Probablistic Graphical Models, Spring 2007
Homework 2 solutions
December 6, 2007
1
Markov Networks
Solution due to Yongjin Park
•
Strong Union:
X
⊥
Y

Z
=
⇒
X
⊥
Y

Z, W
(1)
•
Transitivity:
¬
(
X
⊥
A

Z
)
∧¬
(
A
⊥
Y

Z
)
=
⇒ ¬
(
X
⊥
Y

Z
)
(2)
1.
See Figure 1.
Figure 1: BN
G
of which
I
(
G
) does not satisfy these two properties.
2.
The global Markov assumption defines
X
⊥
Y

Z
if and only if
sep
H
(
X
;
Y

Z
), or there is no other path
between
X
and
Y
that does not pass through
Z
. Due to the monotonicity of separation,
sep
H
(
X
;
Y

Z
) =
⇒
se
H
(
X
;
Y

Z
′
)
,
(3)
where
Z
⊂
Z
′
and
{
X, Y
}
/
∈
Z
.
Then it follows that any set of nodes
Z
′
s.t.
Z
⊂
Z
′
,
X
⊥
Y

Z
=
⇒
X
⊥
Y

Z
′
. Since
{
Z
}⊂{
Z, W
}
, the strong union property holds in
I
(
H
).
As for the transitivity, we can prove by showing the contrapositive proposition is correct:
∀
A,
X
⊥
Y

Z
=
⇒
X
⊥
A

Z
∨
A
⊥
Y

Z
(4)
There can be three cases of
A
without violating
X
⊥
Y

Z
,
•
A
belongs to the only path between
X
and
Y
where
sep
H
(
X
;
Y

Z
). If
A
is between
X
and
Z
then
¬
sep
H
(
X
;
A

Z
) but
sep
H
(
A
;
Y

Z
); thus
A
⊥
Y

Z
. The mirror case is also true.
•
A
is connected to either
X
or
Y
, but not simultaneously. If
∃
a path between
X
and
A
, but not
between
Y
and
A
,
¬
sep
H
(
A
;
X

Z
) but
sep
H
(
A
;
Y
·
).
The mirror case is also true.
Thus, either
X
⊥
A

Z
or
A
⊥
Y

Z
.
•
A
is not connected with either
X
or
Y
at all. Trivially,
X
⊥
A
∧
Y
⊥
A
.
1
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3.
Show
I
l
(
H
) =
⇒
I
p
(
H
). Equivalently we can prove by showing that if
X
⊥
Y

N
H
(
X
), then
X
⊥
Y
X−
{
X, Y
}
, for all disjoint sets
X, Y
. By the definition of Imap,
X
⊥
Y

N
H
(
X
)
⇐⇒
sep
H
(
X
;
Y

N
H
(
X
))
.
(5)
Again exploiting the monotonicity of
d
−
separation (Eq. 3), if
N
H
(
X
)
⊂
Z
,
Z
dseparates
X
and
Y
, i.e.
sep
H
(
X
;
Y

Z
). Inductively, adding each node
A
∈
[
X−{
X, Y
}−
N
H
(
X
)] to
Z
, i.e.
Z
′
←
Z
∪{
A
}
, where
Z
is initially
N
H
(
X
),
sep
H
(
X
;
Y

N
H
(
X
))
=
⇒ ···
=
⇒
sep
H
(
X
;
Y

Z
)
=
⇒
sep
H
(
X
;
Y

Z
′
) =
⇒···
=
⇒
sep
H
(
X
;
Y
X−{
X, Y
}
)
.
(6)
In sum,
X
⊥
Y

N
H
(
X
) =
⇒
X
⊥
Y
X−{
X, Y
}
, and
I
l
(
H
) =
⇒
I
p
(
H
).
2
Constructing Junction Tree
Solution due to Steve Gardiner
.
I will proceed by induction on the length of a path in the tree. All paths mentioned here are assumed
to be in a junction tree.
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 Fall '07
 CarlosGustin
 Trigraph, C0, XY Z, running intersection property

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