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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/3/e/solution
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Past exam of the mathematics course of the University of Cambridge
/
2021
/
iii
/
Paper 219
/
3
/
e
/
Solution
by
Codex
0
2026-09-28
Up to
normalization, the joint posterior is
p
(
M
1
:
N
,
M
0
,
τ
2
∣
D
)
∝
(
τ
2
)
k
∏
s
=
1
N
exp
[
−
2
σ
2
(
D
s
−
M
s
)
2
−
2
τ
2
(
M
s
−
M
0
)
2
]
(
τ
2
)
−
N
/2
.
(1)
A
Gibbs sampler
cycle consists of:
independently draw every
M
s
from the normal conditional in part
a
;
draw
M
0
∣
M
1
:
N
,
τ
2
∼
N
(
M
ˉ
,
τ
2
/
N
)
;
draw
τ
2
∣
M
1
:
N
,
M
0
∼
Inv
-
Gamma
(
2
N
−
k
−
1
,
2
1
∑
s
(
M
s
−
M
0
)
2
)
.
(2)
Each proposal is the exact full conditional and is therefore accepted. If the cycle begins with
density
p
(
θ
t
∣
D
)
, integrating the product of the current posterior and successive conditional kernels over all overwritten coordinates leaves
p
(
θ
t
+
1
∣
D
)
,
(3)
so
a
full Gibbs sweep preserves the joint posterior.
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:
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