• Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H and H can be calculated • Pr c x • Pr • Rejection region can be determined directly based on the posterior . . .. . . .. . . Recap .. . . .. .. . .. . . . . P2 Bayesian Tests is true April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability is true Pr H Pr H Bayesian Intervals x Bayesian Tests P4 P3 . P1 . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . . 6 / 34 . . .. . . .. . . .. . . .. . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ )
• In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H and H can be calculated • Pr c x • Pr • Rejection region can be determined directly based on the posterior . .. . . .. . . . .. . . . . . .. .. . .. . . P1 Recap is true April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability is true Pr H Pr H Bayesian Tests x Bayesian Tests P4 P3 P2 .. Bayesian Intervals . . . . .. . . .. . . .. . .. . . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . . .. . .. . . 6 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability
• In Bayesian framework, the probability of H and H can be calculated • Pr c x • Pr • Rejection region can be determined directly based on the posterior . . .. . . .. . . .. . . . . . .. .. . .. . . P1 Recap is true April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability is true Pr H Pr H Bayesian Tests x Bayesian Tests P4 P3 P2 .. Bayesian Intervals . . . . . . . .. . . .. . .. . . . .. . . .. . . . .. 6 / 34 . . .. . .. . . .. . . .. .. .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated.
• Pr c x • Pr • Rejection region can be determined directly based on the posterior . .. .. . . .. . . . . . . . . .. .. . .. . . . Bayesian Intervals Recap is true April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability is true Pr H Pr H Bayesian Tests x Bayesian Tests P4 P3 P2 P1 . .. . . . .. . .. . . .. . .. . . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. 6 / 34 .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H 0 and H 1 can be calculated
c x • Pr • Rejection region can be determined directly based on the posterior .. .. . . .. . . . . . . .. . . .. . . . . . . .. Bayesian Tests April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability is true Pr H P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap .. . .. . .. .. . . . .. . . . . . .. . . .. . . . 6 / 34 .. . .. . . . .. .. . . . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H 0 and H 1 can be calculated • Pr ( θ ∈ Ω 0 | x ) = Pr ( H 0 is true )
• Rejection region can be determined directly based on the posterior . . . .. . . .. . .. .. .. . .. . . .. . . . P3 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability Bayesian Tests P4 P2 . P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. .. . . .. . . . . . . .. . . .. . . .. 6 / 34 . . . .. . . .. .. . .. . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H 0 and H 1 can be calculated • Pr ( θ ∈ Ω 0 | x ) = Pr ( H 0 is true ) • Pr ( θ ∈ Ω c 0 | x ) = Pr ( H 1 is true )
. . . .. . . .. . .. .. .. . .. . . .. . . . P3 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang probability Bayesian Tests P4 P2 . P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. .. . . .. . . . . . . .. . . .. . . .. 6 / 34 . . . .. . . .. .. . .. . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Hypothesis testing problems can be formulated in a Bayesian model • Bayesian model includes • Sampling distribution f ( x | θ ) • Prior distribution π ( θ ) • Bayesian hypothesis testing is based on the posterior probability • In Frequentist’s framework, posterior probability cannot be calculated. • In Bayesian framework, the probability of H 0 and H 1 can be calculated • Pr ( θ ∈ Ω 0 | x ) = Pr ( H 0 is true ) • Pr ( θ ∈ Ω c 0 | x ) = Pr ( H 1 is true ) • Rejection region can be determined directly based on the posterior
• Consequently, a hypothesis is either true of false • If • If • Pr H is true x and Pr H is true x are function of x , between 0 • These probabilities give useful information about the veracity of H Bayesian Intervals Frequentist’s Framework . Bayesian vs Frequentist Framework P4 P3 P2 P1 . . . . Bayesian Tests Recap . .. . . .. . . .. . . . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang and H . and 1. . . . . . . . . Bayesian Framework . and Pr H is true x c , Pr H is true x and Pr H is true x , Pr H is true x . .. .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . 7 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • θ is considered to be a fixed number
• If • If • Pr H is true x and Pr H is true x are function of x , between 0 • These probabilities give useful information about the veracity of H Recap . Bayesian vs Frequentist Framework P4 P3 P2 P1 Bayesian Intervals Bayesian Tests . . . . . . .. . . .. . . .. . Frequentist’s Framework . . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang and H . and 1. . . . . .. . . . Bayesian Framework . and Pr H is true x c , Pr H is true x and Pr H is true x , Pr H is true x . .. . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . 7 / 34 .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • θ is considered to be a fixed number • Consequently, a hypothesis is either true of false
• These probabilities give useful information about the veracity of H . . . . . . . .. . . .. . .. Bayesian Tests . . .. .. . .. . Recap P1 Bayesian Intervals Bayesian Framework April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang and H . and 1. . . . .. . . Frequentist’s Framework . Bayesian vs Frequentist Framework P4 P3 P2 . . . . . . . .. . . .. . .. . . . .. . . .. . . . .. 7 / 34 . . .. . .. . . . .. . . .. .. . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • θ is considered to be a fixed number • Consequently, a hypothesis is either true of false • If θ ∈ Ω 0 , Pr ( H 0 is true | x ) = 1 and Pr ( H 1 is true | x ) = 0 • If θ ∈ Ω c 0 , Pr ( H 0 is true | x ) = 0 and Pr ( H 1 is true | x ) = 1 • Pr ( H 0 is true | x ) and Pr ( H 1 is true | x ) are function of x , between 0
. .. . .. . . .. . . . Recap . .. .. . .. . . . . . . Bayesian Tests . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang and 1. . . Bayesian Framework . Bayesian Intervals . Frequentist’s Framework . Bayesian vs Frequentist Framework P4 P3 P2 P1 .. . . . . .. . .. . . .. . .. . . . .. . . .. . . .. 7 / 34 . . .. . . .. .. . . .. . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • θ is considered to be a fixed number • Consequently, a hypothesis is either true of false • If θ ∈ Ω 0 , Pr ( H 0 is true | x ) = 1 and Pr ( H 1 is true | x ) = 0 • If θ ∈ Ω c 0 , Pr ( H 0 is true | x ) = 0 and Pr ( H 1 is true | x ) = 1 • Pr ( H 0 is true | x ) and Pr ( H 1 is true | x ) are function of x , between 0 • These probabilities give useful information about the veracity of H 0 and H 1 .
c x • Accept H is Pr c x • Reject H is Pr c x • In other words, the rejection region is c x • Reject H is Pr c x • Accept H is Pr . P3 . . Examples of Bayesian hypothesis testing procedure P4 Bayesian Tests P2 P1 Bayesian Intervals Recap . . . . . .. . . .. .. . x x . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . . . . Pr . . A more conservative (smaller size) test in rejecting H . Pr x Pr .. . . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . .. . . .. . . .. . . .. . 8 / 34 . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A neutral test between H 0 and H 1
c x • In other words, the rejection region is c x • Reject H is Pr c x • Accept H is Pr . . P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . P4 .. . . .. . . P3 . Examples of Bayesian hypothesis testing procedure . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . . . . . . . A more conservative (smaller size) test in rejecting H . Pr x . . .. .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 8 / 34 . . . . .. . . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A neutral test between H 0 and H 1 • Accept H 0 is Pr ( θ ∈ Ω 0 | x ) ≥ Pr ( θ ∈ Ω c 0 | x ) • Reject H 0 is Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x )
c x • Reject H is Pr c x • Accept H is Pr . . Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. . P2 .. . . .. . P1 Examples of Bayesian hypothesis testing procedure P3 . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . . . . P4 . . A more conservative (smaller size) test in rejecting H . . . . .. . .. . .. . . . .. . . .. . . . . . .. . . .. . . .. . . . .. 8 / 34 . . .. .. . . .. . .. .. . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A neutral test between H 0 and H 1 • Accept H 0 is Pr ( θ ∈ Ω 0 | x ) ≥ Pr ( θ ∈ Ω c 0 | x ) • Reject H 0 is Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x ) 0 | x ) > 1 • In other words, the rejection region is { x : Pr ( θ ∈ Ω c 2 }
. . . . .. . . .. . .. . . . .. . . .. . .. . . . . .. . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . . . . Recap Examples of Bayesian hypothesis testing procedure P4 P3 P2 P1 Bayesian Intervals Bayesian Tests . .. . .. . .. . . .. . . . .. . .. . . .. . . . . . .. .. . . .. . . . .. . . 8 / 34 . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A neutral test between H 0 and H 1 • Accept H 0 is Pr ( θ ∈ Ω 0 | x ) ≥ Pr ( θ ∈ Ω c 0 | x ) • Reject H 0 is Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x ) 0 | x ) > 1 • In other words, the rejection region is { x : Pr ( θ ∈ Ω c 2 } A more conservative (smaller size) test in rejecting H 0 • Reject H 0 is Pr ( θ ∈ Ω c 0 | x ) > 0 . 99 • Accept H 0 is Pr ( θ ∈ Ω c 0 | x ) ≤ 0 . 99
c x . P4 Pr x rejecting H if and only if Pr Construct a Bayesian test . . Problem . Example: Normal Bayesian Test P3 Solution P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. .. . . . . x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x Pr x Pr We will reject H if and only if n n n . x the posterior is . From previous lectures, versus H Consider testing H . . . . . . .. . . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . 9 / 34 .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n be iid samples N ( θ, σ 2 ) and let the prior distribution of θ be N ( µ, τ r ) , where σ 2 , µ , and τ 2 are known.
. P1 . . . Problem . Example: Normal Bayesian Test P4 P3 P2 Bayesian Intervals . Bayesian Tests Recap . . . . . .. . . .. . . Solution . . x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x Pr x Pr We will reject H if and only if n n n . x the posterior is . From previous lectures, versus H Consider testing H . . . . . .. .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 9 / 34 .. . .. . . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n be iid samples N ( θ, σ 2 ) and let the prior distribution of θ be N ( µ, τ r ) , where σ 2 , µ , and τ 2 are known. Construct a Bayesian test rejecting H 0 if and only if Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x )
. . P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . . .. . . .. .. . .. Example: Normal Bayesian Test Problem . n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x Pr x Pr We will reject H if and only if n . x n x the posterior is . . Solution . . . . .. .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . . . . 9 / 34 . . .. . . .. . . .. . . .. . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n be iid samples N ( θ, σ 2 ) and let the prior distribution of θ be N ( µ, τ r ) , where σ 2 , µ , and τ 2 are known. Construct a Bayesian test rejecting H 0 if and only if Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x ) Consider testing H 0 : θ ≤ θ 0 versus H 1 : θ > θ 0 . From previous lectures,
. .. Bayesian Intervals Bayesian Tests Recap . . . . . .. . . . P2 . .. . . .. . . .. P1 P3 . the posterior is April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x Pr x Pr We will reject H if and only if . P4 . Solution . . . Problem . Example: Normal Bayesian Test . .. .. . . . . . .. . . .. . .. .. . . .. . . .. . . . .. . . . .. . . .. . . .. . . . .. 9 / 34 . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n be iid samples N ( θ, σ 2 ) and let the prior distribution of θ be N ( µ, τ r ) , where σ 2 , µ , and τ 2 are known. Construct a Bayesian test rejecting H 0 if and only if Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x ) Consider testing H 0 : θ ≤ θ 0 versus H 1 : θ > θ 0 . From previous lectures, ( n τ 2 x + σ 2 µ σ 2 τ 2 ) π ( θ | x ) ∼ N n τ 2 + σ 2 , n τ 2 + σ 2
. .. . .. . . .. . . . Recap . .. . . .. . . . . . . Bayesian Tests . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang the posterior is . . Solution . Bayesian Intervals . Problem . Example: Normal Bayesian Test P4 P3 P2 P1 .. .. . . . . .. .. . . .. . .. . . . .. . . .. . . .. 9 / 34 . . . . .. . . .. . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n be iid samples N ( θ, σ 2 ) and let the prior distribution of θ be N ( µ, τ r ) , where σ 2 , µ , and τ 2 are known. Construct a Bayesian test rejecting H 0 if and only if Pr ( θ ∈ Ω 0 | x ) < Pr ( θ ∈ Ω c 0 | x ) Consider testing H 0 : θ ≤ θ 0 versus H 1 : θ > θ 0 . From previous lectures, ( n τ 2 x + σ 2 µ σ 2 τ 2 ) π ( θ | x ) ∼ N n τ 2 + σ 2 , n τ 2 + σ 2 We will reject H 0 if and only if Pr ( θ ≤ θ 0 | x ) < 1 Pr ( θ ∈ Ω 0 | x ) = 2
. . . .. . . .. . .. .. . . .. . . .. . . . .. n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n x n x Solution (cont’d) . . . . P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . .. . .. . .. . . .. . . . . . .. . . .. . . . 10 / 34 .. . . . . .. . . .. . .. .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Because π ( θ | x ) is symmetric, this is true if and only if the mean for π ( θ | x ) is less than or equal to θ 0 . Therefore, H 0 will be rejected if
. . . . .. . . .. . . .. . . .. . . .. .. . . P4 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n x Solution (cont’d) P3 .. P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . . .. .. .. . . . .. . . . . . .. . . .. . . .. 10 / 34 . . . .. .. . . . .. .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Because π ( θ | x ) is symmetric, this is true if and only if the mean for π ( θ | x ) is less than or equal to θ 0 . Therefore, H 0 will be rejected if n τ 2 x + σ 2 µ < θ 0 n τ 2 + σ 2
. . . .. . . .. . .. .. . .. .. . . .. . . . P3 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x Solution (cont’d) P4 P2 . P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . .. . . . 10 / 34 . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Because π ( θ | x ) is symmetric, this is true if and only if the mean for π ( θ | x ) is less than or equal to θ 0 . Therefore, H 0 will be rejected if n τ 2 x + σ 2 µ < θ 0 n τ 2 + σ 2 θ 0 + σ 2 ( θ 0 − µ ) < n τ 2
• We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter • A 95% confidence interval for • ”The probability that • Formally, the interval [.262,1.184] is one of the possible realized Recap Confidence interval and the parameter P4 P3 P2 P1 Bayesian Intervals Bayesian Tests . . . . . Frequentist’s view of intervals .. . . .. . . .. .. . . . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed is in the interval [.262,1.184] is 95%” : is . . . . . . . . Example . .. . . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . .. .. . . .. . . 11 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
• not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter • A 95% confidence interval for • ”The probability that • Formally, the interval [.262,1.184] is one of the possible realized Recap Confidence interval and the parameter P4 P3 P2 P1 Bayesian Intervals Bayesian Tests .. . . . . . Frequentist’s view of intervals . . .. . . .. .. . . . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed is in the interval [.262,1.184] is 95%” : is . . . . . . . . Example . .. . . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . 11 / 34 .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter
• The random quantity is the interval, not the parameter • A 95% confidence interval for • ”The probability that • Formally, the interval [.262,1.184] is one of the possible realized . . . . . P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . .. . . .. . . .. .. . Confidence interval and the parameter . Frequentist’s view of intervals . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed is in the interval [.262,1.184] is 95%” : is . . . . . . . . Example . . .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . 11 / 34 .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose.
• A 95% confidence interval for • ”The probability that • Formally, the interval [.262,1.184] is one of the possible realized . . P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. . . .. .. . Confidence interval and the parameter . Frequentist’s view of intervals . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed is in the interval [.262,1.184] is 95%” : is . . . . . . . . Example . . .. . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. .. . . . . .. . . .. . . .. . . .. 11 / 34 . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter
• ”The probability that • Formally, the interval [.262,1.184] is one of the possible realized . . Recap . . . . . .. . . .. .. . Bayesian Intervals . . .. .. . .. Bayesian Tests P2 P1 . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed is in the interval [.262,1.184] is 95%” : . Example . . . . Frequentist’s view of intervals . Confidence interval and the parameter P4 P3 . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . . 11 / 34 . . . . .. . . .. . .. . . . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter • A 95% confidence interval for θ is . 262 ≤ θ ≤ 1 . 184
• Formally, the interval [.262,1.184] is one of the possible realized . . . . . . . .. . . .. . .. Bayesian Tests . . .. . .. .. . Recap P1 Bayesian Intervals Example April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed . . . .. . . Frequentist’s view of intervals . Confidence interval and the parameter P4 P3 P2 . . . . . .. . . .. . . .. . . .. . . .. . . .. . . . 11 / 34 . . .. . . .. .. . . . .. . . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter • A 95% confidence interval for θ is . 262 ≤ θ ≤ 1 . 184 • ”The probability that θ is in the interval [.262,1.184] is 95%” :
. . . . . . . .. . . .. . .. Bayesian Tests . . .. . .. .. . . Recap Bayesian Intervals . Example April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang values of the random intervals (depending on the observed data) Incorrect, because the parameter is assumed fixed . . . P1 . . Frequentist’s view of intervals . Confidence interval and the parameter P4 P3 P2 .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 11 / 34 .. .. . . .. . .. . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We have carefully said that the interval covers the parameter • not that the parameter is inside the interval, on purpose. • The random quantity is the interval, not the parameter • A 95% confidence interval for θ is . 262 ≤ θ ≤ 1 . 184 • ”The probability that θ is in the interval [.262,1.184] is 95%” : • Formally, the interval [.262,1.184] is one of the possible realized
• Under Bayesian model, • All Bayesian claims of coverage are made with respect to the . . . .. . . .. . . .. .. . .. . . .. . . . . . . . some probability. April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang posterior distribution of the parameter. distribution. is a random variable with a probability Bayesian interpretation of intervals .. P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . .. . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. . . .. . . .. . . .. 12 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Bayesian setup allows us to say that θ is inside [.262, 1.184] with
• All Bayesian claims of coverage are made with respect to the . .. .. . . .. . . . . . .. . . .. .. . . .. . Bayesian interpretation of intervals April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang posterior distribution of the parameter. distribution. some probability. P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . .. .. .. . . .. . . . . . .. . . .. . . . 12 / 34 .. . . . .. . . . .. .. .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Bayesian setup allows us to say that θ is inside [.262, 1.184] with • Under Bayesian model, θ is a random variable with a probability
. .. .. . . .. . . . . . .. . . .. .. . . .. . Bayesian interpretation of intervals April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang posterior distribution of the parameter. distribution. some probability. P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . .. .. .. . .. . . . . . .. . . .. . . . . .. . . . .. . . .. . . . .. . .. . . .. . 12 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Bayesian setup allows us to say that θ is inside [.262, 1.184] with • Under Bayesian model, θ is a random variable with a probability • All Bayesian claims of coverage are made with respect to the
• If • The credible probability of A is Pr • and A is a credible set (or creditable interval ) for • Both the interpretation and construction of the Bayes credible set are . . .. . . .. . .. . Recap . . .. . . .. . . . . . P1 Bayesian Tests A April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang additional assumptions (for Bayesian framework). more straightforward than those of a classical confidence set, but with . x d A x Bayesian Intervals x is a posterior distribution, for any set A rather than confidence sets Credible sets P4 P3 P2 .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . .. . . .. . . 13 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • To distinguish Bayesian estimates of coverage, we use credible sets
• The credible probability of A is Pr • and A is a credible set (or creditable interval ) for • Both the interpretation and construction of the Bayes credible set are . .. .. . . .. . . . . . . . . .. . . .. .. . . Bayesian Intervals Recap A April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang additional assumptions (for Bayesian framework). more straightforward than those of a classical confidence set, but with . x d A x Bayesian Tests rather than confidence sets Credible sets P4 P3 P2 P1 . .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 13 / 34 .. . . . .. . . .. . . . . .. .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • To distinguish Bayesian estimates of coverage, we use credible sets • If π ( θ | x ) is a posterior distribution, for any set A ⊂ Ω
• and A is a credible set (or creditable interval ) for • Both the interpretation and construction of the Bayes credible set are . . . .. . . .. . . .. .. . .. .. . .. . . . . . . . rather than confidence sets April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang additional assumptions (for Bayesian framework). more straightforward than those of a classical confidence set, but with . Credible sets .. P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . .. . . . . .. . . . .. . .. . . .. . . . .. . .. .. . . .. . . .. . . . 13 / 34 . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • To distinguish Bayesian estimates of coverage, we use credible sets • If π ( θ | x ) is a posterior distribution, for any set A ⊂ Ω ∫ • The credible probability of A is Pr ( θ ∈ A | x ) = A π ( θ | x ) d θ
• Both the interpretation and construction of the Bayes credible set are . .. .. . . .. . . . . . .. . .. .. . . . .. . Credible sets April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang additional assumptions (for Bayesian framework). more straightforward than those of a classical confidence set, but with rather than confidence sets P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . .. .. . .. . .. . . . . . .. . . .. . . . 13 / 34 .. . . . .. . . .. . . .. . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • To distinguish Bayesian estimates of coverage, we use credible sets • If π ( θ | x ) is a posterior distribution, for any set A ⊂ Ω ∫ • The credible probability of A is Pr ( θ ∈ A | x ) = A π ( θ | x ) d θ • and A is a credible set (or creditable interval ) for θ .
. .. .. . . .. . . . . . .. . .. .. . . . .. . Credible sets April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang additional assumptions (for Bayesian framework). more straightforward than those of a classical confidence set, but with rather than confidence sets P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . .. .. . .. .. . . . . . .. . . .. . . . . .. . . . .. . . .. . . .. . 13 / 34 . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • To distinguish Bayesian estimates of coverage, we use credible sets • If π ( θ | x ) is a posterior distribution, for any set A ⊂ Ω ∫ • The credible probability of A is Pr ( θ ∈ A | x ) = A π ( θ | x ) d θ • and A is a credible set (or creditable interval ) for θ . • Both the interpretation and construction of the Bayes credible set are
x i n . Problem . . . . . Solution . i.i.d. . . . . Example: Possible credible set P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap .. . .. . The posterior pdf of . confidence interval is April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang a x i nb b a x i nb b Therefore, a . if a is an integer x i a b nb equally between the upper and lower endpoints, If we simply split the b Gamma a x becomes . . . . . .. .. . .. . . .. . . .. . . .. . . . .. . .. . . .. . . .. . . .. . . . . . .. . .. . . .. . . .. . . . 14 / 34 . . .. . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Poisson ( λ ) and assume that λ ∼ Gamma ( a , b ) . Find a 90% credible set for λ .
. P1 i.i.d. . . Problem . Example: Possible credible set P4 P3 P2 Bayesian Intervals Solution Bayesian Tests Recap . . . . . .. . . .. . . . . . nb April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang a x i nb b a x i b . confidence interval is Therefore, a if a is an integer x i a b nb equally between the upper and lower endpoints, If we simply split the Gamma .. .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. . . 14 / 34 . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Poisson ( λ ) and assume that λ ∼ Gamma ( a , b ) . Find a 90% credible set for λ . The posterior pdf of λ becomes ( x i , [ n + (1/ b )] − 1 ) ∑ π ( λ | x ) = a +
. . . . . Example: Possible credible set P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . Problem .. . . .. . . .. . . . . . nb April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang a x i nb b a x i b . confidence interval is Therefore, a b Gamma . . Solution . i.i.d. .. .. . . . .. . .. . .. . . .. . .. .. . . .. . . .. . . .. . . . 14 / 34 . . . . .. . . . .. . . . .. . . .. . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Poisson ( λ ) and assume that λ ∼ Gamma ( a , b ) . Find a 90% credible set for λ . The posterior pdf of λ becomes ( x i , [ n + (1/ b )] − 1 ) ∑ π ( λ | x ) = a + If we simply split the α equally between the upper and lower endpoints, 2( nb + 1) χ 2 λ ∼ ( if a is an integer ) 2( a + ∑ x i )
. .. Bayesian Tests Recap . . . . . .. . . . P1 . .. . . .. . . .. Bayesian Intervals P2 . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang b b b Gamma . Solution P3 . i.i.d. . . Problem . Example: Possible credible set P4 . .. .. . . .. . . .. . . . . .. .. . . .. . . .. . . . .. 14 / 34 . .. . . .. . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Poisson ( λ ) and assume that λ ∼ Gamma ( a , b ) . Find a 90% credible set for λ . The posterior pdf of λ becomes ( x i , [ n + (1/ b )] − 1 ) ∑ π ( λ | x ) = a + If we simply split the α equally between the upper and lower endpoints, 2( nb + 1) χ 2 λ ∼ ( if a is an integer ) 2( a + ∑ x i ) Therefore, a 1 − α confidence interval is { } 2( nb + 1) χ 2 2( nb + 1) χ 2 λ : 2( ∑ x i + a ) , 1 − α /2 ≤ λ ≤ 2( ∑ x i + a ) ,α /2
. .. Bayesian Tests Recap . . . . . .. . . . P1 . .. . . .. . . .. Bayesian Intervals P2 . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang b b b Gamma . Solution P3 . i.i.d. . . Problem . Example: Possible credible set P4 . .. .. . . .. . . .. . . . . .. .. . . .. . . .. . . . .. 14 / 34 . .. . . .. . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Poisson ( λ ) and assume that λ ∼ Gamma ( a , b ) . Find a 90% credible set for λ . The posterior pdf of λ becomes ( x i , [ n + (1/ b )] − 1 ) ∑ π ( λ | x ) = a + If we simply split the α equally between the upper and lower endpoints, 2( nb + 1) χ 2 λ ∼ ( if a is an integer ) 2( a + ∑ x i ) Therefore, a 1 − α confidence interval is { } 2( nb + 1) χ 2 2( nb + 1) χ 2 λ : 2( ∑ x i + a ) , 1 − α /2 ≤ λ ≤ 2( ∑ x i + a ) ,α /2
• Credible probabilities are the Bayes posterior probability, which • A Bayesian assertion of 90% coverage means that the experimenter, • Coverage probability reflects the uncertainty in the sampling • A classical assertion of 90% coverage means that in a long sequence of . . . . . . .. . . .. . . Bayesian Tests .. . . .. . . .. Recap P2 Bayesian Intervals % sure of coverage April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang parameter. identical trials, 90% of the realized confidence sets will cover the true repeated experimental trials. procedure, getting its probability from the objective mechanism of upon combining prior knowledge with data, is P1 distribution. reflects the experimenter’s subjective beliefs, as expressed in the prior probability Remark: Credible probability and coverage probability P4 P3 . . .. .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 15 / 34 . .. . . .. . . .. . . . . . . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • It is important not to confuse credible probability with coverage
• A Bayesian assertion of 90% coverage means that the experimenter, • Coverage probability reflects the uncertainty in the sampling • A classical assertion of 90% coverage means that in a long sequence of . . . . . . .. . . .. . .. . Bayesian Tests . . .. . . .. . Recap P1 Bayesian Intervals % sure of coverage April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang parameter. identical trials, 90% of the realized confidence sets will cover the true repeated experimental trials. procedure, getting its probability from the objective mechanism of upon combining prior knowledge with data, is .. distribution. reflects the experimenter’s subjective beliefs, as expressed in the prior probability Remark: Credible probability and coverage probability P4 P3 P2 . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . . 15 / 34 .. .. .. . . .. . . .. . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • It is important not to confuse credible probability with coverage • Credible probabilities are the Bayes posterior probability, which
• Coverage probability reflects the uncertainty in the sampling • A classical assertion of 90% coverage means that in a long sequence of . .. . . .. . . .. . . . . . . .. . . .. . . . Bayesian Tests Recap distribution. April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang parameter. identical trials, 90% of the realized confidence sets will cover the true repeated experimental trials. procedure, getting its probability from the objective mechanism of reflects the experimenter’s subjective beliefs, as expressed in the prior . probability Remark: Credible probability and coverage probability P4 P3 P2 P1 Bayesian Intervals .. .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. 15 / 34 . . .. .. . . . .. . . . .. .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • It is important not to confuse credible probability with coverage • Credible probabilities are the Bayes posterior probability, which • A Bayesian assertion of 90% coverage means that the experimenter, upon combining prior knowledge with data, is 90 % sure of coverage
• A classical assertion of 90% coverage means that in a long sequence of . . .. . . .. . . .. . . . . . .. . . .. . . .. . Recap . distribution. April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang parameter. identical trials, 90% of the realized confidence sets will cover the true repeated experimental trials. procedure, getting its probability from the objective mechanism of reflects the experimenter’s subjective beliefs, as expressed in the prior Bayesian Tests probability Remark: Credible probability and coverage probability P4 P3 P2 P1 Bayesian Intervals . .. .. . . . .. . . .. . .. .. . . .. . . .. . . . 15 / 34 . . . .. . . . .. . . . .. .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • It is important not to confuse credible probability with coverage • Credible probabilities are the Bayes posterior probability, which • A Bayesian assertion of 90% coverage means that the experimenter, upon combining prior knowledge with data, is 90 % sure of coverage • Coverage probability reflects the uncertainty in the sampling
. . .. . . .. . . .. . . . . . .. . . .. . . .. . Recap . distribution. April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang parameter. identical trials, 90% of the realized confidence sets will cover the true repeated experimental trials. procedure, getting its probability from the objective mechanism of reflects the experimenter’s subjective beliefs, as expressed in the prior Bayesian Tests probability Remark: Credible probability and coverage probability P4 P3 P2 P1 Bayesian Intervals . .. .. . . . .. . . .. . .. .. . . .. . . .. . . . 15 / 34 . .. . .. . . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • It is important not to confuse credible probability with coverage • Credible probabilities are the Bayes posterior probability, which • A Bayesian assertion of 90% coverage means that the experimenter, upon combining prior knowledge with data, is 90 % sure of coverage • Coverage probability reflects the uncertainty in the sampling • A classical assertion of 90% coverage means that in a long sequence of
X n are iid samples from f x . P1 . . Problem . Practice Problem 1 (from last lecture) P4 P3 P2 Bayesian Intervals . Bayesian Tests Recap . . . . . .. . . .. . .. . . . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang ? estimator for , what is the Bayes rule a a (b) If we use the loss function L (a) Derive the posterior distribution of . are known. where e is the prior distribution of x . Suppose exp Suppose X . . .. . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . 16 / 34 .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . .. . . .. . .. Bayesian Tests . . .. .. . .. . . Recap Bayesian Intervals . a April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang ? estimator for , what is the Bayes rule a (b) If we use the loss function L P1 . . Problem . Practice Problem 1 (from last lecture) P4 P3 P2 .. . . . .. . .. . . .. . .. . . . .. . . .. . . .. . . . . . .. . . .. . . .. . 16 / 34 .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid samples from f ( x | θ ) = θ exp ( − θ x ) . Suppose the prior distribution of θ is 1 Γ( α ) β α θ α − 1 e − θ / β π ( θ ) = where α, β are known. (a) Derive the posterior distribution of θ .
. .. .. . . .. . . . . . .. . . .. . . . .. . Practice Problem 1 (from last lecture) April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . Problem . P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. .. . .. .. . . .. .. . . . . . .. . . .. . . . 16 / 34 .. . . . .. . . . . . .. .. .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid samples from f ( x | θ ) = θ exp ( − θ x ) . Suppose the prior distribution of θ is 1 Γ( α ) β α θ α − 1 e − θ / β π ( θ ) = where α, β are known. (a) Derive the posterior distribution of θ . (b) If we use the loss function L ( θ, a ) = ( a − θ ) 2 , what is the Bayes rule estimator for θ ?
n exp . n P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. . . .. . . e n . x i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n Gamma x i i n n Gamma x i i n exp n x i .. .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . 17 / 34 .. .. . . .. . . . . . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) Posterior distribution of θ f ( x , θ ) = π ( θ ) f ( x | θ ) π ( θ ) 1 ∏ Γ( α ) β α θ α − 1 e − θ / β = [ θ exp ( − θ x i )] i =1
. .. P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. n . .. . . .. . . .. . P4 n .. x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n Gamma x i x i i n n Gamma x i i n exp n . . . .. . .. . . .. . . .. . . . . . .. . . .. . . .. . . . 17 / 34 . . .. . . .. . . .. .. . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) Posterior distribution of θ f ( x , θ ) = π ( θ ) f ( x | θ ) π ( θ ) 1 ∏ Γ( α ) β α θ α − 1 e − θ / β = [ θ exp ( − θ x i )] i =1 ( ) 1 Γ( α ) β α θ α − 1 e − θ / β θ n exp ∑ = − θ i =1
. . P1 Bayesian Intervals Bayesian Tests .. . . . . . .. . .. P3 . . .. . . .. . . P2 P4 . x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n Gamma x i n i n n Gamma x i n x i n .. Recap . .. . . .. .. . . .. . . . . . .. . . .. . . .. . . .. 17 / 34 . .. . . .. . . .. . . .. . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) Posterior distribution of θ f ( x , θ ) = π ( θ ) f ( x | θ ) π ( θ ) 1 ∏ Γ( α ) β α θ α − 1 e − θ / β = [ θ exp ( − θ x i )] i =1 ( ) 1 Γ( α ) β α θ α − 1 e − θ / β θ n exp ∑ = − θ i =1 [ ( )] 1 Γ( α ) β α θ α + n − 1 exp ∑ = − θ 1/ β + i =1
. . Recap . . . . . .. . . .. . Bayesian Tests .. . . .. . . .. . .. Bayesian Intervals .. x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n Gamma Gamma P1 x i n x i n n P4 P3 P2 . 17 / 34 . . . . .. .. . . . .. . . . .. . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) Posterior distribution of θ f ( x , θ ) = π ( θ ) f ( x | θ ) π ( θ ) 1 ∏ Γ( α ) β α θ α − 1 e − θ / β = [ θ exp ( − θ x i )] i =1 ( ) 1 Γ( α ) β α θ α − 1 e − θ / β θ n exp ∑ = − θ i =1 [ ( )] 1 Γ( α ) β α θ α + n − 1 exp ∑ = − θ 1/ β + i =1 ( 1 ) ∝ α + n − 1 , β − 1 + ∑ n i =1 x i
. . .. . . .. . . .. . . . . . .. . . .. . . .. . Recap . x i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang Gamma Gamma x i n n Bayesian Tests n P4 P3 P2 P1 Bayesian Intervals .. . 17 / 34 .. .. . . .. . . . . .. . . . . .. . . .. .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) Posterior distribution of θ f ( x , θ ) = π ( θ ) f ( x | θ ) π ( θ ) 1 ∏ Γ( α ) β α θ α − 1 e − θ / β = [ θ exp ( − θ x i )] i =1 ( ) 1 Γ( α ) β α θ α − 1 e − θ / β θ n exp ∑ = − θ i =1 [ ( )] 1 Γ( α ) β α θ α + n − 1 exp ∑ = − θ 1/ β + i =1 ( 1 ) ∝ α + n − 1 , β − 1 + ∑ n i =1 x i ( ) 1 π ( θ | x ) = α + n − 1 , β − 1 + ∑ n i =1 x i
. . P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . .. P3 . . .. . . .. . .. P2 P4 . E April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n x x (b) Bayes’ rule estimator with squared error loss E x i i n n Gamma x Bayes’ rule estimator with squared error loss is posterior mean. Note that .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . 18 / 34 .. . . .. . . .. . . .. . . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the mean of Gamma ( α, β ) is αβ .
. . . . . . . .. . . .. . .. Bayesian Tests . . .. . . .. . . Recap Bayesian Intervals . x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n E P1 x E Gamma Bayes’ rule estimator with squared error loss is posterior mean. Note that (b) Bayes’ rule estimator with squared error loss P4 P3 P2 .. .. . . . .. . .. . . .. . .. . . . .. . . .. . . .. 18 / 34 . . .. . . .. . . .. . .. . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the mean of Gamma ( α, β ) is αβ . ( 1 ) π ( θ | x ) α + n − 1 , = β − 1 + ∑ n i =1 x i
. .. . . .. . . .. . . .. . . .. . . .. .. . . P4 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang Gamma Bayes’ rule estimator with squared error loss is posterior mean. Note that (b) Bayes’ rule estimator with squared error loss P3 .. P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . . . .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . . .. . . .. 18 / 34 . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the mean of Gamma ( α, β ) is αβ . ( 1 ) π ( θ | x ) α + n − 1 , = β − 1 + ∑ n i =1 x i E [ θ | x ] E [ π ( θ | x )] = α + n − 1 = β − 1 + ∑ n i =1 x i
. .. P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . . Practice Problem 2 . .. . .. .. . . .. . P4 . .. , where April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . vs. H test for H (c) Derive the UMP level . vs. H Problem test for H (b) Derive the UMP level . H vs. LRT for testing H (a) Derive the asymptotic size . . . . . .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . . .. . 19 / 34 . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples from Gamma distribution with parameter (3 , θ ) , which has the pdf 1 2 θ 3 x 2 e − x / θ f ( x | θ ) = ( x > 0) i =1 X i / θ ∼ χ 2 You may use the result that 2 ∑ n 6 n .
. .. Bayesian Tests Recap . . . . . .. . . . P1 . .. . . .. . . .. Bayesian Intervals P2 . , where April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . vs. H test for H (c) Derive the UMP level . vs. H P3 test for H (b) Derive the UMP level . . Problem . Practice Problem 2 P4 . .. .. . . . . . .. . . .. . .. .. . . .. . . .. . . . .. 19 / 34 . . . . .. . . .. . .. . . . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples from Gamma distribution with parameter (3 , θ ) , which has the pdf 1 2 θ 3 x 2 e − x / θ f ( x | θ ) = ( x > 0) i =1 X i / θ ∼ χ 2 You may use the result that 2 ∑ n 6 n . (a) Derive the asymptotic size α LRT for testing H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 .
. . .. . . .. . . .. .. . . . . .. . . .. . . .. . Recap . . April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . vs. H test for H (c) Derive the UMP level . Bayesian Tests Problem . Practice Problem 2 P4 P3 P2 P1 Bayesian Intervals . . .. .. . .. . . .. . . . .. . .. . . .. . . . . . .. . .. . . .. . . .. . . 19 / 34 . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples from Gamma distribution with parameter (3 , θ ) , which has the pdf 1 2 θ 3 x 2 e − x / θ f ( x | θ ) = ( x > 0) i =1 X i / θ ∼ χ 2 You may use the result that 2 ∑ n 6 n . (a) Derive the asymptotic size α LRT for testing H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . (b) Derive the UMP level α test for H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , where θ 1 > θ 0 .
. .. .. . . .. . . . . . .. . . .. . . . .. . Practice Problem 2 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang . . Problem . P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. .. . .. .. . . .. .. . . . . . .. . . .. . . . 19 / 34 .. . . . .. . . .. . . .. . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suppose X 1 , · · · , X n are iid random samples from Gamma distribution with parameter (3 , θ ) , which has the pdf 1 2 θ 3 x 2 e − x / θ f ( x | θ ) = ( x > 0) i =1 X i / θ ∼ χ 2 You may use the result that 2 ∑ n 6 n . (a) Derive the asymptotic size α LRT for testing H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . (b) Derive the UMP level α test for H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , where θ 1 > θ 0 . (c) Derive the UMP level α test for H 0 : θ ≤ θ 0 vs. H 1 : θ > θ 0 .
x i e . P2 n x l x i i n Solution (a) - Obtaining MLEs P4 P3 P1 log Bayesian Intervals Bayesian Tests Recap . . . . . .. . . .. . i log .. n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n x i i n x log x i l x i i n log x i i n n log n log x i . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. . . .. . . .. . . .. . . 20 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L ( θ | x ) =
. . P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. n .. . .. . . .. . . .. Solution (a) - Obtaining MLEs n . n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n x i i n n log x l x i i n log x i i n n log . . .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . 20 / 34 . .. . .. .. . . .. . . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [ 1 ] ∏ 2 θ 3 x 2 i e − x i / θ L ( θ | x ) = i =1 [ ] ∑ l ( θ | x ) = − log 2 − 3 log θ + 2 log x i − x i θ i =1
. . P1 Bayesian Intervals Bayesian Tests Recap . . . . .. .. . .. P3 . . .. . . .. . . P2 P4 . i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n x i n Solution (a) - Obtaining MLEs n x l x i n n n n .. . . .. . . .. .. . . .. . . . . . .. . . .. . . .. . . .. 20 / 34 . . .. . . .. . .. . . .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [ 1 ] ∏ 2 θ 3 x 2 i e − x i / θ L ( θ | x ) = i =1 [ ] ∑ l ( θ | x ) = − log 2 − 3 log θ + 2 log x i − x i θ i =1 log x i − 1 ∑ ∑ − n log 2 − 3 n log θ + 2 = θ i =1 i =1
. . .. . .. . . .. . .. Bayesian Tests . . .. . . .. . . Recap Bayesian Intervals . n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i i n n x i P1 n n n n Solution (a) - Obtaining MLEs P4 P3 P2 .. . . . . . . . . .. . . .. .. . .. . . . .. . . .. . . .. 20 / 34 . . . . .. .. . . . .. . . . . . .. .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [ 1 ] ∏ 2 θ 3 x 2 i e − x i / θ L ( θ | x ) = i =1 [ ] ∑ l ( θ | x ) = − log 2 − 3 log θ + 2 log x i − x i θ i =1 log x i − 1 ∑ ∑ − n log 2 − 3 n log θ + 2 = θ i =1 i =1 − 3 n θ + 1 ∑ l ′ ( θ | x ) = x i = 0 θ 2 i =1
. . .. . . .. . . .. . . . . . .. . . .. . . .. . Recap . n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i n n x i n .. n n Solution (a) - Obtaining MLEs P4 P3 P2 P1 Bayesian Intervals . Bayesian Tests .. . . .. . . .. . . .. .. . . .. . . .. . . . 20 / 34 . . . .. . .. . .. . . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [ 1 ] ∏ 2 θ 3 x 2 i e − x i / θ L ( θ | x ) = i =1 [ ] ∑ l ( θ | x ) = − log 2 − 3 log θ + 2 log x i − x i θ i =1 log x i − 1 ∑ ∑ − n log 2 − 3 n log θ + 2 = θ i =1 i =1 − 3 n θ + 1 ∑ l ′ ( θ | x ) = x i = 0 θ 2 i =1 1 ˆ ∑ = θ 3 n i =1
. .. Recap . . . . . .. . . .. . Bayesian Intervals .. . . .. . . .. . Bayesian Tests P1 .. as April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i . i n n approaches zero or infinity, x P2 Because L n n x i n Solution (a) - Obtaining MLEs P4 P3 . . . . .. . . . .. . . .. . . .. . . .. . . .. . . . 21 / 34 . . . . . .. . . .. . .. .. .. . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . � 3 n θ 2 − 2 � ∑ � l ′′ ( θ | x ) = � θ =ˆ � θ 3 θ � i =1 � θ =ˆ θ
. . . . . . . .. . .. .. . .. Bayesian Tests . . .. . . .. . . Recap Bayesian Intervals . approaches zero or infinity, April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i . i n n as P1 x Because L x i n Solution (a) - Obtaining MLEs P4 P3 P2 .. . . . . . .. . .. . .. . .. . . . .. . . .. . . .. 21 / 34 . . .. . .. . . . .. . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . � 3 n θ 2 − 2 � ∑ � l ′′ ( θ | x ) = � θ =ˆ � θ 3 θ � i =1 � θ =ˆ θ 3 n θ 2 − 6 n = θ 2 < 0 ˆ ˆ
. . . . .. . . .. . . .. . . .. . . .. .. . . P4 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i n Solution (a) - Obtaining MLEs P3 .. P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . . .. .. .. .. . . .. . . . . . . . . .. . . . .. .. . . .. . . . . . .. .. 21 / 34 . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . � 3 n θ 2 − 2 � ∑ � l ′′ ( θ | x ) = � θ =ˆ � θ 3 θ � i =1 � θ =ˆ θ 3 n θ 2 − 6 n = θ 2 < 0 ˆ ˆ Because L ( θ | x ) → 0 as θ approaches zero or infinity, ˆ 1 ∑ n θ = 3 n i =1 x i .
. P3 n log x i n log x l x l x log P4 P2 n log P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . .. .. x i x i . x April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n log n x i log n x i n n log log n x i n log x i x R n x i n . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . .. . . .. . . .. . . . . .. . .. . 22 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is
. Bayesian Tests x i n log x i n log P4 P3 P2 P1 Bayesian Intervals Recap x i . . . . . .. . .. .. . . .. n log n log . x i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n log n x i log n x n n log n x i n log x i x R n x i . . .. .. . .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . 22 / 34 . . . . .. . . .. . .. . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is [ ] l ( θ 0 | x ) − l (ˆ − 2 log λ ( x ) = − 2 θ | x )
. . . . . n log x i P4 P3 P2 P1 Bayesian Intervals Bayesian Tests Recap .. n log .. . . .. . . .. . . x i n . x i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n log n x i log n x x i n log n x i n log x i x R n .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . 22 / 34 . . . . .. . . .. . .. . .. .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is [ ] l ( θ 0 | x ) − l (ˆ − 2 log λ ( x ) = − 2 θ | x ) 6 n log θ 0 + 2 θ − 2 ∑ x i − 6 n log ˆ ∑ = ˆ θ 0 θ
. .. P2 P1 Bayesian Intervals Bayesian Tests .. Recap . . . . . . P4 . .. . . .. . . .. . P3 x i .. n April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n log n x i log x i x i x n log n x i n log x i x R . 22 / 34 . . . . . .. . . .. . .. .. . . .. . . . .. . . .. .. . .. . . .. . . .. . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is [ ] l ( θ 0 | x ) − l (ˆ − 2 log λ ( x ) = − 2 θ | x ) 6 n log θ 0 + 2 θ − 2 ∑ x i − 6 n log ˆ ∑ = ˆ θ 0 θ ( 1 6 n log θ 0 + 2 ) ∑ ∑ − 6 n > χ 2 x i − 6 n log = 1 ,α 3 n θ 0
. . Recap . . . . . .. . . .. . .. .. . . .. . . .. . Bayesian Tests Bayesian Intervals .. log April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n log n x i n P1 x i x R x i x i P4 P3 P2 . 22 / 34 . .. . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is [ ] l ( θ 0 | x ) − l (ˆ − 2 log λ ( x ) = − 2 θ | x ) 6 n log θ 0 + 2 θ − 2 ∑ x i − 6 n log ˆ ∑ = ˆ θ 0 θ ( 1 6 n log θ 0 + 2 ) ∑ ∑ − 6 n > χ 2 x i − 6 n log = 1 ,α 3 n θ 0 { x : 2 } ∑ ∑ x i > χ 2 = x i − 6 n log 1 ,α + 6 n [1 − log (3 n θ 0 )] θ 0
. . . . .. . . .. . . .. . . .. . . .. .. . . P4 April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang R x i x i P3 .. P2 P1 .. Bayesian Tests Recap . . . . . . Bayesian Intervals .. . . . .. . . .. . . . .. . . .. . . .. 22 / 34 . . .. . . .. . . . . .. .. .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (a) - Constructing asymptotic size α LRT The rejection region of asymptotic size α LRT is [ ] l ( θ 0 | x ) − l (ˆ − 2 log λ ( x ) = − 2 θ | x ) 6 n log θ 0 + 2 θ − 2 ∑ x i − 6 n log ˆ ∑ = ˆ θ 0 θ ( 1 6 n log θ 0 + 2 ) ∑ ∑ − 6 n > χ 2 x i − 6 n log = 1 ,α 3 n θ 0 { x : 2 } ∑ ∑ x i > χ 2 = x i − 6 n log 1 ,α + 6 n [1 − log (3 n θ 0 )] θ 0 { } x i > θ 0 ∑ ∑ 2 χ 2 = x : x i − 3 n θ 0 log 1 ,α + 3 n θ 0 [1 − log (3 n θ 0 )]
n exp n exp . .. Bayesian Tests Recap . . . . . .. . . . P1 . .. . . .. . .. .. Bayesian Intervals P3 P2 x i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i exp n x i n . x i x i n x L x L P4 . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . .. . . . .. . . .. . . .. . . .. . .. . . .. . 23 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test for simple hypothesis For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 ,
. .. .. . . .. . . . . . .. . . .. . . . .. . i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i exp n i P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. .. . .. .. . . .. . .. . . . .. . . .. . . . . .. . . . .. . . .. . . . .. 23 / 34 .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test for simple hypothesis For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , [ ] ∏ x 2 ∑ x i 1 − L ( θ 1 | x ) 2 n θ 3 n θ 1 1 exp = L ( θ 0 | x ) [ ] ∏ x 2 ∑ x i 1 − 2 n θ 3 n θ 0 0 exp
. .. .. . . .. . . . . . .. . . .. . . . .. . i April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang x i exp i P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. .. . . .. . . .. . . .. .. . . .. . . .. . . . 23 / 34 .. . .. . . .. . . . .. . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test for simple hypothesis For H 0 : θ = θ 0 vs. H 1 : θ = θ 1 , [ ] ∏ x 2 ∑ x i 1 − L ( θ 1 | x ) 2 n θ 3 n θ 1 1 exp = L ( θ 0 | x ) [ ] ∏ x 2 ∑ x i 1 − 2 n θ 3 n θ 0 0 exp ) 3 n ( θ 0 [ θ 1 − θ 0 ] ∑ = θ 1 θ 0 θ 1
. . Recap . . . . . .. . . .. . Bayesian Intervals .. . . .. .. . .. . Bayesian Tests P1 .. So, the rejection region is April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n x i T x x R k P2 Pr T k T exp n Pr P4 P3 . . . . .. . . .. . . .. . .. . . . .. . . .. . . . . .. . .. . . .. . . .. . . .. . . .. . .. . . 24 / 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test (cont’d) θ 0 T ∼ χ 2 2 Let T = ∑ X i . Then under H 0 , 6 n .
. . .. . . .. . . .. . . . . . .. . . .. . . .. . Recap . R April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n x i T x x So, the rejection region is Bayesian Tests T exp Pr P4 P3 P2 P1 Bayesian Intervals . .. .. . . .. . . . .. . .. .. . . .. . . .. . . . 24 / 34 .. .. . . .. . . . .. . . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test (cont’d) θ 0 T ∼ χ 2 2 Let T = ∑ X i . Then under H 0 , 6 n . [( θ 0 ] ) 3 n [ θ 1 − θ 0 ] = α > k θ 1 θ 0 θ 1 Pr ( T > k ∗ ) =
. .. .. . .. .. . . . . . .. . . .. . . . .. . Pr April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang So, the rejection region is T exp P4 . P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . .. . . .. .. . . .. . . .. . . . .. . . .. . . . 24 / 34 .. . .. . . .. . . . .. . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (b) - UMP level α test (cont’d) θ 0 T ∼ χ 2 2 Let T = ∑ X i . Then under H 0 , 6 n . [( θ 0 ] ) 3 n [ θ 1 − θ 0 ] = α > k θ 1 θ 0 θ 1 Pr ( T > k ∗ ) = { } x i > θ 0 ∑ 2 χ 2 R = x : T ( x ) = 6 n ,α
n y n f Y y f T t f T t f T t . n t n n y e n P4 t P3 P2 P1 Bayesian Intervals Bayesian Tests Recap . . . . . .. . . e t n t April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang is an increasing function of t . This T has MLR property. t exp n t e n n . t e n t n , For arbitrary t e n .. .. . .. .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . 25 / 34 . . . .. . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - UMP level α test for composite hypothesis We need to check whether T has MLR. Because Y = 2 T / θ ∼ χ 2 6 n .
f T t f T t . .. P1 Bayesian Intervals Bayesian Tests Recap . . . . . . P3 . .. . . .. . . .. . P2 For arbitrary P4 e April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang is an increasing function of t . This T has MLR property. t exp n t n .. t n t e n t n , . .. . .. . .. . . .. . . .. . . . . . .. . . .. . . .. . . . 25 / 34 .. . . . .. . . .. . . .. . . . .. . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - UMP level α test for composite hypothesis We need to check whether T has MLR. Because Y = 2 T / θ ∼ χ 2 6 n . 1 2 3 n Γ(3 n ) y 3 n − 1 e − y /2 f Y ( y | θ ) = ) 3 n − 1 ) 3 n − 1 1 ( 2 t 1 ( t e − t / θ = e − t / θ f T ( t | θ ) = 2 3 n − 1 Γ(3 n ) θ θ Γ(3 n ) θ θ
. .. .. . . .. . . . . . .. . . .. . . . .. . t April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang is an increasing function of t . This T has MLR property. t exp t P4 . P3 P2 P1 Bayesian Intervals .. Recap . . . . .. Bayesian Tests . .. .. . . .. . . .. . . . .. . . .. . . . 25 / 34 .. . . .. . . .. .. . . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - UMP level α test for composite hypothesis We need to check whether T has MLR. Because Y = 2 T / θ ∼ χ 2 6 n . 1 2 3 n Γ(3 n ) y 3 n − 1 e − y /2 f Y ( y | θ ) = ) 3 n − 1 ) 3 n − 1 1 ( 2 t 1 ( t e − t / θ = e − t / θ f T ( t | θ ) = 2 3 n − 1 Γ(3 n ) θ θ Γ(3 n ) θ θ For arbitrary θ 1 < θ 2 , ) 3 n − 1 ( 1 e − t / θ 2 ) 3 n f T ( t | θ 2 ) ( θ 1 [ θ 2 − θ 1 ] Γ(3 n ) θ 2 θ 2 = = ) 3 n − 1 f T ( t | θ 1 ) θ 2 θ 1 θ 2 ( 1 e − t / θ 1 Γ(3 n ) θ 1 θ 1
. .. .. . . .. . . . . . .. . . .. . . . .. . t April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang is an increasing function of t . This T has MLR property. t exp t P4 . P3 P2 P1 Bayesian Intervals .. Recap . . . . .. Bayesian Tests . .. .. . . .. . . .. . . . .. . . .. . . . 25 / 34 .. . . .. . . .. .. . . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - UMP level α test for composite hypothesis We need to check whether T has MLR. Because Y = 2 T / θ ∼ χ 2 6 n . 1 2 3 n Γ(3 n ) y 3 n − 1 e − y /2 f Y ( y | θ ) = ) 3 n − 1 ) 3 n − 1 1 ( 2 t 1 ( t e − t / θ = e − t / θ f T ( t | θ ) = 2 3 n − 1 Γ(3 n ) θ θ Γ(3 n ) θ θ For arbitrary θ 1 < θ 2 , ) 3 n − 1 ( 1 e − t / θ 2 ) 3 n f T ( t | θ 2 ) ( θ 1 [ θ 2 − θ 1 ] Γ(3 n ) θ 2 θ 2 = = ) 3 n − 1 f T ( t | θ 1 ) θ 2 θ 1 θ 2 ( 1 e − t / θ 1 Γ(3 n ) θ 1 θ 1
. . .. . . .. . . .. . . . . . .. . . .. . .. .. . Recap . R April 18th, 2013 Biostatistics 602 - Lecture 25 Hyun Min Kang n x i T x x whose rejection is Bayesian Tests test is identical to the answer of part (b), Therefore, the UMP level P4 P3 P2 P1 Bayesian Intervals . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 26 / 34 . . . . .. . .. . . . . .. .. . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution (c) - Constructing UMP level α test Because T has MLR property, UMP level α test for H 0 : θ ≤ θ 0 vs. H 1 : θ > θ 0 has a rejection region T > k , and Pr ( T > k ) = α .
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