S TAT 6 32-600
October5 , 2 011
Exam f l
o
Instructions: W rite a ll y our a nswers n t he t est p aper p rovided.C ircle w hat y ou t hink i s
q
o
o n m ultiple c hoiceq uestions,a nd b e s peci'fic n s hort-answer uest'ions.
the c orrecta nswer
G
Point
Sketch of Solutions for Homework #4
Exer. 2, p. 113
The inverse of the information matrix is
ni x2 ai
4
i
i=1 (1+ai )2
4
ni xi ai
i=1 (1+ai )2
4
ni xi ai
i=1 (1+ai )2
4
ni ai
i=1 (1+ai )2
1
,
B
where
4
B=
i=1
and
ni ai
(1 + ai )2
4
i=1
ni x2 ai
i
(1 + ai
Sketch of Solutions for Homework #5
Exer. 1, p. 440
One term of the likelihood is
1
L(yi |i , ) L(yi |i , ) L(yi |i , ) + (i i )L (yi |i , ) + (i i )2 L (yi |i , ),
2
where the derivatives are partial derivatives with respect to i . The last expression ha
Sketch of Solutions for Homework #6
Problem 1
Below are plots of sample autocorrelation functions obtained after generating 10,000 values
from each of the two poposal distributions. There is very little correlation among the observations generated using t
Sketch of Solutions for Homework #7
Exercise 1, p. 152
20
10
10
0
theta1
30
40
50
(a) I used MCMC to generate values from the posterior and thereby estimate the quantities of
interest. The only reason I didnt generate directly from the posterior is becaus
STAT 6 32
October 1 1, 2 010
Midterm
E xam f l-
Instructions: W rite a ll y our a nswerso n t he t est p aper p rovided. C ircle w hat y ou t hink i s
o
the c orrect a nswer o n m ultiple c hoice q uestions, a nd b e s pecr,fic n s hort-answer q uestzons.
S TAT 6 32
November 1 9' 2 o1o
M idt
e rrn , ,xarn f t2
Instructions: W rite a ll y our a nswerso n t he t est p aper p rovided. C ircle w hat y ou t hink
is t he c orrect a nswer o n m ultiple c hoice q uestions, b e a s s pecific a s p ossible o n s hor
STAT 632
October 11, 2010
Midterm Exam #1
Instructions: Write all your answers on the test paper provided. Circle what you think is
the correct answer on multiple choice questions, and be specic on short-answer questions.
Point values are indicated in par
STAT 632
November 19, 2010
Midterm Exam #2
Instructions: Write all your answers on the test paper provided. Circle what you think
is the correct answer on multiple choice questions, be as specic as possible on short answer questions, and show your work on
STAT 8640
BAYESIAN STATISTICS
SOLUTION for HW9
1
1.
(a) posterior mean and variance
We can write the model as
i < 0 + xi1 1 + xi2 2 + xi3 3
yi N (i , 2 )
0 N (20, 10000)
j N (1, 100), f or j = 1, 2, 3
2 Inv Gamma(0.5, 0.5)
By using DoodleBugs, we have th
STAT 8640
BAYESIAN STATISTICS
SOLUTION for HW1
1.3
(a)
Let
A = cfw_brown-eyed parents
B = cfw_brown-eyed children
C = cfw_children are heterozygotes
A1 = cfw_parents are (Xx, Xx)
A2 = cfw_parents are (Xx, XX)
A3 = cfw_parents are (XX, XX)
So,
P (AB) = P (
STAT 8640
BAYESIAN STATISTICS
SOLUTION for HW4
1
35 points in total
3.1
(a)(5 points) Marginal posterior distribution of .
Label the prior distribution p() as Dirichlet(a1 , . . . , an ), so p(1 , . . . , J ) 1a1 1 . . . JaJ 1 . The
likelihood is p(y1 , .
STAT 8640
10.1
BAYESIAN STATISTICS
SOLUTION for HW7
1
Number of simulation draws
a) Approximate standard deviation
2
1 |y N (8, 42 ), according to the large sample Central Limiting Theorem, 1 |y N (8, 4n ), and
q
42
= 0.4
sd(1 |y) = 100
b)
p
42 /n = 0.1,
STAT 8640
BAYESIAN STATISTICS
SOLUTION for HW3
1
y
2.12 p(y | ) = e y!
log p(y | ) = + y log log y!
log p(y | ) = y
Then,
2
J() = E
log p(y | )
y 2
1
= E
= 2 V ar(y | )
1
1
=
=
2
So the Jeffreys prior is
p
J() =
1
corresponding to the Gamma(1/2,0).
2.
Sketch of Solutions for Homework #3
Exer. 1, p. 95
(a) First show that the joint posterior of 1 and 2 is Dirichlet with parameters y1 + 1 , y2 + 2
and n y1 y2 + 3 + + J , where 1 , . . . , J are the parameters of the prior. Now make
the change of variable
Sketch of Solutions for Homework #2
Exer. 16, p. 70
(a) The marginal distribution of Y is obtained by determining
1
m(y ) =
f (y |) () d.
0
(b) Let n > 1. Use proof by contradiction. Suppose that at least one of and is not 1.
There are three ways this can
Homework #1: Answers for Selected Exercises
Exer. 1, p. 67
In this case our data consist of the knowledge that Y < 3, where Y is the number of
heads in 10 tosses. The posterior is thus (|Y < 3). Use Bayes theorem to nd this
density.
Exer. 5, p. 67
(a) The
S TAT 6 32
November , 2 011
9
Midterm E lxarn f f2
o
Instructions: W rite a ll y our a nswers n t he t est p aper p rovided. C ircle w hat y ou t hink
is the correct answeron multiple choicequestions,be as specificas possibleon short answer questions,and
STAT 632-600
December 9, 2008
Final Exam
Instructions: Answer each question on the test paper provided. Circle the correct answer
on multiple choice questions. Point values are indicated in parentheses. Have a happy
holiday!
1. (12 points) Let Y be an obs
Homework #1: Answers for Selected Exercises
Exer. 1, p. 67
In this case our data consist of the knowledge that Y < 3, where Y is the number of
heads in 10 tosses. The posterior is thus (|Y < 3). Use Bayes theorem to nd this
density.
Exer. 5, p. 67
(a) The
Sketch of Solutions for Homework #2
Exer. 16, p. 70
(a) The marginal distribution of Y is obtained by determining
1
m(y ) =
f (y |) () d.
0
(b) Let n > 1. Use proof by contradiction. Suppose that at least one of and is not 1.
There are three ways this can
Sketch of Solutions for Homework #3
Exer. 1, p. 95
(a) First show that the joint posterior of 1 and 2 is Dirichlet with parameters y1 + 1 , y2 + 2
and n y1 y2 + 3 + + J , where 1 , . . . , J are the parameters of the prior. Now make
the change of variable
Sketch of Solutions for Homework #4
Exer. 2, p. 113
The inverse of the information matrix is
ni x2 ai
4
i
i=1 (1+ai )2
4
ni xi ai
i=1 (1+ai )2
4
ni xi ai
i=1 (1+ai )2
4
ni ai
i=1 (1+ai )2
1
,
B
where
4
B=
i=1
and
ni ai
(1 + ai )2
4
i=1
ni x2 ai
i
(1 + ai
Sketch of Solutions for Homework #5
Exer. 1, p. 440
One term of the likelihood is
1
L(yi |i , ) L(yi |i , ) L(yi |i , ) + (i i )L (yi |i , ) + (i i )2 L (yi |i , ),
2
where the derivatives are partial derivatives with respect to i . The last expression ha
Sketch of Solutions for Homework #6
Problem 1
Below are plots of sample autocorrelation functions obtained after generating 10,000 values
from each of the two poposal distributions. There is very little correlation among the observations generated using t
Sketch of Solutions for Homework #7
Exercise 1, p. 152
20
10
10
0
theta1
30
40
50
(a) I used MCMC to generate values from the posterior and thereby estimate the quantities of
interest. The only reason I didnt generate directly from the posterior is becaus
STAT 6 32
October 1 1, 2 010
Midterm
E xam f l-
Instructions: W rite a ll y our a nswerso n t he t est p aper p rovided. C ircle w hat y ou t hink i s
o
the c orrect a nswer o n m ultiple c hoice q uestions, a nd b e s pecr,fic n s hort-answer q uestzons.