quadratic mean differentiability example
TRANSCRIPT
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8/18/2019 Quadratic mean differentiability Example
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g(x) = 1
2 exp(−|x|) g(x − θ)
g(x − θ)
sθ+h(x)
−sθ(x) =
1
0
1
2
h
·sign(x
−θ
−uh) g(x −
θ
−uh)du
sθ(x) =
gθ(x) sθ(x) x
θ
θ = x
ṡθ(x) := 2
− 32 sign(x − θ)exp( |x−θ|
2 )
sθ(x)
sθ(x) θ = x sx(x) := 0
(δ n) ⊆ R δ n → 0 (hn) ⊆ R |hn| = 1
hn → h hn
f n := sθ+δnhn − sθ
δ n
f := hṡθ.
(f n − f )2dx → 0
(f n(x) − hnṡθ(x))2dx =
(f n(x) − f (x) + (hn − h)sθ(x))2dx
≤ 2
(f n(x) − f (x))2dx + 2
((hn − h)ṡθ(x))2dx
((hn − h)ṡθ(x))2dx hn − h
f n(x) f (x) x = θ
f 2n = 1
δ 2n(sθ+δnhn − sθ)2
= 1
δ 2n
10
1
2δ nhn · sign(x − θ − uδ nhn)
g(x − θ − uδ nhn)du
2≤ 1
4
10
g(x − θ − uδ nhn)du
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8/18/2019 Quadratic mean differentiability Example
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8/18/2019 Quadratic mean differentiability Example
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||∆||2 :=
∆(x)2dx 1
2
L2
h
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8/18/2019 Quadratic mean differentiability Example
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ṡθ(z )
I θ
I θ =
R×{0,1}
f X (x)f
xT θ2 y
F 1
2 (xT θ)+
1 − y(1 − F (xT θ)) 12
2xxT dµ(x, y)
= R
f X (x)f xT θ2 1F (xT θ)
+ 11 − F (xT θ)
xxT dx=
R
f X (x) f
xT θ
2F (xT θ)(1 − F (xT θ))
≤M
xxT dx
E(XX T ) I θ
(f λ)λ>0 f λ(x) = λe−λx x ≥ 0 f λ(x) = 0
x
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8/18/2019 Quadratic mean differentiability Example
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I (λ0) = E
(1 − xλ0)2
4λ20
=
1
4λ20> 0
√ n(λ̂n − λ0)
4λ20