As above,
P
(
X >
1
/
2 and
Z
≤
1) =
R
1
1
/
2
1

x
2
dx
= 5
/
24, so
P
(
X >
1
/
2

Z
≤
1) =
5
/
24
2
/
3
=
5
16
.
(d) [8 points]
(Deleted from exam.)
Find the c.d.f. of
Z
.
Solution:
For
z
∈
(0
,
1],
F
Z
(
z
) =
P
(
Z
≤
z
) =
Z
√
z
0
z

x
2
dx
=
2
z
3
/
2
3
.
Page 4
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Math 431: Exam 2
For
z
∈
(1
,
2),
F
Z
(
z
) =
P
(
Z
≤
z
) =
Z
√
z

1
0
dx
+
Z
1
√
z

1
z

x
2
dx
=
z

2(
z

1)
3
/
2
+ 1
3
.
6. [12 points] Let
X
have exponential distribution with rate
λ
= 3. Find the density of
Y
= 1
/X
.
Solution:
Note that
Y
has range (0
,
∞
). The c.d.f. of
Y
is
F
Y
(
y
) =
P
(
Y
≤
y
) =
P
(1
/X
≤
y
) =
P
(
X
≥
1
/y
) =
Z
∞
1
/y
3
e

3
t
dt.
(We could integrate, but since we are about to differentiate to find the density, it is
not necessary.) The density of
Y
is
f
Y
(
y
) =
d
dy
F
Y
(
y
) =
d
dy
Z
∞
1
/y
3
e

3
t
dt
=

3
e

3(1
/y
)
d
dy
(1
/y
) =
3
e

3
/y
y
2
.
Alternate Solution:
We could also apply the change of variable formula from the
text (instead of essentially rederiving it). Again, not that the range of
Y
is (0
,
∞
).
We have
Y
=
g
(
x
) where
g
(
x
) = 1
/x
. If
y >
0, then
y
=
g
(
x
) = 1
/x
for some
x >
0.
Then
dy/dx
=

1
/x
2
=

y
2
and
x
= 1
/y
, so
f
Y
(
y
) =
f
X
(
x
)

dy/dx

=
f
X
(1
/y
)
y
2
=
3
e

3
/y
y
2
.
7. [15 points] Assume that three points are picked independently and uniformly at random
from a disk of radius one. Let
R
1
,
R
2
and
R
3
be the distances of these points from the
center of the disk. Let
U
= min
{
R
1
, R
2
, R
3
}
and
V
= max
{
R
1
, R
2
, R
3
}
. Find the joint
density of
U
and
V
.
Solution:
Let 0
< u, v <
1 and let Δ
u
and Δ
v
be very small. Then
P
(
U
∈
(
u, u
+ Δ
u
)
, V
∈
(
v, v
+ Δ
v
))
≈
3
·
2
P
(
R
1
∈
(
u, u
+ Δ
u
)
, R
2
∈
(
u, v
)
, R
3
∈
(
v, v
+ Δ
v
))
= 6
P
(
R
1
∈
(
u, u
+ Δ
u
))
P
(
R
2
∈
(
u, v
))
P
(
R
3
∈
(
v, v
+ Δ
v
))
,
Page 5
Math 431: Exam 2
where the 3
·
2 counts the number of ways to choose the least and greatest radius.
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 Spring '12
 Miller
 Normal Distribution, Probability, Probability theory, Exponential distribution

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