University Solutions: Problem Set 10 ECE 313
of Illinois Fall 1998
1. We havea random variable
X
U
(0
;
1)
.
(a) Define the RV
Y
=
,
ln(
X
)
. This is clearly a continuous RV, and hence has a pdf. The graph of
v
=
,
ln(
u
)
is shown below.
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
1
2
3
4
5
6
7
u −−>
v −−>
v = −ln(u)
As can be seen from the graph of
,
ln(
u
)
, the ran-
dom variable
Y
takes on all values in
(0
;
1
)
.
The pdf of
Y
can be given as
f
Y
(
v
)=
f
X
(
u
1
)
=
j
g
0
(
u
1
)
j
,where
u
1
is the single root of the equation
v
=
,
ln(
u
)
, and is equal to
exp(
,
v
)
. Accordingly, we have
g
0
(
u
1
)=
,
exp(
v
)
and
f
Y
(
v
)=
e
,
v
; v
0
0
;v<
0
(b) You don’t need to calculatethe pdf of
p
X
for this part! Let
Z
2
be the RV corresponding to the second
decimal digit of the square root of
X
,and
Z
1
be the RV corresponding to its first digit. Both
Z
1
and
Z
2
are discrete RVs, taking on values in
f
0
;
1
;:::;
9
g
.If
Z
1
=
m
, then the probability that
Z
2
=
k
is
given by the probabilitythat
f
10
m
+
k
100
p
X
<
10
m
+
k
+1
g
(why is this true?). Therefore, the
probability that
Z
1
=
m
and
Z
2
=
k
is obtained as
P
f
Z
1
=
m;
Z
2
=
k
g
=
P
f
10
m
+
k
100
p
X
<
10
m
+
k
+1
g
=
P
1
10
4
(10
m
+
k
)
2
X
<
1
10
4
(10
m
+
k
+1)
2
a
=
1
10
4
,
(10
m
+
k
+1)
2
,
(10
m
+
k
)
2
=
20
m
+2
k
+1
10000
Summing over all values that
Z
1
can take, we get the pmf of
Z
2
as
p
Z
2
(
k
)=
9
X
m
=0
20
m
+2
k
+1
10000
=0
:
091 + 0
:
002
k:
(Check and see that this sums to
1
over all values of
k
.)
1
