Stat 5101 (Geyer) 2002 Midterm 2
Problem 1
This is like Problem 3.6.4 in DeGroot and Schervish.
Since
g
1
(
x

y
) =
f
(
x, y
)
f
2
(
y
)
(1)
First we need to find the marginal of
y
.
f
2
(
y
) =
Z
1
0
f
(
x, y
)
dx
=
2
2

log 2
·
1
1 +
y
Z
1
0
(
x
+
y
)
dx
=
2
2

log 2
·
1
1 +
y
•
x
2
2
+
xy
‚
1
0
=
2
2

log 2
·
1
2
+
y
1 +
y
Plugging back into (1) we get
g
1
(
x

y
) =
2
2

log 2
·
x
+
y
1+
y
2
2

log 2
·
1
2
+
y
1+
y
=
x
+
y
1
2
+
y
,
0
< x <
1
Alternate Solution (Conditional Probability is Renormalization)
The
conditional density
g
1
(
x

y
) is a probability density though of as a function
of the variable(s) in front of the bar (here
x
) and is proportional to the joint
density, thus
g
1
(
x

y
) =
c
(
x
+
y
)
,
0
< x <
1
.
(2)
where
c
is some constant that does not depend on what we are thinking of as
the variable (the stuff in front of the bar
x
), though it may (and generally does)
depend on the stuff behind the bar, which is being treated as constant.
All we need to do is choose the constant
c
so that (2) integrates to one,
integrating with respect to the variable
x
(the stuff behind the bar
y
being
constant).
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 Fall '02
 Staff
 Variance, Probability theory, probability density function, GEYER

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