θ, the MLE of θ (b) Find the asymptotic distribution of . (c) Compute ARE X . Determine whether is asymptotically more efficient than X or not. You may use the following fact: Var X .

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Practice Problem 4

.

Problem

. . A random sample X1, · · · , Xn is drawn from a population N(θ, θ) where θ > 0. (a) Find the ˆ θ, the MLE of θ (b) Find the asymptotic distribution of ˆ θ. (c) Compute ARE X . Determine whether is asymptotically more efficient than X or not. You may use the following fact: Var X .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 25 / 31

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Practice Problem 4

.

Problem

. . A random sample X1, · · · , Xn is drawn from a population N(θ, θ) where θ > 0. (a) Find the ˆ θ, the MLE of θ (b) Find the asymptotic distribution of ˆ θ. (c) Compute ARE(ˆ θ, X). Determine whether ˆ θ is asymptotically more efficient than X or not. You may use the following fact: Var X .

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 25 / 31

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Practice Problem 4

.

Problem

. . A random sample X1, · · · , Xn is drawn from a population N(θ, θ) where θ > 0. (a) Find the ˆ θ, the MLE of θ (b) Find the asymptotic distribution of ˆ θ. (c) Compute ARE(ˆ θ, X). Determine whether ˆ θ is asymptotically more efficient than X or not. You may use the following fact: Var(X2) = 4θ3 + 2θ2.

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l x n log n log

n i

xi n log n log xi xi n l x n xi n n xi n n n xi xi n n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ n log n log xi xi n l x n xi n n xi n n n xi xi n n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ = n 2 log(2π) + n 2 log θ − ∑ x2

i

2θ + ∑ xi − nθ 2 l x n xi n n xi n n n xi xi n n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ = n 2 log(2π) + n 2 log θ − ∑ x2

i

2θ + ∑ xi − nθ 2 l′(θ|x) = n 2θ + ∑ x2

i

2θ2 − n 2 = nθ − ∑ x2

i − nθ2

2θ2 = 0 n n xi xi n n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ = n 2 log(2π) + n 2 log θ − ∑ x2

i

2θ + ∑ xi − nθ 2 l′(θ|x) = n 2θ + ∑ x2

i

2θ2 − n 2 = nθ − ∑ x2

i − nθ2

2θ2 = 0 nθ2 + nθ − ∑ x2

i

= xi n n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ = n 2 log(2π) + n 2 log θ − ∑ x2

i

2θ + ∑ xi − nθ 2 l′(θ|x) = n 2θ + ∑ x2

i

2θ2 − n 2 = nθ − ∑ x2

i − nθ2

2θ2 = 0 nθ2 + nθ − ∑ x2

i

= ˆ θ = −1 + √ 1 + 4 ∑ x2

i /n

2 n xi

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(a) - MLE of θ

L(θ|x) = (2πθ)n/2 exp [ − ∑n

i=1(xi − θ)2

2θ ] l(θ|x) = n 2 log(2π) + n 2 log θ − ∑n

i=1(xi − θ)2

2θ = n 2 log(2π) + n 2 log θ − ∑ x2

i

2θ + ∑ xi − nθ 2 l′(θ|x) = n 2θ + ∑ x2

i

2θ2 − n 2 = nθ − ∑ x2

i − nθ2

2θ2 = 0 nθ2 + nθ − ∑ x2

i

= ˆ θ = −1 + √ 1 + 4 ∑ x2

i /n

2 1 n ∑ x2

i

= ˆ θ2 + ˆ θ

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(b) - Asymptotic distribution of MLE

By CLT, Let W = 1

n

∑ X2

i , then

W ∼ AN ( EX2, Var(X2) n ) n The asymptotic distribution of MLE n for some function and we would like to find using the asymptotic distribution of W.

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(b) - Asymptotic distribution of MLE

By CLT, Let W = 1

n

∑ X2

i , then

W ∼ AN ( EX2, Var(X2) n ) = AN ( θ + θ2, 4θ3 + 2θ2 n ) The asymptotic distribution of MLE n for some function and we would like to find using the asymptotic distribution of W.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 27 / 31

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(b) - Asymptotic distribution of MLE

By CLT, Let W = 1

n

∑ X2

i , then

W ∼ AN ( EX2, Var(X2) n ) = AN ( θ + θ2, 4θ3 + 2θ2 n ) The asymptotic distribution of MLE ˆ θ ˆ θ ∼ AN ( θ, σ2(θ) n ) for some function and we would like to find using the asymptotic distribution of W.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 27 / 31

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(b) - Asymptotic distribution of MLE

By CLT, Let W = 1

n

∑ X2

i , then

W ∼ AN ( EX2, Var(X2) n ) = AN ( θ + θ2, 4θ3 + 2θ2 n ) The asymptotic distribution of MLE ˆ θ ˆ θ ∼ AN ( θ, σ2(θ) n ) for some function σ2(θ) and we would like to find σ2(θ) using the asymptotic distribution of W.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 27 / 31

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(b) - Asymptotic distribution of MLE (cont’d)

Let g(y) = y2 + y, then g′(y) = (2y + 1) and g(ˆ θ) = W. Then by the Delta Method, the asymptotic distribution of W can be written as W g g g n n n

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(b) - Asymptotic distribution of MLE (cont’d)

Let g(y) = y2 + y, then g′(y) = (2y + 1) and g(ˆ θ) = W. Then by the Delta Method, the asymptotic distribution of W can be written as W = g(ˆ θ) ∼ AN ( g(θ), g′(θ)σ2(θ) n ) n n

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 28 / 31

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(b) - Asymptotic distribution of MLE (cont’d)

Let g(y) = y2 + y, then g′(y) = (2y + 1) and g(ˆ θ) = W. Then by the Delta Method, the asymptotic distribution of W can be written as W = g(ˆ θ) ∼ AN ( g(θ), g′(θ)σ2(θ) n ) = AN ( θ2 + θ, (2θ + 1)2σ2(θ) n ) n

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 28 / 31

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(b) - Asymptotic distribution of MLE (cont’d)

Let g(y) = y2 + y, then g′(y) = (2y + 1) and g(ˆ θ) = W. Then by the Delta Method, the asymptotic distribution of W can be written as W = g(ˆ θ) ∼ AN ( g(θ), g′(θ)σ2(θ) n ) = AN ( θ2 + θ, (2θ + 1)2σ2(θ) n ) = AN ( θ2 + θ, 4θ3 + 2θ2 n )

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(b) - Asymptotic distribution of MLE (cont’d)

Let g(y) = y2 + y, then g′(y) = (2y + 1) and g(ˆ θ) = W. Then by the Delta Method, the asymptotic distribution of W can be written as W = g(ˆ θ) ∼ AN ( g(θ), g′(θ)σ2(θ) n ) = AN ( θ2 + θ, (2θ + 1)2σ2(θ) n ) = AN ( θ2 + θ, 4θ3 + 2θ2 n ) σ2(θ) = 4θ3 + 2θ2 (2θ + 1)2 = 2θ2(2θ + 1) (2θ + 1)2 = 2θ2 2θ + 1

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(b) - Asymptotic distribution of MLE (cont’d)

The asymptotic distribution of MLE ˆ θ ˆ θ ∼ AN ( θ, σ2(θ) n ) = AN ( θ, 2θ2 n(2θ + 1) ) Note that you cannot use CR-bound for the asymptotic variance of MLE because the regularity condition does not hold (open set criteria).

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(b) - Asymptotic distribution of MLE (cont’d)

The asymptotic distribution of MLE ˆ θ ˆ θ ∼ AN ( θ, σ2(θ) n ) = AN ( θ, 2θ2 n(2θ + 1) ) Note that you cannot use CR-bound for the asymptotic variance of MLE because the regularity condition does not hold (open set criteria).

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 29 / 31

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(c) - ARE of MLE compared to X

By CLT, the asymptotic distribution of X is X ∼ AN ( θ, θ n ) Then, ARE X is ARE X Therefore, is more efficient estimator than X.

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(c) - ARE of MLE compared to X

By CLT, the asymptotic distribution of X is X ∼ AN ( θ, θ n ) Then, ARE(ˆ θ, X) is ARE(ˆ θ, X) = θ

2θ2 2θ+1

Therefore, is more efficient estimator than X.

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(c) - ARE of MLE compared to X

By CLT, the asymptotic distribution of X is X ∼ AN ( θ, θ n ) Then, ARE(ˆ θ, X) is ARE(ˆ θ, X) = θ

2θ2 2θ+1

= 2θ + 1 2θ = 1 + 1 2θ > 1 Therefore, is more efficient estimator than X.

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(c) - ARE of MLE compared to X

By CLT, the asymptotic distribution of X is X ∼ AN ( θ, θ n ) Then, ARE(ˆ θ, X) is ARE(ˆ θ, X) = θ

2θ2 2θ+1

= 2θ + 1 2θ = 1 + 1 2θ > 1 Therefore, ˆ θ is more efficient estimator than X.

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Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

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Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 31 / 31

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Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 31 / 31

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Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 31 / 31

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

. . . . . . . . Review . . . . P1 . . . . . . P2 . . . . . P3 . . . . . . P4 . Wrap-up

Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 31 / 31

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

. . . . . . . . Review . . . . P1 . . . . . . P2 . . . . . P3 . . . . . . P4 . Wrap-up

Wrapping Up

. . 1 Many thanks for your attentions and feedbacks. . . 2 Please complete your teaching evaluations, which will be very helpful

for further improvement in the next year.

. . 3 Final exam will be Thursday April 25th, 4:00-6:00pm. . . 4 The last office hour will be held Wednesday April 24th, 4:00-5:00pm. . . 5 The grade will be posted during the weekend. . . 6 Don’t forget the materials we have learned, because they are the key

topics for your candidacy exam.

Hyun Min Kang Biostatistics 602 - Lecture 26 Apil 23rd, 2013 31 / 31