Homework 5 solutions of STAT 5101
1
Page 151
1.1
1.1.1
Problem 7.
(a)
We have
f1 (x) =
=
=
Therefore, we have g2 (y |x) =
1.1.2
42x
3
(4 2x y )dy
16
0
3
3
3
y xy y 2 |42x
4
8
32 0
3
(x 2)2
8
42xy
2(x22 )
f (x,y )
f1 (x)
=
3y
2
= (3 y )
9
for X = x, y < 4
Homework 6 solutions of STAT 5101
1
1.1
Page 187
Problem 4.
Dene Z = X1 , hence, x1 = z , x2 = y . We have
z
(
)
(
)
x1
x1
1
01
y
z
J = det
= det 1 y
=
x2
x2
z
z
z2
y
z
Hence g (y, z ) = 1 +
g (y, z ). Then
y
.
z2
The pdf of Y is the marginal pdf of Y bas
Homework 7 solutions of STAT 5101
1
1.1
Page 240
Problem 3.
E (X ) = 1, E (X 2 ) = 2, E (X 3 ) = 5. Hence we have
E (X )3 = E (X 1)3
= E (X 3 3X 2 + 3X 1)
= 532+31
=1
1.2
Problem 7.
3
We have (t) = 1 (3et + et ), then (t) = 4 et 1 et and (t) = 3 et + 1 et
Homework 4 solutions of STAT 5101
1
Page 129
1.1
1.1.1
Problem 3.
(a)
If we sum f (x, y ) over the 25 possible pairs of values (x, y ), we obtain 40c.
Since this sum must be equal to 1, it follows that c = 1/40.
1.1.2
(b)
f (0. 2) = (1/40) 2 = 1/20
1.1.3
Homework 4 solutions of STAT 5101
1
Page 129
1.1
1.1.1
Problem 3.
(a)
If we sum f (x, y ) over the 25 possible pairs of values (x, y ), we obtain 40c.
Since this sum must be equal to 1, it follows that c = 1/40.
1.1.2
(b)
f (0. 2) = (1/40) 2 = 1/20
1.1.3
Homework 1 solutions of STAT 5101
1
1.1
Page 41
Problem 10.
()
The sample space contains 24 outcomes. If the person select both the 2
10
defective bulbs, then the left 8 bulbs will be selected from 22 bulbs. Hence
the probability is
()
22
8
(24) = 0.163
1
Homework 3 solutions of STAT 5101
1
Page 85
1.1
Problem 4.
Let A denote the event that the response to the test is positive; B1 denotes
the event that he has the disease, B2 denotes the event he doesnt have the
disease.Then we have
P (B1 |A) =
1.2
1.2.1
P
Homework 1 solutions of STAT 5101
1
Page 15
1.1
Problem 5.
1.2
Problem 6.
To prove the rst result, let x ( i Ai )c . This means that x is not in i Ai .
IN other words, for every i I , x is not in Ai Hence for every i I , x Ac .
.
i
So x i Ac . This prover