UCSD ECE 153
Handout #28
Prof. Young-Han Kim
Wednesday, December 10, 2008
Solutions to Final Exam
(Total: 200 points)
There are 4 problems. The Frst 3 problems have 4 parts each, while the last problem has 8
parts. Each part is uniformly worth 10 points.
Your answer should be as clear and readable as possible. In particular, if the answer involves
a pmf or pdf, make sure to identify the values or intervals for which the pmf or pdf is nonzero.
1.
Order statistics.
Let
X
1
,X
2
,X
3
be independent and uniformly drawn from the interval [0
,
1]. Let
Y
1
be
the smallest of
X
1
,X
2
,X
3
, let
Y
2
be the median (second smallest) of
X
1
,X
2
,X
3
, and
let
Y
3
be the largest of
X
1
,X
2
,X
3
. ±or example, if
X
1
=
.
3
,X
2
=
.
1
,X
3
=
.
7, then
Y
1
=
.
1
,Y
2
=
.
3
,Y
3
=
.
7. The random variables
Y
1
,Y
2
,Y
3
are called the
order statistics
of
X
1
,X
2
,X
3
.
(a) What is the probability P
{
X
1
≤
X
2
≤
X
3
}
?
(b) ±ind the pdf of
Y
1
.
(c) ±ind the pdf of
Y
3
.
(d) (Di²cult.) ±ind the pdf of
Y
2
.
(Hint:
Y
2
≤
y
if and only if at least two among
X
1
,X
2
,X
3
are
≤
y
.)
Solution:
(a) By symmetry, P
{
X
i
≤
X
j
≤
X
k
}
should be identical for all
i
n
=
j
n
=
k
. Since
there are 3! = 6 such (
i,j,k
), the probability should be 1
/
6.
(b) We have P
{
Y
1
> y
}
= P
{
X
1
,X
2
,X
3
> y
}
= P
{
X
1
> y
}
P
{
X
2
> y
}
P
{
X
3
> y
}
=
(1
−
y
)
3
. Hence,
f
Y
1
(
y
) =
d
dy
(1
−
(1
−
y
)
3
) = 3(1
−
y
)
2
for 0
≤
y
≤
1.
(c) We can use the similar steps to part (b) to Fnd
f
Y
3
(
y
) = 3
y
2
,
0
≤
y
≤
1.
Alternatively, we can see that by symmetry 1