Probability Theory and Statistics (EE/TE 3341)
Homework 5 Solutions
Problem Solutions
: Yates and Goodman,
3.1.1 3.2.1 3.2.4 3.3.2 3.3.4 3.3.6 3.4.3 3.4.5
3.4.11 and 3.5.4
Problem 3.1.1 Solution
The CDF of
X
is
F
X
(
x
) =
0
x <

1
(
x
+ 1)
/
2

1
≤
x <
1
1
x
≥
1
(1)
Each question can be answered by expressing the requested probability in terms of
F
X
(
x
).
(a)
P
[
X >
1
/
2] = 1

P
[
X
≤
1
/
2] = 1

F
X
(1
/
2) = 1

3
/
4 = 1
/
4
(2)
(b) This is a little trickier than it should be. Being careful, we can write
P
[

1
/
2
≤
X <
3
/
4] =
P
[

1
/
2
< X
≤
3
/
4] +
P
[
X
=

1
/
2]

P
[
X
= 3
/
4]
(3)
Since the CDF of
X
is a continuous function, the probability that
X
takes on any
specific value is zero. This implies
P
[
X
= 3
/
4] = 0 and
P
[
X
=

1
/
2] = 0. (If this is
not clear at this point, it will become clear in Section 3.6.) Thus,
P
[

1
/
2
≤
X <
3
/
4] =
P
[

1
/
2
< X
≤
3
/
4] =
F
X
(3
/
4)

F
X
(

1
/
2) = 5
/
8
(4)
(c)
P
[

X
 ≤
1
/
2] =
P
[

1
/
2
≤
X
≤
1
/
2] =
P
[
X
≤
1
/
2]

P
[
X <

1
/
2]
(5)
Note that
P
[
X
≤
1
/
2] =
F
X
(1
/
2) = 3
/
4. Since the probability that
P
[
X
=

1
/
2] = 0,
P
[
X <

1
/
2] =
P
[
X
≤
1
/
2]. Hence
P
[
X <

1
/
2] =
F
X
(

1
/
2) = 1
/
4. This implies
P
[

X
 ≤
1
/
2] =
P
[
X
≤
1
/
2]

P
[
X <

1
/
2] = 3
/
4

1
/
4 = 1
/
2
(6)
(d) Since
F
X
(1) = 1, we must have
a
≤
1. For
a
≤
1, we need to satisfy
P
[
X
≤
a
] =
F
X
(
a
) =
a
+ 1
2
= 0
.
8
(7)
Thus
a
= 0
.
6.
1
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Problem 3.2.1 Solution
f
X
(
x
) =
braceleftbigg
cx
0
≤
x
≤
2
0
otherwise
(1)
(a) From the above PDF we can determine the value of
c
by integrating the PDF and
setting it equal to 1.
integraldisplay
2
0
cx dx
= 2
c
= 1
(2)
Therefore
c
= 1
/
2.
(b)
P
[0
≤
X
≤
1] =
integraltext
1
0
x
2
dx
= 1
/
4
(c)
P
[

1
/
2
≤
X
≤
1
/
2] =
integraltext
1
/
2
0
x
2
dx
= 1
/
16
(d) The CDF of
X
is found by integrating the PDF from 0 to
x
.
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 Spring '09
 Variance, Probability theory, CDF

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