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2.
2
1
1
[
( )]
( )
( )
( )
( )
[
( )] 
2
2
X
X
X
X
X
X
E F
x
F
x f
x dx
F
x dF
x
F
x
+∞
+∞
+∞
∞
∞
∞
=
=
=
=
∫
∫
3.
The PDF of X is a piecewise function
(
)
( )
(
2
)
2
0
X
h
x a
a
x
b
b a
h
f
x
x
b
a
b
x
b a
b a
otherwise

≤ ≤

= 

+
≤ ≤


a.
By definition of a PDF, we have
( )
1
X
f
x dx
+∞
∞
=
∫
Put in the piecewise function, and we get sum of two integrals equals to 1.
2
(
)
(
2
)
1
b
b a
a
b
h
h
x a dx
x
b a dx
b a
b a


+


+
=


∫
∫
In order to simplify computation, we can have y=xb, hence
0
0
(
)
(
)
1
b a
a b
h
h
y
b a dy
y b
a dy
b a
b a


+ 
+

 +
=


∫
∫
Do another substitution u= y for the first integral, hence
0
0
(
)(
)
(
)
1
b a
b a
h
h
u b a
du
y b a dy
b a
b a


 + 

+

 +
=


∫
∫
Now the first integral is the same as the second, hence
0
1
(
)
2
b a
h
y b a dy
b a


 +
=

∫
0
1
(
)
2
b a
a b
y b a dy
h


 +
=
∫
2
0
1
1
[
(
) ]
2
2
b a
a b
y
b a y
h




=
2
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This note was uploaded on 04/21/2008 for the course EE 351k taught by Professor Bard during the Fall '07 term at University of Texas at Austin.
 Fall '07
 BARD

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