Mathematics Department, UCLA
T. Richthammer
winter 09, sheet 8
Feb 20, 2009
Homework assignments: Math 170A Probability, Sec. 1
093. The following functions are CDFs of RVs. Determine the corresponding PDF:
(a)
F
(
x
) =
1
x
2
+1
1
(
∞
,
0)
(
x
) + 1
[0
,
∞
)
(
x
)
(b)
F
(
x
) =
e
x
2
1
(
∞
,
0)
(
x
) + (1

e

x
2
)1
[0
,
∞
)
(
x
)
Answer:
We use
f
(
x
) =
F
X
(
x
), so
(a)
f
(
x
) =

2
x
(
x
2
+1)
2
1
(
∞
,
0)
(
x
)
(b)
f
(
x
) =
1
2
e

x

.
094. Let
X
be a RV with PDF
f
(
x
) =
1
18
x
2
1
[

3
,
3]
(
x
). Calculate
R
(
Y
) and the PDF of
Y
, where
(a)
Y
=
X
+ 2
(b)
Y
= 3
X
(c)
Y
=
e
X
(d)
Y
=
X
2
(e)
Y
=

X
Answer:
R
(
X
) = [

3
,
3] and for
c
∈
R
(
X
) we have
F
X
(
c
) =
1
18
R
c

3
x
2
=
c
3
+27
54
.
(a)
R
(
Y
) = [

1
,
5] and for
c
∈
[

1
,
5] we have
F
Y
(
c
) =
P
(
X
+ 2
≤
c
) =
P
(
X
≤
c

2) =
F
X
(
c

2), so
f
Y
(
c
) =
f
X
(
c

2) =
1
18
(
c

2)
2
1
[

1
,
5]
(
c
)
(b)
R
(
Y
) = [

9
,
9] and for
c
∈
[

9
,
9] we have
F
Y
(
c
) =
P
(3
X
≤
c
) =
P
(
X
≤
c/
3) =
F
X
(
c/
3), so
f
Y
(
c
) =
f
X
(
c/
3)
/
3 =
1
486
c
2
1
[

9
,
9]
(
c
)
(c)
R
(
Y
) = [
e

3
, e
3
] and for
c
∈
R
(
Y
) we have
F
Y
(
c
) =
P
(
e
X
≤
c
) =
P
(
X
≤
ln
c
) =
F
X
(ln
c
), so
f
Y
(
c
) =
f
X
(ln
c
)
/c
=
1
18
(ln
c
)2
c
1
[
e

3
,e
3
]
(
c
)
(d)
R
(
Y
) = [0
,
9] and for
c
∈
R
(
Y
) we have
F
Y
(
c
) =
P
(
X
2
≤
c
) =
P
(

X
 ≤
√
c
) =
F
X
(
√
c
)

F
X
(

√
c
), so
f
Y
(
c
) =
f
X
(
√
c
)
1
2
√
c
+
f
X
(

√
c
)
1
2
√
c
=
√
c
18
1
[0
,
9]
(
c
)
(e)
R
(
Y
) = [

3
,
3] and for
c
∈
R
(
Y
) we have
F
Y
(
c
) =
P
(

X
≤
c
) =
P
(
X
≥ 
c
) =
1

F
X
(

c
), so
f
Y
(
c
) =
f
X
(

c
) =
1
18
x
2
1
[

3
,
3]
(
c
) =
f
(
c
)
095. For
a >
0,
b
∈
R
express the PDF of
Y
=
aX
+
b
in terms of the PDF
f
X
of the RV
X
.
Answer:
F
Y
(
c
) =
P
(
aX
+
b
≤
c
) =
P
(
X
≤
c

b
a
) =
F
X
(
c

b
a
), so
f
Y
(
c
) =
1
a
f
X
(
c

b
a
).
096. Compute
P
(0
< X <
3),
E
(
X
) and
V
(
X
) if
X
has a PDF given by
(a)
f
(
x
) = 5
x
4
1
[0
,
1]
(
x
)
(b)
f
(
x
) =
8
x
3
1
[2
,
∞
)
(
x
)
(c)
f
(
x
) =
3
4
(1

x
2
)1
(

1
,
1)
(
x
)
Answer:
(a)
R
3
0
f
(
x
)
dx
= 1,
E
(
X
) =
5
6
,
E
(
X
2
) =
5
7
, so
V
(
X
) =
5
252
(b)
R
3
0
f
(
x
)
dx
=
5
9
,
E
(
X
) = 4,
E
(
X
2
) =
∞
, so
V
(
X
) =
∞
(c)
R
3
0
f
(
x
)
dx
=
1
2
,
E
(
X
) = 0, so
V
(
X
) =
E
(
X
2
) =
1
5
.
097. If the density function of
X
is
f
(
x
) = (
a
+
bx
2
)1
[0
,
1]
(
x
) and
E
(
X
) =
3
5
, determine
a
and
b
.
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 Winter '10
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 Natural logarithm

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