10-705: Intermediate Statistics
Fall 2012
Test I Solutions
Lecturer: Larry Wasserman
TA: Wanjie Wang, Haijie Gu
Problem 1 Let X1 , X2 be iid U nif (0, 4) . Find the density of Y = X1 + X2 .
Solution
f (x1 , x2 ) = f (x1 )f (x2 ) =
1
16 ,
0,
0 x1 4 , 0 x2
36-705 Intermediate Statistics HW9
Problem 1
(a) Solution:
Note that p = X = n1
n
i=1
Xi .
Following Lecture Notes 13, X1 , ., Xn Bernoulli()
p
Then, np =
n
i=1
Xi Binomial(n, X ).
(b) Solution:
E( |X1 , ., Xn ) = n1 E[Y Binomial(n, X )] = n1 nX = X
p
Var
36-705 Intermediate Statistics HW5
Problem 1 (C& B 10.31)
Solution:
(a) By CLT,
n1
and
n2
p1 p1
p1 (1 p1 )
p2 p2
p2 (1 p2 )
d
N (0, 1),
d
N (0, 1).
So if p1 and p2 are independent, under H0 : p1 = p2 = p,
p1 p2
n1 +n2
p(1
n1 n2
d
N (0, 1).
p)
where we
705 Fall 2012
Test III Solutions
(1) Let X1 , ., Xn Bernoulli(p).
(a) Find the score function and the Fisher information function.
Solution: The likelihood function is
L(p; X n ) = p
n
i=1
so, l(p; X n ) = log L(p; X n ) =
So, the score function is
l
S (
36-705 Intermediate Statistics Test 2 Solution
(1) Let X1 , . . . , Xn Bernoulli() where 0 < < 1. Let Yi = e3Xi . Let
1
Wn =
n
n
Yi .
i=1
(a) Show that there is a number such that Wn converges in probability .
Solution: As Xi is either 0 or 1, Yi is eithe
36-705 Intermediate Statistics HW1
Problem 2.31
Since the mgf is dened as MX (t) = EetX , we necessarily have MX (0) = Ee0 = 1. But
t/(1 t) is 0 at t = 0, therefore it cannot be an mgf.
Problem 2.36
Problem 2.38
(a)
r+x1 r
p (1 p)x etx
x
MX (t) =
x=0
= pr
10-705: Intermediate Statistics
Fall 2012
Homework 8 Solutions
Lecturer: Larry Wasserman
TA: Wanjie Wang, Haijie Gu
Problem 1
Given x, say that x Bj , then the estimator is p(x) =
j
h
=
1
nh
n
i=1
I (Xi Bj ).
For any Xi , the distribution of I (Xi Bj ) is
10-705: Intermediate Statistics
Fall 2012
Homework 6 Solutions
Lecturer: Larry Wasserman
TA: Wanjie Wang, Haijie Gu
Problem 1
The mle for Binomial(k, p) is pmle = X n /k . By the asymptotic eciency of the mle, we
have:
n(mle p)
p
N (0, p(1 p)/k )
By delta
36-705 Intermediate Statistics HW5
Problem 1 (C& B 10.1)
First calculate some moments for this distribution.
E [X 2 ] = 1/3,
E [X ] = /3,
V ar(X ) =
So 3Xn is an unbiased estimator of with variance
V ar(3Xn ) = 9V ar(X )/n = (3 2 )/n 0
1 2
.
3
9
as n .
So
b. Using the normal approximation, we have v = r(1 p)/p = 20(.3)/.7 = 8.57 and
2
v = r(1 p)/p = (20)(.3)/.49 = 3.5.
Then,
P (Vn = 0) = 1 P (Vn 1) = 1 (
!"#$%&'()*+),(-)(.#/012(3)(.4"1(45(-6(.4"1($7P
Vn 8.57
18.57
3.5
3.5
10-705: Intermediate Statistics
An
Problem 1
Show that
Sn (A7052 )Fall(A1 ) + sn (A2 )
sn 2010
1A
Homework 3 Solutions
Solution:
10-705: Intermediate Statistics
Fall 2012
Since A1 A2 = cfw_A : A A1 or A A2 , therefore if A1 A2 picks out some subset G of
Problem 1
Homework 3 either A1
F ,
36-705 Intermediate Statistics HW2
Problem 1
The proof is similar as the proof for Theorem 6 on lecture 2.
For any u > 0, we have, from Morkovs inequality, that
1
P(
n
n
i=1
1
Xi
n
n
n
i t) = P (
i=1
(Xi i ) nt) = P (e
i=1
P
u n (Xi i )
i=1
= P (e
= eunt