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Solutions to the EE351K homework assignment 8
2.
a. Since X and Y are independent, we can have their join PDF as
3 2
,
1
0,
0
( , )
( )
( )
6
0
x
y
X Y
X
Y
e
x
y
f
x y
f
x f
y
otherwise
 
≥
≥
=
=
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2
,
0 0
3
3
2
2
0
0
0
0
5
3
6
3
2
0
0
1
(
)
( , )
6
1
1
( 2
 )
6
6
1
1
6
(
1)
(
3
) 
0.6
3
3
5
x
x
y
X Y
X Y
x
x
x
y
y
x
x
x
x
x
P X
Y
f
x y dxdy
e e dydx
e
e dydx
e
e
dx
e
e
dx
e
e
+∞


>
+∞
+∞




+∞




+∞
>
=
=
=
=

= 

= 

+
=
∫∫
∫ ∫
∫
∫
∫
∫
b.
,
0
0
0
0
(
)
( , )
( )
( )
(
) ( )
X Y
X
Y
E XY
xyf
x y dxdy
xf
x dx
yf
y dy
E X E Y
+∞ +∞
+∞
+∞
=
=
=
∫ ∫
∫
∫
Let’s use Z to represent a variable whose PDF is
1
( )
z
Z
f
z
e
λ

=
when
0
z
≥
, and ZOW, hence
0
0
0
0
0
0
0
1
( )
( )
(
) 
(
)
0 0
(
) 
z
z
z
z
Z
z
z
E Z
zf
z dz
z
e
dz
zde
ze
e
d
z
e
dz
e
+∞
+∞
+∞
+∞




+∞
+∞


+∞
=
=
=

= 


=  +
= 
=
∫
∫
∫
∫
∫
So E(XY) = E(X) E(Y) = 2*3 =6
c.
Cov(X,Y) = E(XY)E(X)E(Y) = 62*3=0
3.
a.
,
2,0,2
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 Fall '07
 BARD

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