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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-221/3/i/solution
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Past exam of the mathematics course of the University of Cambridge
/
2023
/
iii
/
Paper 221
/
3
/
i
/
Solution
by
Codex
0
2026-09-28
The
graph
factorizes
as
p
(
x
)
=
p
(
x
2
)
p
(
x
1
∣
x
2
)
p
(
x
4
∣
x
2
)
p
(
x
3
∣
x
1
,
x
4
)
p
(
x
5
∣
x
2
)
p
(
x
6
∣
x
4
,
x
5
)
.
(1)
Conditioning on all variables except
X
1
, terms not involving
x
1
cancel, leaving
p
(
x
1
∣
x
2
,
x
3
,
x
4
,
x
5
,
x
6
)
∝
p
(
x
1
∣
x
2
)
p
(
x
3
∣
x
1
,
x
4
)
.
(2)
This depends only on
(
x
2
,
x
3
,
x
4
)
, so
X
1
⊥
(
X
5
,
X
6
)
∣
(
X
2
,
X
3
,
X
4
)
.
(3)
Thus
(
X
2
,
X
3
,
X
4
)
is
a
Markov blanket
of
X
1
.
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:
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