[PPT] - Biostatistics 602 - Statistical Inference April 18th, 2013 PowerPoint Presentation

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. .

Biostatistics 602 - Statistical Inference Lecture 25 Bayesian Test & Practice Problems

Hyun Min Kang April 18th, 2013

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 1 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Last Lecture

What is an E-M algorithm?
When would the E-M algorithm be useful?
Is MLE via E-M algorithm always guaranteed to converge?
What are the practical limitations of the E-M algorithm?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 2 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Last Lecture

What is an E-M algorithm?
When would the E-M algorithm be useful?
Is MLE via E-M algorithm always guaranteed to converge?
What are the practical limitations of the E-M algorithm?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 2 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Last Lecture

What is an E-M algorithm?
When would the E-M algorithm be useful?
Is MLE via E-M algorithm always guaranteed to converge?
What are the practical limitations of the E-M algorithm?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 2 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Last Lecture

What is an E-M algorithm?
When would the E-M algorithm be useful?
Is MLE via E-M algorithm always guaranteed to converge?
What are the practical limitations of the E-M algorithm?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 2 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Last Lecture

What is an E-M algorithm?
When would the E-M algorithm be useful?
Is MLE via E-M algorithm always guaranteed to converge?
What are the practical limitations of the E-M algorithm?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 2 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Overview of E-M Algorithm (cont’d)

.

Objective

. .

Maximize L(θ|y) or l(θ|y).
Let f(y, z|θ) denotes the pdf of complete data. In E-M algorithm,

rather than working with l(θ|y) directly, we work with the surrogate function Q(θ|θ(r)) = E [ log f(y, Z|θ)|y, θ(r)] where θ(r) is the estimation of θ in r-th iteration.

Q(θ|θ(r)) is the expected log-likelihood of complete data, conditioning
n the observed data and θ(r).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 3 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Key Steps of E-M algorithm

.

Expectation Step

. .

Compute Q(θ|θ(r)).
This typically involves in estimating the conditional distribution Z|Y,

assuming θ = θ(r).

After computing Q(θ|θ(r)), move to the M-step

.

Maximization Step

. .

Maximize Q(θ|θ(r)) with respect to θ.
The arg maxθ Q(θ|θ(r)) will be the (r + 1)-th θ to be fed into the

E-step.

Repeat E-step until convergence

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 4 / 34

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Does E-M iteration converge to MLE?

.

Theorem 7.2.20 - Monotonic EM sequence

. . The sequence {ˆ θ(r)} defined by the E-M procedure satisfies L ( ˆ θ(r+1)|y ) ≥ L ( ˆ θ(r)|y ) with equality holding if and only if successive iterations yield the same value of the maximized expected complete-data log likelihood, that is E [ log L ( ˆ θ(r+1)|y, Z ) |ˆ θ(r), y ] = E [ log L ( ˆ θ(r)|y, Z ) |ˆ θ(r), y ] Theorem 7.5.2 further guarantees that L(ˆ θ(r)|y) converges monotonically to L(ˆ θ|y) for some stationary point ˆ θ.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 5 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

Hypothesis testing problems can be formulated in a Bayesian model
Bayesian model includes
Sampling distribution f x
Prior distribution
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

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

Hypothesis testing problems can be formulated in a Bayesian model
Bayesian model includes
Sampling distribution f x
Prior distribution
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

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

Hypothesis testing problems can be formulated in a Bayesian model
Bayesian model includes
Sampling distribution f(x|θ)
Prior distribution
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

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 and H can be calculated
Pr

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 and H can be calculated
Pr

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 and H can be calculated
Pr

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 H0 and H1 can be calculated
Pr

x Pr H is true

Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 H0 and H1 can be calculated
Pr(θ ∈ Ω0|x) = Pr(H0 is true)
Pr

c x

Pr H is true

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 H0 and H1 can be calculated
Pr(θ ∈ Ω0|x) = Pr(H0 is true)
Pr(θ ∈ Ωc

0|x) = Pr(H1 is true)

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian Tests

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 H0 and H1 can be calculated
Pr(θ ∈ Ω0|x) = Pr(H0 is true)
Pr(θ ∈ Ωc

0|x) = Pr(H1 is true)

Rejection region can be determined directly based on the posterior

probability

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 6 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian vs Frequentist Framework

.

Frequentist’s Framework

. .

θ is considered to be a fixed number
Consequently, a hypothesis is either true of false
If

, Pr H is true x and Pr H is true x

If

c, Pr H is true x

and Pr H is true x

.

Bayesian Framework

. . . . . . . .

Pr H is true x and Pr H is true x are function of x, between 0

and 1.

These probabilities give useful information about the veracity of H

and H .

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 7 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian vs Frequentist Framework

.

Frequentist’s Framework

. .

θ is considered to be a fixed number
Consequently, a hypothesis is either true of false
If

, Pr H is true x and Pr H is true x

If

c, Pr H is true x

and Pr H is true x

.

Bayesian Framework

. . . . . . . .

Pr H is true x and Pr H is true x are function of x, between 0

and 1.

These probabilities give useful information about the veracity of H

and H .

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 7 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian vs Frequentist Framework

.

Frequentist’s Framework

. .

θ is considered to be a fixed number
Consequently, a hypothesis is either true of false
If θ ∈ Ω0, Pr(H0 is true|x) = 1 and Pr(H1 is true|x) = 0
If θ ∈ Ωc

0, Pr(H0 is true|x) = 0 and Pr(H1 is true|x) = 1

.

Bayesian Framework

. .

Pr(H0 is true|x) and Pr(H1 is true|x) are function of x, between 0

and 1.

These probabilities give useful information about the veracity of H

and H .

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 7 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian vs Frequentist Framework

.

Frequentist’s Framework

. .

θ is considered to be a fixed number
Consequently, a hypothesis is either true of false
If θ ∈ Ω0, Pr(H0 is true|x) = 1 and Pr(H1 is true|x) = 0
If θ ∈ Ωc

0, Pr(H0 is true|x) = 0 and Pr(H1 is true|x) = 1

.

Bayesian Framework

. .

Pr(H0 is true|x) and Pr(H1 is true|x) are function of x, between 0

and 1.

These probabilities give useful information about the veracity of H0

and H1.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 7 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Examples of Bayesian hypothesis testing procedure

.

A neutral test between H0 and H1

. .

Accept H is Pr

x Pr

c x

Reject H is Pr

x Pr

c x

In other words, the rejection region is

x Pr

c x

.

A more conservative (smaller size) test in rejecting H

. . . . . . . .

Reject H is Pr

c x

Accept H is Pr

c x

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 8 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Examples of Bayesian hypothesis testing procedure

.

A neutral test between H0 and H1

. .

Accept H0 is Pr(θ ∈ Ω0|x) ≥ Pr(θ ∈ Ωc

0|x)

Reject H0 is Pr(θ ∈ Ω0|x) < Pr(θ ∈ Ωc

0|x)

In other words, the rejection region is

x Pr

c x

.

A more conservative (smaller size) test in rejecting H

. . . . . . . .

Reject H is Pr

c x

Accept H is Pr

c x

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 8 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Examples of Bayesian hypothesis testing procedure

.

A neutral test between H0 and H1

. .

Accept H0 is Pr(θ ∈ Ω0|x) ≥ Pr(θ ∈ Ωc

0|x)

Reject H0 is Pr(θ ∈ Ω0|x) < Pr(θ ∈ Ωc

0|x)

In other words, the rejection region is {x : Pr(θ ∈ Ωc

0|x) > 1 2}

.

A more conservative (smaller size) test in rejecting H

. . . . . . . .

Reject H is Pr

c x

Accept H is Pr

c x

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 8 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Examples of Bayesian hypothesis testing procedure

.

A neutral test between H0 and H1

. .

Accept H0 is Pr(θ ∈ Ω0|x) ≥ Pr(θ ∈ Ωc

0|x)

Reject H0 is Pr(θ ∈ Ω0|x) < Pr(θ ∈ Ωc

0|x)

In other words, the rejection region is {x : Pr(θ ∈ Ωc

0|x) > 1 2}

.

A more conservative (smaller size) test in rejecting H0

. .

Reject H0 is Pr(θ ∈ Ωc

0|x) > 0.99

Accept H0 is Pr(θ ∈ Ωc

. . Consider testing H0 : θ ≤ θ0 versus H1 : θ > θ0. From previous lectures, the posterior is π(θ|x) ∼ N (nτ 2x + σ2µ nτ 2 + σ2 , σ2τ 2 nτ 2 + σ2 ) We will reject H0 if and only if Pr(θ ∈ Ω0|x) = Pr(θ ≤ θ0|x) < 1 2

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 9 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (cont’d)

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 θ0. Therefore, H0 will be rejected if n x n x n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 10 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (cont’d)

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 θ0. Therefore, H0 will be rejected if nτ 2x + σ2µ nτ 2 + σ2 < θ0 x n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 10 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (cont’d)

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 θ0. Therefore, H0 will be rejected if nτ 2x + σ2µ nτ 2 + σ2 < θ0 x < θ0 + σ2(θ0 − µ) nτ 2

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 10 / 34

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. . . . . . . .

A 95% confidence interval for

is

”The probability that

is in the interval [.262,1.184] is 95%” : Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 37

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. . . . . . . .

A 95% confidence interval for

is

”The probability that

is in the interval [.262,1.184] is 95%” : Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 38

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. . . . . . . .

A 95% confidence interval for

is

”The probability that

is in the interval [.262,1.184] is 95%” : Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 39

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. . . . . . . .

A 95% confidence interval for

is

”The probability that

is in the interval [.262,1.184] is 95%” : Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 40

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. .

A 95% confidence interval for θ is .262 ≤ θ ≤ 1.184
”The probability that

is in the interval [.262,1.184] is 95%” : Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 41

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. .

A 95% confidence interval for θ is .262 ≤ θ ≤ 1.184
”The probability that θ is in the interval [.262,1.184] is 95%” :

Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 42

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Confidence interval and the parameter

.

Frequentist’s view of intervals

. .

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

.

Example

. .

A 95% confidence interval for θ is .262 ≤ θ ≤ 1.184
”The probability that θ is in the interval [.262,1.184] is 95%” :

Incorrect, because the parameter is assumed fixed

Formally, the interval [.262,1.184] is one of the possible realized

values of the random intervals (depending on the observed data)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 11 / 34

SLIDE 43

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian interpretation of intervals

Bayesian setup allows us to say that θ is inside [.262, 1.184] with

some probability.

Under Bayesian model,

is a random variable with a probability distribution.

All Bayesian claims of coverage are made with respect to the

posterior distribution of the parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 12 / 34

SLIDE 44

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian interpretation of intervals

Bayesian setup allows us to say that θ is inside [.262, 1.184] with

some probability.

Under Bayesian model, θ is a random variable with a probability

distribution.

All Bayesian claims of coverage are made with respect to the

posterior distribution of the parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 12 / 34

SLIDE 45

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Bayesian interpretation of intervals

Bayesian setup allows us to say that θ is inside [.262, 1.184] with

some probability.

Under Bayesian model, θ is a random variable with a probability

distribution.

All Bayesian claims of coverage are made with respect to the

posterior distribution of the parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 12 / 34

SLIDE 46

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Credible sets

To distinguish Bayesian estimates of coverage, we use credible sets

rather than confidence sets

If

x is a posterior distribution, for any set A

The credible probability of A is Pr

A x

A

x d

and A is a credible set (or creditable interval) for

.

Both the interpretation and construction of the Bayes credible set are

more straightforward than those of a classical confidence set, but with additional assumptions (for Bayesian framework).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 13 / 34

SLIDE 47

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Credible sets

To distinguish Bayesian estimates of coverage, we use credible sets

rather than confidence sets

If π(θ|x) is a posterior distribution, for any set A ⊂ Ω
The credible probability of A is Pr

A x

A

x d

and A is a credible set (or creditable interval) for

.

Both the interpretation and construction of the Bayes credible set are

more straightforward than those of a classical confidence set, but with additional assumptions (for Bayesian framework).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 13 / 34

SLIDE 48

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Credible sets

To distinguish Bayesian estimates of coverage, we use credible sets

rather than confidence 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)dθ

and A is a credible set (or creditable interval) for

.

Both the interpretation and construction of the Bayes credible set are

more straightforward than those of a classical confidence set, but with additional assumptions (for Bayesian framework).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 13 / 34

SLIDE 49

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Credible sets

To distinguish Bayesian estimates of coverage, we use credible sets

rather than confidence 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)dθ

and A is a credible set (or creditable interval) for θ.
Both the interpretation and construction of the Bayes credible set are

more straightforward than those of a classical confidence set, but with additional assumptions (for Bayesian framework).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 13 / 34

SLIDE 50

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Credible sets

To distinguish Bayesian estimates of coverage, we use credible sets

rather than confidence 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)dθ

and A is a credible set (or creditable interval) for θ.
Both the interpretation and construction of the Bayes credible set are

more straightforward than those of a classical confidence set, but with additional assumptions (for Bayesian framework).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 13 / 34

SLIDE 51

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Example: Possible credible set

.

Problem

. . Let X1, · · · , Xn

i.i.d.

∼ Poisson(λ) and assume that λ ∼ Gamma(a, b). Find

a 90% credible set for λ. .

Solution

. . . . . . . . The posterior pdf of becomes x Gamma a xi n b If we simply split the equally between the upper and lower endpoints, nb b

a xi

if a is an integer Therefore, a confidence interval is b nb

xi a

b nb

xi a

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 14 / 34

SLIDE 52

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Example: Possible credible set

.

Problem

. . Let X1, · · · , Xn

i.i.d.

∼ Poisson(λ) and assume that λ ∼ Gamma(a, b). Find

a 90% credible set for λ. .

Solution

. . The posterior pdf of λ becomes π(λ|x) = Gamma ( a + ∑ xi, [n + (1/b)]−1) If we simply split the equally between the upper and lower endpoints, nb b

a xi

if a is an integer Therefore, a confidence interval is b nb

xi a

b nb

xi a

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 14 / 34

SLIDE 53

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Example: Possible credible set

.

Problem

. . Let X1, · · · , Xn

i.i.d.

∼ Poisson(λ) and assume that λ ∼ Gamma(a, b). Find

a 90% credible set for λ. .

Solution

. . The posterior pdf of λ becomes π(λ|x) = Gamma ( a + ∑ xi, [n + (1/b)]−1) If we simply split the α equally between the upper and lower endpoints, 2(nb + 1) b λ ∼ χ2

2(a+∑ xi)

(if a is an integer) Therefore, a confidence interval is b nb

xi a

b nb

xi a

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 14 / 34

SLIDE 54

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Example: Possible credible set

.

Problem

. . Let X1, · · · , Xn

i.i.d.

∼ Poisson(λ) and assume that λ ∼ Gamma(a, b). Find

a 90% credible set for λ. .

Solution

. . The posterior pdf of λ becomes π(λ|x) = Gamma ( a + ∑ xi, [n + (1/b)]−1) If we simply split the α equally between the upper and lower endpoints, 2(nb + 1) b λ ∼ χ2

2(a+∑ xi)

(if a is an integer) Therefore, a 1 − α confidence interval is { λ : b 2(nb + 1)χ2

2(∑ xi+a),1−α/2 ≤ λ ≤

b 2(nb + 1)χ2

2(∑ xi+a),α/2

}

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 14 / 34

SLIDE 55

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Example: Possible credible set

.

Problem

. . Let X1, · · · , Xn

i.i.d.

∼ Poisson(λ) and assume that λ ∼ Gamma(a, b). Find

a 90% credible set for λ. .

Solution

. . The posterior pdf of λ becomes π(λ|x) = Gamma ( a + ∑ xi, [n + (1/b)]−1) If we simply split the α equally between the upper and lower endpoints, 2(nb + 1) b λ ∼ χ2

2(a+∑ xi)

(if a is an integer) Therefore, a 1 − α confidence interval is { λ : b 2(nb + 1)χ2

2(∑ xi+a),1−α/2 ≤ λ ≤

b 2(nb + 1)χ2

2(∑ xi+a),α/2

}

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 14 / 34

SLIDE 56

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Remark: Credible probability and coverage probability

It is important not to confuse credible probability with coverage

probability

Credible probabilities are the Bayes posterior probability, which

reflects the experimenter’s subjective beliefs, as expressed in the prior distribution.

A Bayesian assertion of 90% coverage means that the experimenter,

upon combining prior knowledge with data, is % sure of coverage

Coverage probability reflects the uncertainty in the sampling

procedure, getting its probability from the objective mechanism of repeated experimental trials.

A classical assertion of 90% coverage means that in a long sequence of

identical trials, 90% of the realized confidence sets will cover the true parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 15 / 34

SLIDE 57

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Remark: Credible probability and coverage probability

It is important not to confuse credible probability with coverage

probability

Credible probabilities are the Bayes posterior probability, which

reflects the experimenter’s subjective beliefs, as expressed in the prior distribution.

A Bayesian assertion of 90% coverage means that the experimenter,

upon combining prior knowledge with data, is % sure of coverage

Coverage probability reflects the uncertainty in the sampling

procedure, getting its probability from the objective mechanism of repeated experimental trials.

A classical assertion of 90% coverage means that in a long sequence of

identical trials, 90% of the realized confidence sets will cover the true parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 15 / 34

SLIDE 58

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Remark: Credible probability and coverage probability

It is important not to confuse credible probability with coverage

probability

Credible probabilities are the Bayes posterior probability, which

reflects the experimenter’s subjective beliefs, as expressed in the prior distribution.

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

procedure, getting its probability from the objective mechanism of repeated experimental trials.

A classical assertion of 90% coverage means that in a long sequence of

identical trials, 90% of the realized confidence sets will cover the true parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 15 / 34

SLIDE 59

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Remark: Credible probability and coverage probability

It is important not to confuse credible probability with coverage

probability

Credible probabilities are the Bayes posterior probability, which

reflects the experimenter’s subjective beliefs, as expressed in the prior distribution.

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

procedure, getting its probability from the objective mechanism of repeated experimental trials.

A classical assertion of 90% coverage means that in a long sequence of

identical trials, 90% of the realized confidence sets will cover the true parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 15 / 34

SLIDE 60

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Remark: Credible probability and coverage probability

It is important not to confuse credible probability with coverage

probability

Credible probabilities are the Bayes posterior probability, which

reflects the experimenter’s subjective beliefs, as expressed in the prior distribution.

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

procedure, getting its probability from the objective mechanism of repeated experimental trials.

A classical assertion of 90% coverage means that in a long sequence of

identical trials, 90% of the realized confidence sets will cover the true parameter.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 15 / 34

SLIDE 61

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 1 (from last lecture)

.

Problem

. . . . . . . . Suppose X Xn are iid samples from f x exp x . Suppose the prior distribution of is e where are known. (a) Derive the posterior distribution of . (b) If we use the loss function L a a , what is the Bayes rule estimator for ?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 16 / 34

SLIDE 62

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 1 (from last lecture)

.

Problem

. . Suppose X1, · · · , Xn are iid samples from f(x|θ) = θ exp(−θx). Suppose the prior distribution of θ is π(θ) = 1 Γ(α)βα θα−1e−θ/β where α, β are known. (a) Derive the posterior distribution of θ. (b) If we use the loss function L a a , what is the Bayes rule estimator for ?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 16 / 34

SLIDE 63

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 1 (from last lecture)

.

Problem

. . Suppose X1, · · · , Xn are iid samples from f(x|θ) = θ exp(−θx). Suppose the prior distribution of θ is π(θ) = 1 Γ(α)βα θα−1e−θ/β 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?

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 16 / 34

SLIDE 64

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(a) Posterior distribution of θ

f(x, θ) = π(θ)f(x|θ)π(θ) = 1 Γ(α)βα θα−1e−θ/β

n

∏

i=1

[θ exp (−θxi)] e

n exp n i

xi

n

exp

n i

xi Gamma n

n i

xi x Gamma n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 17 / 34

SLIDE 65

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(a) Posterior distribution of θ

f(x, θ) = π(θ)f(x|θ)π(θ) = 1 Γ(α)βα θα−1e−θ/β

n

∏

i=1

[θ exp (−θxi)] = 1 Γ(α)βα θα−1e−θ/βθn exp ( −θ

n

∑

i=1

xi )

n

exp

n i

xi Gamma n

n i

xi x Gamma n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 17 / 34

SLIDE 66

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(a) Posterior distribution of θ

f(x, θ) = π(θ)f(x|θ)π(θ) = 1 Γ(α)βα θα−1e−θ/β

n

∏

i=1

[θ exp (−θxi)] = 1 Γ(α)βα θα−1e−θ/βθn exp ( −θ

n

∑

i=1

xi ) = 1 Γ(α)βα θα+n−1 exp [ −θ ( 1/β +

n

∑

i=1

xi )] Gamma n

n i

xi x Gamma n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 17 / 34

SLIDE 67

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(a) Posterior distribution of θ

f(x, θ) = π(θ)f(x|θ)π(θ) = 1 Γ(α)βα θα−1e−θ/β

n

∏

i=1

[θ exp (−θxi)] = 1 Γ(α)βα θα−1e−θ/βθn exp ( −θ

n

∑

i=1

xi ) = 1 Γ(α)βα θα+n−1 exp [ −θ ( 1/β +

n

∑

i=1

xi )] ∝ Gamma ( α + n − 1, 1 β−1 + ∑n

i=1 xi

) x Gamma n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 17 / 34

SLIDE 68

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(a) Posterior distribution of θ

f(x, θ) = π(θ)f(x|θ)π(θ) = 1 Γ(α)βα θα−1e−θ/β

n

∏

i=1

[θ exp (−θxi)] = 1 Γ(α)βα θα−1e−θ/βθn exp ( −θ

n

∑

i=1

xi ) = 1 Γ(α)βα θα+n−1 exp [ −θ ( 1/β +

n

∑

i=1

xi )] ∝ Gamma ( α + n − 1, 1 β−1 + ∑n

i=1 xi

) π(θ|x) = Gamma ( α + n − 1, 1 β−1 + ∑n

i=1 xi

)

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 17 / 34

SLIDE 69

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

(b) Bayes’ rule estimator with squared error loss

Bayes’ rule estimator with squared error loss is posterior mean. Note that the mean of Gamma(α, β) is αβ. x Gamma n

n i

xi E x E x n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 18 / 34

(a) Derive the asymptotic size LRT for testing H vs. H . (b) Derive the UMP level test for H

vs. H

, where . (c) Derive the UMP level test for H

vs. H

.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 19 / 34

SLIDE 73

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples from Gamma distribution with parameter (3, θ), which has the pdf f(x|θ) = 1 2θ3 x2e−x/θ (x > 0) You may use the result that 2 ∑n

i=1 Xi/θ ∼ χ2 6n.

(a) Derive the asymptotic size α LRT for testing H0 : θ = θ0 vs. H1 : θ ̸= θ0. (b) Derive the UMP level test for H

vs. H

, where . (c) Derive the UMP level test for H

vs. H

.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 19 / 34

SLIDE 74

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples from Gamma distribution with parameter (3, θ), which has the pdf f(x|θ) = 1 2θ3 x2e−x/θ (x > 0) You may use the result that 2 ∑n

i=1 Xi/θ ∼ χ2 6n.

(a) Derive the asymptotic size α LRT for testing H0 : θ = θ0 vs. H1 : θ ̸= θ0. (b) Derive the UMP level α test for H0 : θ = θ0 vs. H1 : θ = θ1, where θ1 > θ0. (c) Derive the UMP level test for H

vs. H

.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 19 / 34

SLIDE 75

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Practice Problem 2

.

Problem

. . Suppose X1, · · · , Xn are iid random samples from Gamma distribution with parameter (3, θ), which has the pdf f(x|θ) = 1 2θ3 x2e−x/θ (x > 0) You may use the result that 2 ∑n

i=1 Xi/θ ∼ χ2 6n.

(a) Derive the asymptotic size α LRT for testing H0 : θ = θ0 vs. H1 : θ ̸= θ0. (b) Derive the UMP level α test for H0 : θ = θ0 vs. H1 : θ = θ1, where θ1 > θ0. (c) Derive the UMP level α test for H0 : θ ≤ θ0 vs. H1 : θ > θ0.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 19 / 34

SLIDE 76

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

L(θ|x) =

n i

xi e

xi

l x

n i

log log log xi xi n log n log

n i

log xi

n i

xi l x n

n i

xi n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 20 / 34

SLIDE 77

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

L(θ|x) =

n

∏

i=1

[ 1 2θ3 x2

i e−xi/θ

] l(θ|x) =

n

∑

i=1

[ − log 2 − 3 log θ + 2 log xi − xi θ ] n log n log

n i

log xi

n i

xi l x n

n i

xi n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 20 / 34

SLIDE 78

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

L(θ|x) =

n

∏

i=1

[ 1 2θ3 x2

i e−xi/θ

] l(θ|x) =

n

∑

i=1

[ − log 2 − 3 log θ + 2 log xi − xi θ ] = −n log 2 − 3n log θ + 2

n

∑

i=1

log xi − 1 θ

n

∑

i=1

xi l x n

n i

xi n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 20 / 34

SLIDE 79

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

L(θ|x) =

n

∏

i=1

[ 1 2θ3 x2

i e−xi/θ

] l(θ|x) =

n

∑

i=1

[ − log 2 − 3 log θ + 2 log xi − xi θ ] = −n log 2 − 3n log θ + 2

n

∑

i=1

log xi − 1 θ

n

∑

i=1

xi l′(θ|x) = −3n θ + 1 θ2

n

∑

i=1

xi = 0 n

n i

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 20 / 34

SLIDE 80

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

L(θ|x) =

n

∏

i=1

[ 1 2θ3 x2

i e−xi/θ

] l(θ|x) =

n

∑

i=1

[ − log 2 − 3 log θ + 2 log xi − xi θ ] = −n log 2 − 3n log θ + 2

n

∑

i=1

log xi − 1 θ

n

∑

i=1

xi l′(θ|x) = −3n θ + 1 θ2

n

∑

i=1

xi = 0 ˆ θ = 1 3n

n

∑

i=1

xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 20 / 34

SLIDE 81

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

l′′(θ|x)

θ=ˆ

θ

= 3n θ2 − 2 θ3

n

∑

i=1

xi

θ=ˆ

θ

n n Because L x as approaches zero or infinity,

n n i

xi.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 21 / 34

SLIDE 82

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

l′′(θ|x)

θ=ˆ

θ

= 3n θ2 − 2 θ3

n

∑

i=1

xi

θ=ˆ

θ

= 3n ˆ θ2 − 6n ˆ θ2 < 0 Because L x as approaches zero or infinity,

n n i

xi.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 21 / 34

SLIDE 83

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Obtaining MLEs

l′′(θ|x)

θ=ˆ

θ

= 3n θ2 − 2 θ3

n

∑

i=1

xi

θ=ˆ

θ

= 3n ˆ θ2 − 6n ˆ θ2 < 0 Because L(θ|x) → 0 as θ approaches zero or infinity, ˆ θ =

1 3n

∑n

i=1 xi.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 21 / 34

SLIDE 84

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is log x l x l x n log xi n log xi n log xi n log n xi n R x xi n log xi n log n x xi n log xi n log n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 85

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is −2 log λ(x) = −2 [ l(θ0|x) − l(ˆ θ|x) ] n log xi n log xi n log xi n log n xi n R x xi n log xi n log n x xi n log xi n log n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 86

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is −2 log λ(x) = −2 [ l(θ0|x) − l(ˆ θ|x) ] = 6n log θ0 + 2 θ0 ∑ xi − 6n log ˆ θ − 2 ˆ θ ∑ xi n log xi n log n xi n R x xi n log xi n log n x xi n log xi n log n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 87

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is −2 log λ(x) = −2 [ l(θ0|x) − l(ˆ θ|x) ] = 6n log θ0 + 2 θ0 ∑ xi − 6n log ˆ θ − 2 ˆ θ ∑ xi = 6n log θ0 + 2 θ0 ∑ xi − 6n log ( 1 3n ∑ xi ) − 6n > χ2

1,α

R x xi n log xi n log n x xi n log xi n log n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 88

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is −2 log λ(x) = −2 [ l(θ0|x) − l(ˆ θ|x) ] = 6n log θ0 + 2 θ0 ∑ xi − 6n log ˆ θ − 2 ˆ θ ∑ xi = 6n log θ0 + 2 θ0 ∑ xi − 6n log ( 1 3n ∑ xi ) − 6n > χ2

1,α

R = { x : 2 θ0 ∑ xi − 6n log ∑ xi > χ2

1,α + 6n[1 − log(3nθ0)]

} x xi n log xi n log n

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 89

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (a) - Constructing asymptotic size α LRT

The rejection region of asymptotic size α LRT is −2 log λ(x) = −2 [ l(θ0|x) − l(ˆ θ|x) ] = 6n log θ0 + 2 θ0 ∑ xi − 6n log ˆ θ − 2 ˆ θ ∑ xi = 6n log θ0 + 2 θ0 ∑ xi − 6n log ( 1 3n ∑ xi ) − 6n > χ2

1,α

R = { x : 2 θ0 ∑ xi − 6n log ∑ xi > χ2

1,α + 6n[1 − log(3nθ0)]

} = { x : ∑ xi − 3nθ0 log ∑ xi > θ0 2 χ2

1,α + 3nθ0[1 − log(3nθ0)]

}

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 22 / 34

SLIDE 90

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (b) - UMP level α test for simple hypothesis

For H0 : θ = θ0 vs. H1 : θ = θ1, L x L x

n n exp

xi

n n exp

xi

n

exp xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 23 / 34

SLIDE 91

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (b) - UMP level α test for simple hypothesis

For H0 : θ = θ0 vs. H1 : θ = θ1, L(θ1|x) L(θ0|x) =

1 2nθ3n

1 exp

[ −

∑ xi θ1

] ∏ x2

i 1 2nθ3n

0 exp

[ −

∑ xi θ0

] ∏ x2

i n

exp xi

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 23 / 34

SLIDE 92

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (b) - UMP level α test for simple hypothesis

For H0 : θ = θ0 vs. H1 : θ = θ1, L(θ1|x) L(θ0|x) =

1 2nθ3n

1 exp

[ −

∑ xi θ1

] ∏ x2

i 1 2nθ3n

0 exp

[ −

∑ xi θ0

] ∏ x2

i

= (θ0 θ1 )3n exp [θ1 − θ0 θ0θ1 ∑ xi ]

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 23 / 34

SLIDE 93

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (b) - UMP level α test (cont’d)

Let T = ∑ Xi. Then under H0,

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (c) - UMP level α test for composite hypothesis

We need to check whether T has MLR. Because Y = 2T/θ ∼ χ2

6n.

fY y

n

n y n e

y

fT t

n

n t

n

e

t

n t

n

e

t

For arbitrary , fT t fT t

n t n

e

t n t n

e

t n

exp t is an increasing function of t. This T has MLR property.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 25 / 34

SLIDE 97

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (c) - UMP level α test for composite hypothesis

We need to check whether T has MLR. Because Y = 2T/θ ∼ χ2

6n.

fY(y|θ) = 1 23nΓ(3n)y3n−1e−y/2 fT(t|θ) = 1 23n−1Γ(3n)θ (2t θ )3n−1 e−t/θ = 1 Γ(3n)θ ( t θ )3n−1 e−t/θ For arbitrary , fT t fT t

n t n

e

t n t n

e

t n

exp t is an increasing function of t. This T has MLR property.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 25 / 34

SLIDE 98

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (c) - UMP level α test for composite hypothesis

We need to check whether T has MLR. Because Y = 2T/θ ∼ χ2

6n.

fY(y|θ) = 1 23nΓ(3n)y3n−1e−y/2 fT(t|θ) = 1 23n−1Γ(3n)θ (2t θ )3n−1 e−t/θ = 1 Γ(3n)θ ( t θ )3n−1 e−t/θ For arbitrary θ1 < θ2, fT(t|θ2) fT(t|θ1) =

1 Γ(3n)θ2

(

t θ2

)3n−1 e−t/θ2

1 Γ(3n)θ1

(

t θ1

)3n−1 e−t/θ1 = (θ1 θ2 )3n exp [θ2 − θ1 θ1θ2 t ] is an increasing function of t. This T has MLR property.

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 25 / 34

SLIDE 99

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

Solution (c) - UMP level α test for composite hypothesis

We need to check whether T has MLR. Because Y = 2T/θ ∼ χ2

6n.

fY(y|θ) = 1 23nΓ(3n)y3n−1e−y/2 fT(t|θ) = 1 23n−1Γ(3n)θ (2t θ )3n−1 e−t/θ = 1 Γ(3n)θ ( t θ )3n−1 e−t/θ For arbitrary θ1 < θ2, fT(t|θ2) fT(t|θ1) =

1 Γ(3n)θ2

(

t θ2

)3n−1 e−t/θ2

1 Γ(3n)θ1

(

t θ1

)3n−1 e−t/θ1 = (θ1 θ2 )3n exp [θ2 − θ1 θ1θ2 t ] is an increasing function of t. This T has MLR property.

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Solution (c) - Constructing UMP level α test

Because T has MLR property, UMP level α test for H0 : θ ≤ θ0 vs. H1 : θ > θ0 has a rejection region T > k, and Pr(T > k) = α. Therefore, the UMP level test is identical to the answer of part (b), whose rejection is R x T x xi

n

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SLIDE 101

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Solution (c) - Constructing UMP level α test

Because T has MLR property, UMP level α test for H0 : θ ≤ θ0 vs. H1 : θ > θ0 has a rejection region T > k, and Pr(T > k) = α. Therefore, the UMP level α test is identical to the answer of part (b), whose rejection is R x T x xi

n

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SLIDE 102

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Solution (c) - Constructing UMP level α test

Because T has MLR property, UMP level α test for H0 : θ ≤ θ0 vs. H1 : θ > θ0 has a rejection region T > k, and Pr(T > k) = α. Therefore, the UMP level α test is identical to the answer of part (b), whose rejection is R = { x : T(x) = ∑ xi > θ0 2 χ2

6n,α

}

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SLIDE 103

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Practice Problem 3

.

Problem

. . Let (X1, Y1), · · · , (Xn, Yn) be a random samples from a bivariate normal ( Xi Yi ) ∼ N ([ µX µY ] , [ σ2

X

ρσXσY ρσXσY σ2

Y

]) We are interested in testing H0 : µX = µY vs. H1 : µX ̸= µY. (a) Show that the random variables Wi Xi Yi are iid

W W .

(b) Show that the above hypothesis can be tested with the statistic TW W SW n where W

n n i

Wi and SW

n n i

Wi W . Furthermore, show that, under H , TW follows the Student’s t distribution with n degrees of freedom.

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Practice Problem 3

.

Problem

. . Let (X1, Y1), · · · , (Xn, Yn) be a random samples from a bivariate normal ( Xi Yi ) ∼ N ([ µX µY ] , [ σ2

X

ρσXσY ρσXσY σ2

Y

]) We are interested in testing H0 : µX = µY vs. H1 : µX ̸= µY. (a) Show that the random variables Wi = Xi − Yi are iid N(µW, σ2

W).

(b) Show that the above hypothesis can be tested with the statistic TW W SW n where W

n n i

Wi and SW

n n i

Wi W . Furthermore, show that, under H , TW follows the Student’s t distribution with n degrees of freedom.

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Practice Problem 3

.

Problem

. . Let (X1, Y1), · · · , (Xn, Yn) be a random samples from a bivariate normal ( Xi Yi ) ∼ N ([ µX µY ] , [ σ2

X

ρσXσY ρσXσY σ2

Y

]) We are interested in testing H0 : µX = µY vs. H1 : µX ̸= µY. (a) Show that the random variables Wi = Xi − Yi are iid N(µW, σ2

W).

(b) Show that the above hypothesis can be tested with the statistic TW W SW n where W

n n i

Wi and SW

n n i

Wi W . Furthermore, show that, under H , TW follows the Student’s t distribution with n degrees of freedom.

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Practice Problem 3

.

Problem

. . Let (X1, Y1), · · · , (Xn, Yn) be a random samples from a bivariate normal ( Xi Yi ) ∼ N ([ µX µY ] , [ σ2

X

ρσXσY ρσXσY σ2

Y

]) We are interested in testing H0 : µX = µY vs. H1 : µX ̸= µY. (a) Show that the random variables Wi = Xi − Yi are iid N(µW, σ2

W).

(b) Show that the above hypothesis can be tested with the statistic TW = W √ S2

W/n

where W

n n i

Wi and SW

n n i

Wi W . Furthermore, show that, under H , TW follows the Student’s t distribution with n degrees of freedom.

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Practice Problem 3

.

Problem

. . Let (X1, Y1), · · · , (Xn, Yn) be a random samples from a bivariate normal ( Xi Yi ) ∼ N ([ µX µY ] , [ σ2

X

ρσXσY ρσXσY σ2

Y

]) We are interested in testing H0 : µX = µY vs. H1 : µX ̸= µY. (a) Show that the random variables Wi = Xi − Yi are iid N(µW, σ2

W).

(b) Show that the above hypothesis can be tested with the statistic TW = W √ S2

W/n

where W = 1

n

∑n

i=1 Wi and S2 W = 1 n−1

∑n

i=1(Wi − W)2. Furthermore,

show that, under H0, TW follows the Student’s t distribution with n − 1 degrees of freedom.

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z Xi Yi T, m

X Y T, and A

. Then AZ Xi Yi Wi Am A AT

X Y X X Y Y W W

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z = [Xi Yi]T, m = [µX µY]T, and A = [1 − 1]. Then AZ Xi Yi Wi Am A AT

X Y X X Y Y W W

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z = [Xi Yi]T, m = [µX µY]T, and A = [1 − 1]. Then AZ = Xi − Yi = Wi Am A AT

X Y X X Y Y W W

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z = [Xi Yi]T, m = [µX µY]T, and A = [1 − 1]. Then AZ = Xi − Yi = Wi ∼ N(Am, AΣAT)

X Y X X Y Y W W

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z = [Xi Yi]T, m = [µX µY]T, and A = [1 − 1]. Then AZ = Xi − Yi = Wi ∼ N(Am, AΣAT) = N(µX − µY, σ2

X − 2ρσXσY + σ2 Y) W W

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Solution (a)

To solve Problem (a), we first need to know that, if Z ∼ N(m, Σ), then AZ ∼ N(Am, AΣAT) Let Z = [Xi Yi]T, m = [µX µY]T, and A = [1 − 1]. Then AZ = Xi − Yi = Wi ∼ N(Am, AΣAT) = N(µX − µY, σ2

X − 2ρσXσY + σ2 Y)

= N(µW, σ2

W)

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Solution (b)

Because µW = µX − µY, testing H0 : µX = µY vs. H1 : µX ̸= µY is equivalent to testing H0 : µW = 0 vs. H1 : µW ̸= 0 When Ui and both mean and variance are unknown, we know that LRT testing H

vs. H

follows that TU U SU n and TU follows Tn under H .

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Solution (b)

Because µW = µX − µY, testing H0 : µX = µY vs. H1 : µX ̸= µY is equivalent to testing H0 : µW = 0 vs. H1 : µW ̸= 0 When Ui ∼ N(µ, σ2) and both mean and variance are unknown, we know that LRT testing H0 : µ = µ0 vs. H0 : µ ̸= µ0 follows that TU U SU n and TU follows Tn under H .

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Solution (b)

Because µW = µX − µY, testing H0 : µX = µY vs. H1 : µX ̸= µY is equivalent to testing H0 : µW = 0 vs. H1 : µW ̸= 0 When Ui ∼ N(µ, σ2) and both mean and variance are unknown, we know that LRT testing H0 : µ = µ0 vs. H0 : µ ̸= µ0 follows that TU = U − µ0 √ S2

U/n

and TU follows Tn−1 under H0.

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Solution (b) (cont’d)

Therefore, the LRT test for the original test, H0 : µW = 0 vs. H1 : µW ̸= 0 is TW = W √ S2

W/n

and TW follows Tn−1 under H0.

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Practice Problem 4

.

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 .

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Practice Problem 4

.

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 .

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Practice Problem 4

.

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 .

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Practice Problem 4

.

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.

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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.

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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.

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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.

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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.

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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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SLIDE 129

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. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

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

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 33 / 34

SLIDE 130

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

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

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 33 / 34

SLIDE 131

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

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

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 33 / 34

SLIDE 132

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

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(x|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 )

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 33 / 34

SLIDE 133

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4

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).

Hyun Min Kang Biostatistics 602 - Lecture 25 April 18th, 2013 34 / 34

SLIDE 134

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

. . . . Recap . . . . . Bayesian Tests . . . . . Bayesian Intervals . . . P1 . . . . . . . . P2 . . . . P3 . . . . P4