Stat 351 Fall 2006
Assignment #9 Solutions
(a) The probability that fewer than two calls come in the rst hour is P (T2 > 1). However,
cfw_T2 > 1 = cfw_X1 < 2 so it is equivalent to calculate either P (T2 > 1) or P (X1 < 2). Since
X1 Po(4), we nd
40 4 41 4
Stat 351 Fall 2006
Solutions to Assignment #5
Problem #13, page 28: Suppose that Y 2 (n) so that the density of Y is given by
fY (y) =
Let U =
1
Y
1
y n/21 2n/2 ey/2 , 0 < y < .
( n )
2
. Since y > 0, the distribution function of U is given by
1
u
Y
FU
Stat 351 Fall 2006
Assignment #6 Solutions
Problem #3, page 115: If 0 y 1/2, then
1y
fY (y) =
y
1y
2 dz = 2(1 2y).
fX(1) ,X(2) (y, z) dz =
y
On the other hand, if 1/2 y 1, then
y
fY (y) =
1y
y
fX(1) ,X(2) (z, 1 y) dz =
2 dz = 2(2y 1).
1y
Problem #6, page
Stat 351 Fall 2006
Assignment #3 Solutions
Problem #5, page 27: Suppose that X C(0, 1) so that the density of X is given by
fX (x) =
1
1
, < x < .
1 + x2
Let Y = X 2 . If y 0, then the distribution function of Y is given by
FY (y) = P (Y y) = P (X y) = P
Stat 351 Fall 2006
Assignment #5 Solutions
Problem #2, page 55: Suppose that X + Y = c. By denition of conditional density,
fX|X+Y =c (x) =
fX,X+Y (x, c)
.
fX+Y (c)
We now nd the joint density fX,X+Y (x, c). Let U = X and V = X + Y so that X = U and
Y = V
Stat 351 Fall 2006
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 ()
P ( B)
=
=0
P (B)
P (B)
since cfw_ B = . Similarly,
Q() = P (|B) =
P ( B)
P (B)
=
Stat 351 Fall 2006
Solutions to Selected Problems
Problem #11, page 56: Since
1
1
1
cx2 dy dx = c
0
x2 (1 x) dx = c
x
0
1
1 3 1 4
x x
3
4
=
0
we conclude that c = 12. The marginal for Y is therefore given by
y
12x2 dx = 4y 3 , 0 < y < 1
fY (y) =
0
and the
Stat 351 Fall 2006
Assignment #8 Solutions
Problem #9, page 144: Note that by Theorem 7.1, in order to show X1 , X2 , and X3 are
independent, it is enough to show that cov(X1 , X2 ) = cov(X1 , X3 ) = cov(X2 , X3 ) = 0. Thus, if X1
and X2 + X3 are independ
Stat 351 Fall 2007
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 ()
P ( B)
=
=0
P (B)
P (B)
since cfw_ B = . Similarly,
Q() = P (|B) =
P ( B)
P (B)
=
Stat 351 Fall 2007
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 2007
Assignment #6 Solutions
Problem #3, page 115: If 0 y 1/2, then
1y
fY (y) =
y
1y
2 dz = 2(1 2y).
fX(1) ,X(2) (y, z) dz =
y
On the other hand, if 1/2 y 1, then
y
fY (y) =
1y
y
fX(1) ,X(2) (z, 1 y) dz =
2 dz = 2(2y 1).
1y
Problem #6, page