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Material Type: Assignment; Professor: Liang; Class: Statistical Learning; Subject: Statistics; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;
Typology: Assignments
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Problem 1
Since yi ∼ N(x
t
i
β, σ
2 ), i ∈ (1 : n), the log-likelihood is:
l(
β) = n log
σ
2 π
n
i=
(y i
− x i
β)
2
2 σ
2
then,
AIC = − 2 l(
β) + 2p
= − 2 n log
σ
2 π
n
i=
(yi − xi
β)
2
σ
2
σ
2
(constant + RSS + 2σ
2
p).
Given Cp = RSS + 2σ
2 p and σ is known, when AIC is minimized, so does Cp, and
vice versa. So, AIC criterion and Mallow’s C p
are equivalent.
Problem 2
(a) By the definition of
β
∗ , we have:
β
∗
= arg min
β
(y
∗
1
− x
t
1
β)
2
n ∑
j=
(y j
− x
t
j
β)
2
.
In order to prove
β
β [1]
, we just need to show:
β [1]
= arg min
β
(y
∗
1
− x
t
1
β)
2
n ∑
j=
(y j
− x
t
j
β)
2 .
As known,
β [1]
= arg min
β
n ∑
j=
(y j
− x
t
j
β)
2
.
Plus,
(y
∗
1
− x
t
1
β [1]
2 = 0.
Then,
β [1]
= arg min
β
(y
∗
1
− x
t
1
β)
2
n ∑
j=
(y j
− x
t
j
β)
2 .
(b) Let Y
∗ = {yˆ [1]
, y 2 , · · · , yn}
T and
∗ = {ˆy
∗
1
, yˆ
∗
2
, · · · , ˆy
∗
n
T
Since
∗ = X
β
∗ , we have
yˆ
∗
1
= x
t
1
β [1]
= ˆy [1]
Additionally, since the modified data remain the same X as the original, the
projection matrix H keeps the same, that is
∗ = HY
∗
. Then, we have
ˆy
∗
1
11
ˆy [1]
n ∑
j=
1 j
y j
Thus,
ˆy [1]
11
yˆ [1]
n ∑
j=
1 j
y j
Therefore,
11
)ˆy [1]
n ∑
j=
1 j
y j
And finally,
yˆ [1]
n
j=
1 j
y j
(c) By (b), we have
yˆ [1]
n
j=
1 j
y j
11
And then,
y 1 − ˆy [1]
y 1
11
y 1
n
j=
1 j
y j
11
y 1
n
j=
1 j
y j
y 1 − x
t
1
β