Stat 351 Fall 2008
Assignment #2 Solutions
Problem 2: We verify that Q(A) is a probability by checking the three conditions.
Since P () = 0, we conclude
Q() = P (|B) =
P ( B)
P ()
=
=0
P (B)
P (B)
since cfw_ B = . Similarly,
Q() = P (|B) =
P (B)
P ( B)
=
Stat 351 Fall 2008
Assignment #3 Solutions
Problem #3, page 27: Suppose that T t(n) so that the density of T is given by
( n+1 )
x2
fT (x) = 2 n 1 +
n
n ( 2 )
(n+1)/2
, < x < .
Let Y = T 2 . If y 0, then the distribution function of Y is given by
FY (y) =
Stat 351 Fall 2008
Solutions to Assignment #4
Problem #3, page 55: Suppose that X + Y = 2. By denition of conditional density,
fX,X+Y (x, 2)
.
fX+Y (2)
fX|X+Y =2 (x) =
We now nd the joint density fX,X+Y (x, 2). Let U = X and V = X + Y so that X = U and
Y
Stat 351 Fall 2008
Assignment #1 Solutions
2. (a) If X Unif[1, 3], then FX (x) =
x1
2
for 1 x 3, and if Y N (0, 1), then
y
1
2
eu /2 du
2
FY (y) =
for < y < . Since X and Y are independent, we conclude that
FX,Y (x, y) = FX (x) FY (y) =
x1
2
y
1
2
eu /2
Stat 351 Fall 2008
Solutions to Assignment #5
1. (a) We begin by calculating E(Y1 ). That is,
E(Y1 ) = 1 P (Y = 1) + (1) P (Y = 1) = p (1 p) = 2p 1.
We now notice that Sn+1 = Sn + Yn+1 . Therefore,
E(Sn+1 |X1 , . . . , Xn ) = E(Sn + Yn+1 |X1 , . . . , Xn
Stat 351 Fall 2008
Assignment #9 Solutions
1. (a) By Denition I, we see that X1 X2 is normally distributed with mean
E(X1 X2 ) = E(X1 ) E(X2 ) = 0
and variance
var(X1 X2 ) = var(X1 ) + 2 var(X2 ) 2 cov(X1 , X2 ) = 1 + 2 22 = 1 2 .
That is, X1 X2 = Y where
Stat 351 Fall 2008
Assignment #8 Solutions
1. Recall from Stat 251 that if X N (0, 1), then X 2 2 (1). Furthermore, recall that if Z1 , . . . , Zn
are independent with Zj 2 (pj ), then
n
Zj 2 (p1 + + pn ).
j=1
(That is, the sum of independent chi-squared
Stat 351 Fall 2008
Assignment #6 Solutions
1. We nd
P (X < Y < Z) =
f (x, y, z)dx dy dz.
cfw_x<y<z
Since f (x, y, z) =
sion are
0
z
0
y
0
x
for 0 < x, y, z < , the six equivalent iterated integrals for this expres
exyz dx dy dz =
0
z
=
0
exyz
y
exyz dy dz
Stat 351 Fall 2007
Assignment #7 Solutions
1. If X1 , X2 are independent N (0, 1) random variables, then by Denition I, Y1 = X1 + 3X2 2 is
normal with mean E(Y1 ) = E(X1 ) + 3E(X2 ) 2 = 2 and variance var(Y1 ) = var(X1 + 3X2 2) =
var(X1 ) + 9 var(X2 ) + 6
Stat 351 Fall 2009
Assignment #1 Solutions
Problem 1:
(a) If X Unif[1, 3], then FX (x) =
x1
2
for 1 x 3, and if Y N (0, 1), then
y
FY (y) =
1
2
eu /2 du
2
for < y < . Since X and Y are independent, we conclude that
FX,Y (x, y) = FX (x) FY (y) =
x1
2
y
1
Stat 351 Fall 2009
Assignment #5 Solutions
1. If X1 , X2 are independent N (0, 1) random variables, then by Denition I, Y1 = X1 + 3X2 2 is
normal with mean E(Y1 ) = E(X1 ) + 3E(X2 ) 2 = 2 and variance var(Y1 ) = var(X1 + 3X2 2) =
var(X1 ) + 9 var(X2 ) + 6
Statistics 351 Fall 2006 Midterm #2 Solutions
1. (a) Recall that a square matrix is strictly positive denite if and only if the determinants
of all of its upper block diagonal matrices are strictly positive. Since
2 2
2 3
=
we see that det( 1 ) = det(2) =
Stat 351 Fall 2009
Assignment #6 Solutions
1. Recall from Stat 251 that if X N (0, 1), then X 2 2 (1). Furthermore, recall that if Z1 , . . . , Zn
are independent with Zj 2 (pj ), then
n
Zj 2 (p1 + + pn ).
j=1
(That is, the sum of independent chi-squared
Stat 351 Fall 2009
Chapter 2 Solutions
Problem #2. Suppose that X + Y = 2. By denition of conditional density,
fX,X+Y (x, 2)
.
fX+Y (2)
fX|X+Y =2 (x) =
We now nd the joint density fX,X+Y (x, 2). Let U = X and V = X + Y so that X = U and
Y = V U . The Jaco
Stat 351 Fall 2009
Chapter 5 Solutions
Problem #2. Let X = (X, Y ) with
0
,
0
XN
1
1
,
and consider the change of variables to polar coordinates (R, ) . The inverse of this transformation
is given by
x = r cos and y = r sin
for 0 < 2, r > 0 so that the
Stat 351 Fall 2009
Chapter 4 Solutions
Problem #3. If 0 y 1/2, then
1y
1y
2 dz = 2(1 2y).
fX(1) ,X(2) (y, z) dz =
fY (y) =
y
y
On the other hand, if 1/2 y 1, then
y
y
fY (y) =
1y
2 dz = 2(2y 1).
fX(1) ,X(2) (y, z) dz =
1y
Problem #5. Since E[F (X(n) ) F (
Stat 351 Fall 2009
Chapter 1 Solutions
Problem #2. Suppose that X C(m, a) so that the density of X is given by
fX (x) =
1
a
2
, < x < .
a + (x m)2
Let Y = 1/X. If X = 0, then for < y < the distribution function of Y is
1/y
FY (y) = P (Y y) = P (1/X y) =