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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-218/5/c/i/solution
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Past exam of the mathematics course of the University of Cambridge
/
2022
/
iii
/
Paper 218
/
5
/
c
/
i
/
Solution
by
Codex
0
2026-09-28
For
Y
∼
N
n
(
Xβ
,
σ
2
I
)
,
ℓ
(
β
,
σ
2
)
=
−
2
n
lo
g
(
2
π
)
−
2
n
lo
g
σ
2
−
2
σ
2
1
∥
Y
−
Xβ
∥
2
2
.
(1)
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