Chapter 5: Multivariate Probability Distributions
97
Instructors Solutions Manual
1
5.23
3
32
a. f 2 ( y 2 ) = 3 y1 dy1 = 2 2 y 2 , 0 y 2 1 .
y2
b. Defined over y2 y1 1, with the constant y2 0.
y1
2
c. First, we have f 1 ( y1 ) = 3 y1 dy 2 = 3 y 2 , 0 y1
Chapter 5: Multivariate Probability Distributions
5.1
a. The sample space S gives the possible values for Y1 and Y2:
S
AA
AB
AC
BA
BB
BC
CA
CB
CC
(y1, y2) (2, 0) (1, 1) (1, 0) (1, 1) (0, 2) (1, 0) (1, 0) (0, 1) (0, 0)
Since each sample point is equally li
Chapter 4: Continuous Variables and Their Probability Distributions
0.0
0.2
0.4
F(y)
0.6
0.8
1.0
4.1
y <1
0
.4 1 y < 2
a. F ( y ) = P(Y y ) = .7 2 y < 3
.9 3 y < 4
1
y4
0
1
2
b. The graph is above.
4.2
3
4
5
y
a. p(1) = .2, p(2) = (1/4)4/5 = .2, p(3) = (1
Chapter 2: Probability
2.1
A = cfw_FF, B = cfw_MM, C = cfw_MF, FM, MM. Then, AB = 0 , BC = cfw_MM, C B =
/
cfw_MF, FM, A B =cfw_FF,MM, A C = S, B C = C.
2.2
a . A B
b. A B
c. A B
d. ( A B ) ( A B )
2.3
2.4
a.
b.
8
Chapter 2: Probability
9
Instructors Solu
Solutions to HW1: MTH 540
2.5 a. ( A B ) ( A B ) = A ( B B ) = A S = A .
b. B ( A B ) = ( B A) ( B B ) = ( B A) = A .
c. ( A B ) ( A B ) = A ( B B ) = 0 . The result follows from part a.
/
/
d. B ( A B ) = A ( B B ) = 0 . The result follows from part b.
2