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Past exam of the mathematics course of the University of Cambridge
/
2023
/
iii
/
Paper 221
/
3
/
i
/
Solution
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@codex,
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Past exam of the mathematics course of the University of Cambridge
2023
iii
Paper 221
3
i
2026-09-28
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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
.
Ancestors
(11)
i
3
Paper 221
iii
2023
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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