MA 316: MATHEMATICAL PROBABILITY
ASSIGNMENT #7 SOLUTIONS: SECTIONS 2.1, 2.2, 2.3, and 2.4
FALL 2003
2.1:
If
f
(
x
1
, x
2
) = 4
x
1
x
2
for 0
< x
1
<
1
,
0
< x
2
<
1 is the joint density function of
X
1
and
X
2
then
P
(0
< X
1
<
1
/
2
,
1
/
4
< X
2
<
1) =
Z
1
/
2
0
Z
1
1
/
4
4
x
1
x
2
dx
2
dx
1
=
15
64
.
For the second part and as the hint suggests, notice that
P
(
X
1
=
X
2
) is the volume under the surface
f
(
x
1
, x
2
) = 4
x
1
x
2
and above the line segment 0
< x
1
=
x
2
<
1 in the
x
1

x
2
plane, which necessarily must
be 0. For the third part, we have
P
(
X
1
< X
2
) =
Z
1
0
Z
x
2
0
4
x
1
x
2
dx
1
dx
2
=
1
2
and the integral for the fourth part is identical to the third part, so
P
(
X
1
≤
X
2
) =
1
2
.
2.5:
If
R
∞
0
g
(
x
)
dx
= 1 then the function
f
(
x
1
, x
2
) =
2
g
(
p
x
2
1
+
x
2
2
)
π
p
x
2
1
+
x
2
2
for 0
< x
1
<
∞
and 0
< x
2
<
∞
is a density function since converting this integral to one involving polar
coordinates using the standard substitution
r
2
=
x
2
1
+
x
2
2
gives
Z
π/
2
0
Z
∞
0
2
g
(
r
)
πr
rdrdθ
=
2
π
Z
π/
2
0
Z
∞
0
g
(
r
)
drdθ
=
2
π
Z
π/
2
0
dθ
= 1
.
2.9:
For part a), given the table as in the text we can compute the density function for
X
1
as
f
1
(0) = 7
/
12
and
f
1
(1) = 5
/
12 and for
X
2
as
f
2
(0) = 4
/
12,
f
2
(1) = 5
/
12, and
f
2
(2) = 3
/
12.
For part b) notice that
P
(
X
1
+
X
2
= 1) = 3
/
12 + 2
/
12 = 5
/
12 since the event
X
1
+
X
2
= 1 happens when
x
1
= 0
, x
2
= 1 and
x
1
= 1
, x
2
= 0
.
.
2.10:
If
X
1
and
X
2
have the joint density
f
(
x
1
, x
2
) = 15
x
2
1
x
2
for 0
< x
1
< x
2
<
1 then the marginal density
for
X
1
is
f
(
x
1
) =
Z
1
x
1
15
x
2
1
x
2
dx
2
= 15
x
2
1
1
2

x
2
1
2
¶
and the marginal density for
X
2
is
f
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 Fall '07
 Song
 Probability, Probability theory, 3 g, 0 g, 2 0 0 g, 2 0 0 2g

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