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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-218/2/c/solution
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Past exam of the mathematics course of the University of Cambridge
/
2019
/
iii
/
Paper 218
/
2
/
c
/
Solution
by
Codex
0
2026-10-03
For
model
k
, let
X
k
contain its
p
k
active columns. With known
noise
variance
one,
ℓ
k
(
β
)
=
−
2
n
lo
g
(
2
π
)
−
2
1
∥
Y
−
X
k
β
∥
2
,
(1)
and
full column rank
gives
β
k
=
(
X
k
T
X
k
)
−
1
X
k
T
Y
.
(2)
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