10708 Graphical Models: Homework 1 Solutions
October 13, 2006
1
Conditional Probability
1.1
Prove
P
(
S
) =
f
(
X,Z
)
g
(
Y,Z
) =
⇒
(
X
⊥
Y

Z
)
P
(
X,Y

Z
) =
P
(
X,Y,Z
)
P
(
Z
)
=
P
(
X,Y,Z
)
∑
x
∑
y
P
(
x,y,Z
)
=
f
(
X,Z
)
g
(
Y,Z
)
∑
x
f
(
x,Z
)
∑
y
g
(
y,Z
)
=
P
(
X

Z
)
P
(
Y

Z
)
Prove (
X
⊥
Y

Z
) =
⇒
P
(
S
) =
f
(
X,Z
)
g
(
Y,Z
)
P
(
S
) =
P
(
X,Y,Z
) =
P
(
X,Y

Z
)
P
(
Z
) =
P
(
X

Z
)
P
(
Y

Z
)
P
(
Z
)
where
f
(
X,Z
) =
P
(
X

Z
) and
f
(
Y,Z
) =
P
(
Y

Z
)
P
(
Z
).
1.2
Assume that the joint can be expressed as the product of marginals,
P
(
X,Y,Z
) =
f
(
X,Z
)
g
(
Y,Z
).
Note that
P
(
Z
) =
∑
x
f
(
x,Z
) =
∑
y
g
(
y,Z
). It follows that
X
x
X
y
P
(
x,y,Z
) =
X
x
X
y
f
(
x,Z
)
g
(
y,Z
)
P
(
Z
) =
P
(
Z
)
P
(
Z
)
which is only true is
P
(
Z
=
z
) = 1 for some
z
. For general distributions, the statement does
not hold.
1