So
P
(
X
= 0

Y
=
y
) =
e

7
y
. Also note that
Y
is exponentially distributed
with rate 5, so
f
Y
(
y
) = 5
e

5
y
. Therefore,
P
(
X
= 0) =
Z
∞
0
P
(
X
= 0

Y
=
y
)
f
Y
(
y
)
dy
=
Z
∞
0
5
e

12
y
dy
=
±

5
12
e

12
y
²
∞
0
=
5
12
.
(We saw another way to solve this problem in class. See Example 2 in Section
5.2 and note that
X
= 0 is just the event that the ﬁrst woman arrives before
the ﬁrst man.)
5. [8 points] Let
U
1
,U
2
,U
3
,U
4
,U
5
be independently and uniformly chosen from (0
,
1). Let
Y
be the largest and
X
be the third largest. (In the book’s notation,
X
=
U
(3)
and
Y
=
U
(5)
.) Find the joint density of (
X,Y
).
Solution:
For 0
< x < y <
1,
P
(
X
∈
dx
and
Y
∈
dy
)
= 5
·
4
·
P
(
U
1
is the third largest and
U
1
∈
dx,U
2
is the largest and
U
2
∈
dy
)
= 20
·
3
·
P
(
U
3
,U
4
∈
(0
,x
)
,U
1
∈
dx,U
5
∈
(
x,y
)
,U
2
∈
dy
)
= 60[
x
2
][1
·
dx
][
y

x
][1
·
dy
] = 60
x
2
(
y

x
)
dx dy.
Therefore, the joint density of
X
and
Y
is
f
(
x,y
) = 60
x
2
(
y

x
), for 0
< x < y <
1.
Page 4
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentMath 431: Final Exam
6. Let
X
be a random variable with exponential
λ
distribution. Given
X
=
x
, let
Y
be
uniformly distributed on (0
,x
).
(a) [3 points] What is
E
(
Y

X
)?
Solution:
The mean of a uniform distribution on an interval is the midpoint,
so
E
(
Y

X
=
x
) =
x/
2. More compactly,
E
(
Y

X
) =
X/
2.
(b) [3 points] What is
E
(
Y
)?
Solution:
The expectation of
X
is 1
/λ
, so
E
(
Y
) =
E
[
E
(
Y

X
)] =
E
±
X
2
²
=
1
2
λ
.
(c) [6 points] What the joint density of
X
and
Y
?
Solution:
f
X
(
x
) =
λe

λx
, for
x >
0,
f
Y
(
y

X
=
x
) =
1
x
, for
y
∈
(0
,x
), so
f
X,Y
(
x,y
) =
f
X
(
x
)
f
Y
(
y

X
=
x
) =
λe

λx
x
, when 0
< y < x
.
7. [8 points] Assume that
X
1
,...,X
5
are independent identically distributed random vari
ables with nonzero variance. Let
Y
=
X
1
+
···
+
X
5
. What is
Corr
(
Y,X
1
)?
Solution:
Cov
(
Y,X
1
) =
Cov
(
X
1
+
···
+
X
5
,X
1
)
=
Cov
(
X
1
,X
1
) +
Cov
(
X
2
,X
1
) +
···
+
Cov
(
X
5
,X
1
) =
V ar
(
X
1
)
,
V ar
(
Y
) =
V ar
(
X
1
) +
···
+
V ar
(
X
5
) = 5
V ar
(
X
1
), so
Corr
(
Y,X
1
) =
Cov
(
Y,X
1
)
SD
(
Y
)
SD
(
X
1
)
=
V ar
(
X
1
)
p
5
V ar
(
X
1
)
p
V ar
(
X
1
)
=
1
√
5
.
Page 5
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '12
 Miller
 Probability, Probability theory, dy

Click to edit the document details