University of Illinois
Fall 2009
ECE 313:
Problem Set 9: Solutions
Continuous Random Variables
1.
[Validity of PDFs]
If
f
(
u
) is a nonnegative (or nonpositive) function with ﬁnite nonzero area
A
, then
A

1
·
f
(
u
) is a valid
pdf. This does not work if
f
(
u
) takes on both positive and negative values.
(a)
f
(
u
) = 2
u,
0
< u <
1 is a valid pdf.
(b)
f
(
u
) =

u

,

u

<
1
2
is not a valid pdf, but 4
·
f
(
u
) is.
(c)
f
(
u
) = 1
 
u

,

u

<
1 is a valid pdf.
(d)
f
(
u
) = ln
u,
0
< u <
1 is not a valid pdf but

f
(
u
) is.
(e)
f
(
u
) = ln
u,
0
< u <
2 is not a valid pdf, nor is
C
·
f
(
u
) a valid pdf for any choice of
C
.
(f)
f
(
u
) =
2
3
(
u

1)
,
0
< u <
3
,
is not a valid pdf, nor is
C
·
f
(
u
) a valid pdf for any choice of
C
.
(g)
f
(
u
) = exp(

2
u
)
, u >
0 is not a valid pdf but 2
·
f
(
u
) is.
(h)
f
(
u
) = 4
e

2
u

e

u
, u >
0
,
is not a valid pdf, nor is
C
·
f
(
u
) a valid pdf for any choice of
C
.
(i)
f
(
u
) = exp(

u

)
,

u

<
1 is not a valid pdf but
e/
[2(
e

1)]
·
f
(
u
) is.
2.
[Calculating probabilities from pdfs]
(a) The pdf is as sketched in the lefthand ﬁgure below.
f
X
(
u
)
f
X
(
u
)
uu
11
2
u
1
2
c
1
1
2
4
3
1
3
f
X
(
u
)
Since the area under the curve is obviously
c/
2 and should equal 1, we conclude that
c
= 2.
(b) Since
X
takes on values only in (0
,
1),
P
{
X
>

0
.
5
}
= 1.
Note: this question had an unintentional typo in it. We meant to ask for
P
{
X
>
0
.
5
}
which is
the shaded area in the middle ﬁgure above and is obviously
1
4
.
(c)
P
{
6
X
2
>
5
X

1
}
=
P
{
6
X
2

5
X
+ 1
>
0
}
=
P
{
(3
X

1)(2
X

1)
>
0
}
.
Now, the product of two quantities is positive only when both are positive or both are negative.
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 Probability theory, valid pdf

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