This preview shows page 1. Sign up to view the full content.
ISyE 2027B
Probability with Applications
Fall 2010
Homework 8 Solutions
1.
Cov(
U, V
)
=
E[(
U

E[
U
])(
V

E[
V
])] = E[
UV
]

E[
U
]E[
V
]
=
0
×
1
4
+ 1
×
1
2
+ 0
×
1
4

(0
×
1
4
+ 1
×
1
2
+ 2
×
1
4
)(0
×
1
2
+ 1
×
1
2
)
=
1
2

1
2
= 0
ρ
(
U, V
)
=
0
2. Similar to Problem 1, compute marginal probabilities ﬁrst. Both X and
Y have marginal probabilities (1
/
4
,
1
/
4
,
1
/
4
,
1
/
4).
E[
X
]
=
1
×
1
4
+ 2
×
1
4
+ 3
×
1
4
+ 4
×
1
4
= 2
.
5
E[
Y
]
=
1
×
1
4
+ 2
×
1
4
+ 3
×
1
4
+ 4
×
1
4
= 2
.
5
E[
XY
]
=
1
×
1
×
16
136
+
. . .
+ 4
×
4
×
1
136
= 6
.
25
Cov(
X, Y
)
=
E[
XY
]

E[
X
]E[
Y
] = 6
.
25

2
.
5
×
2
.
5 = 0
3. (a) E[
X
2
] =
V ar
(
X
) + (E[
X
])
2
= 4 + 2
2
= 8.
(b) E[

2
X
2
+
Y
] =

2E[
X
2
] + E[
Y
] =

2
×
8 + 3 =

13.
4. For constant
a, b, c, d
∈
R
, Cov(
aX
+
b, cY
+
d
) =
ab
Cov(
X, Y
),
ρ
(
aX
+
b, cY
+
d
) =
ρ
(
X, Y
). Since
T
= 9
/
5
X
+ 32,
S
= 9
/
5
Y
+ 32. Thus
Cov(
T, S
) = (9
/
5)
2
Cov(
X, Y
) = 9
.
72 and
ρ
(
T, S
) =
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: ρ ( X, Y ) = 0 . 8. 5. Cov( X, Y ) = E[( XE[ X ])( YE[ Y ])] = E[ XYX E[ Y ]Y E[ X ] + E[ X ]E[ Y ]] = E[ XY ]E[ X ]E[ Y ]E[ X ]E[ Y ] + E[ X ]E[ Y ] = E[ XY ]E[ X ]E[ Y ] 6. (a) Since ∫ ∞∞ ∫ ∞∞ f ( x, y ) dxdy = ∫ 1 ∫ 1 kxdxdy = k/ 2 = 1, we take k = 2. (b) E[ X ] = ± 1 ± 1 2 x 2 dxdy = 2 3 E[ Y ] = ± 1 ± 1 2 xydxdy == 1 2 E[ XY ] = ± 1 ± 1 2 x 2 ydxdy = 1 3 Cov( X, Y ) = E[ XY ]E[ X ]E[ Y ] = 0 1...
View
Full
Document
This note was uploaded on 02/21/2011 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 Zahrn

Click to edit the document details