Supplement to Chapter 1.2:Basic Results on Integration
September 18, 2009
Here we give the details of some of the proofs for some of the basic results on
integration. The development is somewhat out of order with the statements in the
notes, so we use a d
Solutions Homework 9
December 10, 2011
Solution to Exercise 3.2.1: Suppose that > 0, P [ X n X >
] 0. Then, given > 0, we can nd an N such that for all n N ,
P [ X n X > ] < . Simply take = to get the desired conclusion.
Conversely, suppose > 0, N such th
Solutions Homework 8
December 10, 2011
Solution to Exercise 2.4.4
(a) This is trivial by properties of the determinant: the determinant of
a product is the product of the determinant, and taking transpose doesnt
change the determinant. Thus, if V = AAt ,
1
Solution to Exercise 2.1.5 If x 1 and p < q , then xp xq . This
follows since the exponential function is monotone increasing, so xp = exp[p log x]
exp[q log x] = xq since p log x q log x when p < q and log x 0, i.e. x 1.
Thus
E [ X p]
=
E [ X p I[0,1]
Solutions Homework 6
November 17, 2011
Solution to Exercise 5.1: Lets try guring out E [X |Y = y ] using intuition
about conditional distributions. If we observe Y = 1/2, then we know 0 1/2
and nothing more. Using elementary conditional probability:
P [X
Solutions Homework 5
October 4, 2010
Solution to Exercise 3.17: Since the measure of a set is the integral of its
indicator, we have
(1 2 )(A)
=
1 2
=
1
IA (1 , 2 ) d(1 2 )(1 , 2 )
2
IA (1 , 2 ) d2(2 ) d1(1 ) ,
where of course Fubinis theorem was applied
Solutions Homework 4
September 29, 2010
Solution to Supplement Exercise SupEx 2: Assume 0 fn f .
Then f = limn fn = lim inf n fn . Thus, by Fatous Lemma, we have f lim inf n fn .
Now fn is an increasing sequence of extended real numbers by the monotonicit
Solutions Homework 2
September 20, 2010
Solution to Exercise 2.12 The denition given is that f is integrable i
both f are nite. If f is integrable (statement (i), then f = f+ f , and since
both terms in the latter expression are nite, we have that f is de
Solutions Homework 2
September 9, 2009
Solution to Exercise 2.1:
x f 1 (Ac ) f (x) Ac
f (x) A.
/
Now if x f 1 (A), then of course f (x) A, which cant happen if x f 1 (Ac ), i.e.,
by contraposition x f 1 (Ac ) implies x f 1 (A)c , which means f 1 (Ac ) f
Solutions Homework 1
September 7, 2009
Solution to Exercise 1.1: Lets see if we can use logical equivalence (if and
only if, symbolized by , and abbreviated i) to cut the steps in half.
c
x
A
x
/
AA
A
AA
it is not true that A A, x A
A A, x A
/
x
Ac .
AA
I
Final Exam Solutions
Statistics 581
December 22, 2011
1. [15 points] Suppose X is a random variable with Lebesgue density f (x)
and cumulative distribution function F (x). Assume a and b are real numbers with
a < b. Dene the random variable
Y=
a
if X a,
X
Solutions to Midterm Exam
Statistics 581
December 22, 2011
1. [35 points] On the real numbers I , consider the collections of subsets
R
C
=
cfw_A I : A is uncountable,
R
(1)
F
=
cfw_A I : A C or Ac C.
R
(2)
Recall that it was stated in class that I is unc