
3.30. Given the joint probability density function
f
XY
(x, y) =
x(1 + 3y
2
)/4, 0 < x < 2, 0 < y < 1
0, elsewhere,
derive the marginal densities f
X
(x), f
Y
(y), the conditional density f
X|Y
(x|y) and evaluate the prob-
ability
Pr
1
4
< X <
1
2
Y =
1
3
.
Solution: The marginal densities are
f
X
(x) =
1
0
f
XY
(x, y)dy =
1
0
1
4
x(1 + 3y
2
)dy
=
1
4
x
y + y
3
1
0
=
x
2
, 0 < x < 2
f
Y
(y) =
2
0
f
XY
(x, y)dx =
2
0
1
4
x(1 + 3y
2
)dx
=
(1 + 3y
2
)
4
x
2
2
2
0
=
1 + 3y
2
2
, 0 < y < 1
The conditional density is
f
X|Y
(x|y) =
f
XY
(x, y)
f
Y
(y)
=
x(1 + 3y
2
)/4
(1 + 3y
2
) /2
=
x
2
, 0 < x < 2
Then, the probability is
Pr
1
4
< X <
1
2
Y =
1
3
=
1/2
1/4
f
X|Y
(x|y = 1/3)dx
=
x
2
4
1/2
1/4
=
1
4
1
4
−
1
16
=
3
64
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