University of Illinois Fall 1998
ECE 313: Solutions to Problem Set #7
1. (a) By the properties of a pdf we have
1=
Z
+
1
,1
f
X
(
u
)
du
=
c
Z
3
0
udu
+
c
Z
6
3
(6
,
u
)
du
=9
c
Hence
c
=1
=
9
.
(b) We have a continuous random variable, and therefore
F
X
(
a
)=
R
a
,1
f
X
(
u
)
du
. Thus
a<
0:
F
X
(
a
)=0
0
a<
3:
F
X
(
a
)=
1
9
Z
a
0
udu
=
a
2
18
3
a<
6:
F
X
(
a
)=
1
9
Z
3
0
udu
+
1
9
Z
a
3
(6
,
u
)
du
=
1
2
,
1
18
(6
,
u
)
2
a
3
=1
,
(6
,
a
)
2
18
a
6:
F
X
(
a
)=1
This function
F
X
(
u
)
is sketched below.
−2 0 2 4 6 8
0
0.2
0.4
0.6
0.8
1
u −−>
CDF of random variable X
(c) It is easy to see that
F
X
(
u
)
is nondecreasing and that
0
F
X
(
u
)
1
for all
u
, either from the sketch
above or by analyzingthe functions
u
2
=
18
and
1
,
(6
,
u
)
2
=
18
in the intervals
[0
;
3]
and
[3
;
6]
.Further,in
our case
F
X
(
,1
)=
F
X
(0) = 0
and
F
X
(+
1
)=
F
X
(6) = 1
. Finally, we see that
F
X
(
u
)
is continuous
everywhere, including the points
u
=0
,
u
=3
,and
u
=6
.
(d) We have
P
(
A
)=
P
(
X>
3) = 1
,
F
X
(3) = 0
:
5
. Similarly
P
(
B
)=
P
(1
:
5
X
9) =
P
(
X
9)
,
P
(
X<
1
:
5) =
F
X
(9)
,
F
X
(1
:
5
,
)=1
,
0
:
125 = 0
:
875
.
(e) The intersection of
A
and
B
is the event
f
3
<X
9
g
and
P
(
AB
)=
P
(3
<X
9) =
P
(
X>
3) =
0
:
5=
P
(
A
)
6
=
P
(
A
)
P
(
B
)
. Hence, the two events are not independent.
2. Let
X
denote the chosen number. Then
X
is a continuous RV, uniformly distributed on the interval
(0
;
1)
.We
have
(a)
P
(0
:
1
X<
0
:
2) = 0
:
2
,
0
:
1=0
:
1
.
(b)
P
(second digit = 2) =
P
9
k
=0
P
(0
:k
2
X<
0
:k
3) = 10
0
:
01 = 0
:
1
.
(c)
P
(0
:
3
p
X<
0
:
4) =
P
(0
:
09
X<
0
:
16) = 0
:
07
.
3. (a)
P
(
X>
5) =
1
5
R
1
5
e
,
u=
5
du
=
,
e
,
u=
5
1
5
=
e
,
1
.
(b)
P
(
X<
6) =
1
5
R
6
0
e
,
u=
5
du
=
,
e
,
u=
5
6
0
=1
,
e
,
6
=
5
.
