EE 351K PROBABILITY & RANDOM PROCESSES
FALL 2011
Instructor: Sujay Sanghavi
[email protected]
Homework 5 Solution
Problem 1
Let
Y
be a random variable with probability density function (PDF)
f
Y
(
y
) =
1 +
y,

1
≤
y
≤
0
,
y,
0
< y
≤
1
,
0
,
otherwise
.
Find
(a)
P
(

Y

<
0
.
5)
.
(b)
P
(
Y >
0

Y <
0
.
5)
.
(c)
E
[
Y
]
.
Sol
:
The pdf of
Y
is as shown below: So based on this, you have two ways of calculating: one is using the
structure of graph, the other is using a standard way, i.e. integral.
(a) By inspection from the figure,
P
(

Y

<
0
.
5) = 0
.
5
. Or you can get it by
P
(

Y

<
0
.
5) =
P
(

0
.
5
< Y <
0
.
5) =
∫
0

0
.
5
(1 +
y
)
dy
+
∫
0
.
5
0
ydy
= 0
.
5
.
(b) By inspection from the figure or integral,
P
(
Y >
0

Y <
0
.
5) =
P
(0
< Y <
1
2
)
P
(
Y <
1
2
)
=
1
5
.
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(c) We have that
E
[
Y
]
=
∫
0

1
y
(1 +
y
)
dy
+
∫
1
0
y
2
dy
=
[
y
2
2
+
y
3
3
]
0
y
=

1
+
[
y
3
3
]
1
y
=0
=
1
6
.
Problem 2
The random variable
X
has the PDF
f
X
(
x
) =
{
cx

2
,
1
≤
x
≤
2
0
,
otherwise
(a) Determine the value of
c
.
(b) Find the CDF of
X
.
(c) Let
A
be the event
{
X >
1
.
5
}
. Calculate
P
(
A
)
and the conditional PDF of
X
given that event
A
has
occurred.
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 Spring '07
 BARD
 Probability theory, dx, ydy, Sujay Sanghavi

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