Statistics 351 Fall 2006 (Kozdron) Midterm #1 — Solutions
1. (a)
By definition,
f
X

Y
=
y
(
x
) is given by
f
X

Y
=
y
(
x
) =
f
X,Y
(
x, y
)
f
Y
(
y
)
.
We being by calculating
f
Y
(
y
) =
√
4

y
2
0
1
π
dx
=
1
π
4

y
2
for 0
< y <
2. Therefore,
f
X

Y
=
y
(
x
) =
1
π
1
π
4

y
2
=
1
4

y
2
,
0
< x <
4

y
2
.
1. (b)
We find
E
(
X

Y
=
y
) =
∞
∞
x
·
f
X

Y
=
y
(
x
)
dx
=
√
4

y
2
0
x
·
1
4

y
2
dx
=
1
2
4

y
2
2
4

y
2
=
4

y
2
2
.
1. (c)
Solving for
X
and
Y
we find
X
=
U
cos
V
and
Y
=
U
sin
V.
The Jacobian of this transformation is
J
=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
=
cos
v

u
sin
v
sin
v
u
cos
v
=
u
cos
2
v
+
u
sin
2
v
=
u.
The joint density of (
U, V
) is therefore given by
f
U,V
(
u, v
) =
f
X,Y
(
u
cos
v, u
sin
v
)
· 
J

=
u
π
provided that 0
< u <
2 and 0
< v <
π
2
.
1. (d)
From
(c)
, we see that
U
and
V
are independent since we can write
f
U,V
(
u, v
) =
f
U
(
u
)
·
f
V
(
v
) where
f
U
(
u
) =
u
2
, 0
< u <
2, and
f
V
(
v
) =
2
π
, 0
< v <
π
2
.
2.
By the law of total probability,
f
X
(
x
) =
∞
∞
f
X

M
=
m
(
x
)
f
M
(
m
)
dm.
Since
X

M
=
m
∈
U
(0
, m
), we know that
f
X

M
=
m
(
x
) =
1
m
, 0
< x < m
. We also find
that
f
M
(
m
) =
F
M
(
m
) = 12
m
2

12
m
3
= 12
m
2
(1

m
)
,
0
< m <
1
.
Therefore,
f
X
(
x
) =
1
x
1
m
·
12
m
2
(1

m
)
dm
= (6
m
2

4
m
3
)
1
x
= 2

6
x
2
+ 4
x
3
,
0
< x <
1
.
1
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3.
Notice that
P
(
X
(1)
=
X
1
) =
P
(
X
1
< X
2
). Therefore, by the law of total probability,
P
(
X
1
< X
2
) =
∞
0
P
(
X
2
> x

X
1
=
x
)
·
f
X
1
(
x
)
dx.
Since
X
1
and
X
2
are independent, we find
P
(
X
2
> x

X
1
=
x
) =
P
(
X
2
> x
) =
∞
x
2
e

2
y
dy
=
e

2
x
.
Thus,
P
(
X
1
< X
2
) =
∞
0
e

2
x
·
e

x
dx
=

1
3
e

3
x
∞
0
=
1
3
.
An alternative solution can be given by conditioning on the value of
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 Fall '08
 MichaelKozdron
 Statistics, Probability, X1

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