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i = 1,
αiyi = 0. Let
αiyi = 0. In other words, α0 corresponds to
, n+1} and α� corresponds to the support vector
αiyixi� 2. Let α0 = argmaxαw(α) subject to αi ≥ 0 and �
αiyi = 0, and ψ = �
αiyixi. Since the
�
�
�
1
2
1
1
classifier trained from {(xi, yi) : i = 1,
· · ·
, p − 1, p + 1,
· · ·
1
↓
· · · , ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
the margin m(α0),
7
p ·
Lecture 04
Generalization error of SVM.
18.465
and w(α�) ≤ w(α0). On the other hand, the hyperplane determined by α0 − α0 γ might not separate (xi, yi)
for i =� p and corresponds to a equivalent or larger “margin” 1/�ψ(α0 − α0 γ)� than m(α�)).
p ·
p ·
Let us consider the inequality
max... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
yp · ψ� ∗ xp)2
.
�xp�2
2
For the right hand side,
w(α0 − αp
0 · γ) =
=
�
0 − αp
αi
0 −
1 �
�
2
�
0 yixi −α0 ypxp�2
αi
p
��
ψ0
�
�
i − α0
α0
p −
1
�ψ0�2 + αp
2
�
�2
α0
p
1
2
�xp�2
0 ypψ0 ∗ xp −
�
αp
0�2
1
2
�xp�2
= w(α0) − α0
p(1 − yp · ψ0 ∗ xp) −
1 �
2
= w(α0) −
�xp�2 .
0�2 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
18.465
For a fixed f ∈ F, if we observe 1
n
the Law of Large Numbers,
�
n I (f (Xi) = Yi) is small, can we say that P (f (X) = Y ) is small? By
i=1
�
n1 �
n
i=1
I (f (Xi) = Yi) EI(f (X) = Y ) = P (f (X) = Y ) .
→
�
�
The Central Limit Theorem says
n � �
√
1
n
n
i=1
I (f (Xi) = Yi) − EI(f (X) = Y ) �
√
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, EZ 2 = σ2 , |Z| < M = const, Z1, · · · , Zn independent copies of
Z, and t ≥ 0. Then
where φ(x) = (1 + x) log(1 + x) − x.
i=1
Proof. Since Zi are i.i.d.,
�
n
�
P
�
�
Zi ≥ t ≤ exp −
�
nσ2
M 2
φ
��
tM
nσ2
,
�
n
�
P
�
Zi ≥ t ≤ e−λtEe λ
i=1 Zi = e−λt
P
n
n
�
Ee λZi = e−λt Ee λZ
�
�n
i=1
i=1
.... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤ e−λt exp
�
nσ2
M 2
�
�
e λM − 1 − λM
�
�
= exp −λt +
nσ2
M 2
�
e λM − 1 − λM �
�
Now, minimize the above bound with respect to λ. Taking derivative w.r.t. λ and setting it to zero:
−t +
nσ2
M 2
�
M eλM − M = 0
�
e λM =
+ 1
tM
nσ2
�
log 1 +
�
.
tM
nσ2
λ =
1
M
The bound becomes
�
n
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
copies of X, and t ≥ 0. Then
�
n
�
P
i=1
�
�
Xi ≥ t ≤ exp −
�
nσ2
M 2
φ
��
tM
nσ2
,
where φ(x) = (1 + x) log(1 + x) − x.
2
If X is small, φ(x) = (1 + x)(x − x
2 + · · · ) − x = x + x2 − x
2 − x + · · · = x
2 + · · · .
2
2
If X is large, φ(x) ∼ x log x.
We can weaken the bound by decreasing φ(x). Ta... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
� e−u
≥ 1 − e−u
i=1
√
b,
�
P
n
�
i=1
Xi ≤
√
2nσ2u +
�
2uM
3
≥ 1 − e−u
For non-centered Xi, replace Xi with Xi − EX or EX − Xi. Then |Xi − EX| ≤ 2M and so with high
probability
Normalizing by n,
and
�
(Xi − EX) ≤
√
2nσ2u +
4uM
3
.
1 �
n
Xi − EX ≤
�
2σ2u
n
+
4uM
3n
EX −
1 �
n
Xi ≤
�
2σ2u
n... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
In Bernstein’s inequality take ��Xi
i
�� to
1
�
�
�
EI(f (Xi) = Yi) −
n1 �
n
i=1
I (f (Xi) = Yi) ≤
�
�
2P (f (Xi) = Yi) (1 − P (f (Xi) = Yi))u
n
�
�
+
2u
3n
because EI(f (Xi) = Yi) = P (f (Xi) = Yi) = EI 2 and therefore Var(I) = σ2 = EI 2 − (EI)2 . Thus,
�
�
P (f (Xi) �= Yi) ≤
I (f (Xi) �= Yi) +
n1 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
t ≥ 0,
�
P
n
�
i=1
�
�
εiai ≥ t ≤ exp −
�
t2
�
.
n
2
2
i=1 ai
Proof. Similarly to the proof of Bennett’s inequality (Lecture 5),
�
n
�
ε
P
i=1
�
t≥a
i
i
≤
�
λ
exp
−
e
λt
E
� �
n
ε
i
a
i
n
= e−λt �
E
exp (
λε
i .
a
)
i
i=1
i=1
Using inequality
x
−x
+e
e
2
2
x≤ e
/2
(from Taylor expansion),... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
− e−u
i
.
�
n
�
i=1 ai = Var( n
2
i=1
i=1
i=1
εiai).
Rademacher sums will play important role in future. Consider again the problem of estimating
Ef .
We will see that by the Symmetrization technique,
1
n
n
�
i=1
f (Xi) − Ef ∼
1
n
n
�
i=1
f (Xi) −
1
n
n
�
f (Xi
�).
i=1
�
n
i=1
1
n
f (Xi)−
I... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
i=1
n
�
f (Xi
�) − Ef
i=1
�
�
�
�
�
13
Lecture 07
Hoeffding, Hoeffding-Chernoff, and Khinchine inequalities.
18.465
while for the first inequality we use Jensen’s inequality:
�
�
1
E
�
�
�
n
n
�
i=1
�
�
f (Xi) − Ef
�
�
�
�
�
1
= E
�
�
�
n
n
�
f (Xi) −
1
n
n
�
i=1
Ef (Xi
�
�
�)
�
�
�
≤ EX EX � ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= eλ(x·1+(1−x)·0) ≤ xeλ + (1 − x)eλ·0 = 1 − x + xeλ. Hence,
Ee λX = 1 − EX + EXeλ = 1 − µ + µe λ .
Again, we minimize the following bound with respect to
λ >
0:
�
P
n
�
X ≥ n µ
(
)+ t
i
�
i=1
Take derivative w.r.t. λ:
e−λn(µ+t)Ee λ Xi
P
e−λn(µ+t) Ee λX
�
�n
�
e−λn(µ+t) 1 − µ + µe λ
�n
≤
=
≤
−n(µ + ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
��, and Khinchine inequalities.
18.465
completing the proof. Moreover,
�
P µ −
1
n
n
�
�
Xi ≥ t = P
i=1
�
1
n
n
�
i=1
�
Zi − µZ ≥ t ≤ e−nD(µz +t,µZ ) = e−nD(1−µX +t,1−µX )
where Zi = 1 − Xi (and thus µZ = 1 − µX ).
�
If 0 < µ ≤ 1/2,
Hence, we get
Solving for t,
D(1 − µ + t, 1 − µ) ≥
t2
.
2µ(1 − µ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
i = 1) = P(�i = −1) = 0 5, and 0 < p < ∞. Then
.
· · ·
, an ∈ R, �i,
· · ·
, �n be i.i.d. Rademacher random variables:
�
n
�
i=1
�1/2
2
≤
� �
n
�
�
E
�
�
�
i=1
�
p�1/p
�
�
ai�i
�
�
≤ Bp ·
2
|ai|
�
n
�
i=1
�1/2
Ap ·
|ai|
for some constants Ap and Bp depending on p.
Proof. Let
�
|ai| 2 = 1 wi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
�
� �
2
≤ E
�
�
ai�i
�
�
3
p+(2− 2
2
�
� �
E
�
�
ai�i
�
�
=
3
p)
≤
≤
� � �
2
� �
� �
� �
p 3
6−2p
E
E
�
�
�
�
ai�i
ai�i
�
�
�
�
�
p 2
� �
� �
3
E
�
�
·
ai�i
�
�
(B6−2p
)2− p
2
3
.
Thus E | ai�i|
p ≤ (B6−2p)6−2p, completing the proof.
�
�
1
3
, Holder’s inequality
�
16
Lecture 08 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
of sets
uniformGC(F).
Let C = {C ⊆ X}, fC (x) = I(x ∈ C). The most pessimistic value is
sup E sup Pn (C) − P (C)
P
C∈C
|
| →
0.
For any sample {x1, . . . , xn}, we can look at the ways that C intersects with the sample:
{C ∩ {x1, . . . , xn} : C ∈ C}.
Let
�n(C, x1, . . . , xn) = card {C ∩ {x1, . . . , xn} : C... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
V.
Hence, C will pick out only very few subsets out of 2n (because V
�
en
�V
∼ nV ).
Lemma 8.3. The number �n(C, x1, . . . , xn) of subsets picked out by C is bounded by the number of subsets
shattered by C.
Proof. Without loss of generality, we restrict C to C := {C ∩ {x1, . . . , xn} : C ∈ C}, and we have card(C... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
means that xi ∈ Ti(C), and that C\{xi} ∈ C. Thus both B {xi} and B\{xi} are picked out by C.
Since either B = B {xi} or B = B\{xi}, B is picked out by C. Thus A is shattered by C.
Apply the operator T = T1 ◦ . . . Tn until T k+1(C) = T k(C). This will happen for at most
�
C∈C
card(C)
times, since
card(Ti(C)... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
is at most d.
Proof. Observation: For any {x1, . . . , xd+1} if we cannot shatter {x1, . . . , xd+1} ←→ ∃I ⊆ {1 . . . d + 1} s.t.
we cannot pick out {xi, i ∈ I}. If we can pick out {xi, i ∈ I}, then for some C ∈ C there are α1, . . . , αd s.t.
�d
�d
k=1 αkfk(x) > 0 for i ∈ I and
k=1 αkfk(x) ≤ 0 for i /∈ I.
Denot... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
d
�
αkfk(x1) + . . . + φd+1
d
�
αkfk(xd+1) = 0.
φ1
k=1
k=1
Hence,
Contradiction.
d
�
� �
φi
�
k=1
i∈I
�
�
i /∈I
αkfk(xi) =
��
>0
�
(−φi)
� �� �
≥0 � k=1
αkfk(xi)
��
≤0
�
d
�
�
.
�
�
• Half-spaces in Rd: {{α1x1 + . . . + αdxd + αd+1 > 0} : α1, . . . , αd+1 ∈ R}.
By setting f1 = x1, .... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
. . . , xn) ≤ �n(C, x1, . . . , xn)�n(C ∩ D, x1, . . . , xn)
� �V �
�V
≤
en
V
C
en
V
C
D D
≤ 2n
for large enough n.
(b) (C ∪ D) = (C c ∩ Dc)c, and the result follows from (1) and (2).
Example 9.1. Decision trees on Rd with linear decision rules: {C1 ∩ . . . C�} is VC and
leaves{C1 ∩ . . . C�}
is VC.
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
sup
�
�
n
C∈C
�
occurs. Let X = (X1, . . . , Xn) ∈ {supC∈C
�
1
n
n
�
I(Xi ∈ C) − P (C) ≥ t
�
�
�
�
�
�
�
I(Xi ∈ C) − P (C)
i=1
�
n
i=1
n
�
�
�
1
�
�
�
n
i=1
Then
≥ t}.
�
�
I(Xi ∈ CX ) − P (CX )
�
�
�
≥ t.
∃CX such that
For a fixed C,
PX �
��
�
1
�
�
�
n
n
�
i=1
I(Xi
�
� ∈ C) − P (C) ≥ t/2... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
1
�
�
�
n
n
�
i=1
I(Xi
� ∈ CX ) − P (CX ) ≤ t/2 ∃CX
≥ 1/2
�
�
�
�
�
�
�
�
�
�
�
2/n. Assume that the event
�
�
1
�
�
�
n
n
�
i=1
I(Xi
� ∈ CX ) − P (CX ) ≤ t/2
�
�
1
�
�
�
n
n
�
i=1
I(Xi ∈ CX ) − P (CX ) ≥ t.
�
�
�
�
�
�
�
�
�
�
occurs. Recall that
Hence, it must be that
�
�
1
�
�
�
n... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
n
C∈C
n
�
i=1
I(Xi ∈ C) −
1
n
Theorem 10.1.
If
V C
C
( ) =
V
, then
�
�
�
�
�
�
I(Xi ∈ C) − P (C) ≥ t
n
�
i=1
I(Xi
�
�
� ∈ C) ≥ t/2
�
�
�
�
.
P
Proof.
�
�
�
1
�
sup
�
�
n
C∈C
n
�
i=1
�
I(Xi ∈ C) − P (C) ≥ t
�
�
�
�
�
�
�V
2en
V
≤ 4
2nt
8
e−
.
I(Xi ∈ C) −
�
n1
n
i=1
I(Xi
εi (I(Xi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
names (i.e. introducing
The first equality is due to the fact that Xi and Xi
random signs εi, P (εi = ±1) = 1/2) does not have any effect. In the last line, it’s important to see that the
probability is taken with respect to εi’s, while Xi and Xi
By Sauer’s lemma,
�’s are fixed.
�2n (C, X1, . . . , Xn, X1
� , . . . , ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
∈ Ck)) ≥ t/2
�
�
�
�
�
≤ 2E
2 exp
−
N
�
k=1
Hoeffding’s inequality
−n2t2
(I(Xi ∈ C) − I(Xi
�
n
8
i=1
� ∈ C))2
N
� �
2 exp −
≤ 2E
k=1
�
2t2
−n
8n
�
�V
2en
V
≤ 2
2nt
2e− 8 .
�
2en
V
�V
. Hence,
�
23
Lecture 11
Optimistic VC inequality.
18.465
Last time we proved the Pessimistic VC in... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= ±1. These examples are labeled according to some unknown C0 such
that Y = 1 if X = C0 and Y = 0 if X /∈ C0.
Let C = {C : C ⊆ X }, a set of classifiers. C makes a mistake if
Similarly to last lecture, we can derive bounds on
X ∈ C \ C0 ∪ C0 \ C = C�C0.
�
�
n1
�
�
sup
�
�
n
C
i=1
I(Xi ∈ C�C0) − P (C�C0)
�
�
�
�... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≥
1
n
� ∈/ C) = (1 − P (C))n can be as close to 0 as we want.
n I(Xi
since
i=1
1
n
n P (Xi
i=1
n
�
i=1
�
� ∈ C) ≥ nP (C) ≥ 1. Otherwise P (
�
n
i=1 I(Xi
� ∈ C) = 0) =
Similarly to the proof of the previous lecture, let
�
(Xi) ∈
sup
C
P (C) −
I(Xi ∈ C)
�
n
1
i=1
n
�
P (C)
�
.
≥ t
24
Lecture 11
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
i=1
�
εi (I(Xi
�
�
� �
1
t
�
� ∈ Ck) − I(Xi ∈ Ck)) ≥ √
n
2
�
i=1
n
≤ E
exp
−
N
�
k=1
�
t2 1
n
1
2 2
2
n
n
i=1
�
(I(Xi ∈ Ck) + I(Xi
(I(Xi
� ∈ Ck))
� ∈ Ck) − I(Xi ∈ Ck))2
(I(Xi ∈ Ck) + I(Xi
⎞
� ∈ Ck))⎠
since upper sum in the exponent is bigger than the lower sum (compare term-by-term)
≤ E
2
e... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
C ∈ C}. Then F(C) is VC-subgraph class if and only
if C is a VC class of sets.
Assume d functions are fixed: {f1, . . . , fd} : X �→ R. Let
�
d
�
F =
αifi(x) : α1, . . . , αd ∈ R
�
.
Then V C(F) ≤ d + 1. To prove this, it’s easier to use the second definition.
i=1
Packing and covering numbers
Let f, g ∈ F and... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤ i ≤ N such that d(f, fi) ≤ ε.
Definition 12.5. The ε-covering number, N (F, ε, d), is the minimal cardinality of an ε-cover of F.
27
Lecture 12
VC subgraph classes of functions. Packing and covering numbers.
18.465
Lemma 12.1.
D(F, 2ε, d) ≤ N (F, ε, d) ≤ D(F, ε, d).
Proof. To prove the first inequality, assume t... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Example 12.3. Consider the L1-ball {x ∈ Rd , |x| ≤ 1} = B1(0) and d(x, y) = |x − y|1. Then
D(B1(0), ε, d) ≤
�
2 + ε
ε
�d � �d
≤
,
3
ε
where ε ≤ 1. Indeed, let f1, . . . , fD be optimal ε-packing. Then the volume of the ball with ε/2-fattening
(so that the center of small balls fall within the boundary) is
More... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
ε
log
�V
7
ε
.
for some δ.)
(which is ≤ K
ε
�
�V +δ
Proof. Let m = D(F, ε, d) and f1, . . . , fm be ε-separated, i.e.
1
n
n
�
|fr(xi) − f�(xi)| > ε.
i=1
Let (z1, t1), . . . , (zk, tk) be constructed in the following way: zi is chosen uniformly from x1, . . . , xn and ti is
uniform on [−1, 1].
Consider fr... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
graph classes.
18.465
Substituting,
P (Cfr and Cf� pick out different subsets of (z1, t1), . . . , (zk, tk))
= 1 − P ((z1, t1) is picked by both Cfr , Cf� or by neither)k
�
≥ 1 − e−ε/2
�k
= 1 − e−kε/2
There are 2 ways to choose fr and f�, so
� �
m
P (All pairs Cfr and Cf� pick out different subsets of (z1, t1),... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
e
ε
log m 1/V
s ≤
4e
ε
log s.
Note that
s
log s
is increasing for s ≥ e and so for large enough s, the inequality will be violated. We now
e log 7
check that the inequality is violated for s� = 8
ε
� �2
7
ε
4e
ε
log
>
4e
ε
�
log
8e
ε
log
�
7
ε
ε . Indeed, one can show that
since
Hence, m1/V = s ≤ ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
that
(1) ∀f, g ∈ Fj , d(f, g) > 2−j
(2) ∀f ∈ F , we can find g ∈ Fj such that d(f, g) ≤ 2−j
How to construct Fj+1 if we have Fj :
• Fj+1 := Fj
• Find f ∈ F , d(f, g) > 2−(j+1) for all g ∈ Fj+1
• Repeat until you cannot find such f
Define projection πj : F �→ Fj as follows: for f ∈ F find g ∈ Fj with d(f, g) ≤ 2−j an... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
j )2
e−u
j=1
�
≥ 1 −
1
22
+
1
32
+
1
42
�
e−u
= 1 − (π2/6 − 1)e−u ≥ 1 − e−u
Recall that R(f ) = ∞ R(πj (f ) − πj−1(f )). If f is close to 0, −2k+1 < d(0, f ) ≤ 2−k . Find such a k.
�
j=1
Then π0(f ) = . . . = πk(f ) = 0 and so
33
Lecture 14
Kolmogorov’s chaining method. Dudley’s entropy integral.
18.46... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, d)dε
�
��
�
Dudley’s entropy integral
since 2−(k+1) < d(0, f ).
�
34
log D1/2−(κ+2)22−(κ+1) Lecture 15
More symmetrization. Generalized VC inequality.
18.465
Lemma 15.1. Let ξ, ν - random variables. Assume that
P (ν ≥ t) ≤ Γe−γt
where Γ ≥ 1, t ≥ 0, and γ > 0. Furthermore, for all a > 0 assume that
Eφ(ξ) ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
)+
=
·
Γ e e−γt
·
1
= Γ · e · e−γt
where we chose optimal a = t − 1
γ to minimize Γe−γa
γ
.
�
35
(x-a)+a Lecture 15
More symmetrization. Generalized VC inequality.
18.465
Lemma 15.2. Let x = (x1, . . . , xn), x� = (x1
� , . . . , xn
� ). If for functions ϕ1(x, x�), ϕ2(x, x�), ϕ3(x, x�)
then
�
�
P ϕ1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
� ϕ2 + Ex� ϕ3t} = {sup(Ex� ϕ1 − Ex� ϕ2 − δEx� ϕ3)4δ ≥ t}.
δ>0
�
�
��
ξ
By assumption, P (ν ≥ t) ≤ Γe−γt . We want to prove P (ξ ≥ t) ≤ Γ e e−γt . By the previous lemma, we
only need to check whether Eφ(ξ) ≤ Eφ(ν).
·
·
ξ = sup Ex� (ϕ1 − ϕ2 − δϕ3)4δ
δ>0
≤ Ex� sup(ϕ1 − ϕ2 − δϕ3)4δ
δ>0
= Ex� ν
φ(ξ) ≤ φ(Ex� ν)... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x� ))))2
i
i
.
�1/2
n
1 �
n
i=1
In Lecture 14, we proved
�
Pε ∀f ∈ F,
n
1 �
n
i=1
εi(f (xi) − f (x�
i)) ≤ √
29/2 � d(0,f )
n 0
log1/2 D(F, ε, d)dε
Complement of the above is
�
�
Pε ∃f ∈ F,
εi(f (xi) − f (xi
� )) ≥ √
Taking expectation with respect to x, x�, we get
P ∃f ∈ F,
εi(f (xi) − f (xi
� )) ≥ √
+2... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
To the right of ”≥” sign, only distance d(f, g) depends
By Lemma 15.2 (minus technical detail ”∃f ”),
�
P ∃f ∈ F, Ex�
n
1 �
n
i=1
(f (xi) − f (x� )) ≥ Ex� √
i
log1/2 D(F, ε, d)dε
� d(0,f )
29/2
n 0
�
+27/2
�
, f )2t
≤ e · e−t ,
Ex� d(0
n
where
and
Ex�
1
n
n
�
(f (xi) − f (xi
� )) =
i=1
1
n
n
�
f... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x1, . . . , xn), D(F, ε, dx) ≤ D(F, ε)
where dx(f, g) =
� �
n
1
i=1
n
(f (xi) − g(xi))2
�1/2
Lemma 16.1. If F satisfies uniform entropy condition, then
� d(0,f )
Ex�
0
log1/2 D(F, ε, d)dε ≤
E
x� d(0,f )2
� √
0
log1/2 D(F, ε/2)dε
Proof. Using inequality (a + b)2 ≤ 2(a2 + b2),
n
1 �
n
i=1
�
�
d(f, g) =
≤... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, ε, d) ≤ D(F, ε/2, dx,x� ).
38
Lecture 16
Consequences of the generalized VC inequality.
18.465
� d(0,f )
Ex�
0
log1/2 D(F, ε, d)dε ≤ Ex�
� d(0,f )
log1/2 D(F, ε/2, dx,x� )dε
0
� d(0,f )
≤
log1/2 D(F, ε/2)dε
0
Let φ(x) = x log1/2 D(F, ε)dε. It is concave because φ�(x) = log1/2 D(F, ε/2) is decreasing w... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
.1. If F satisfies Uniform Entropy Condition and F = {f : X → [0, 1]}. Then
� √
�
�
2Ef
�
P ∀f ∈ F, Ef −
log1/2 D(F, ε/2)dε + 27/2
≥ 1 − e−t .
n
1 �
n
i=1
f (xi) ≤ √
29/2
n
0
2Ef t
·
n
Proof. If Ef ≥ 1
n
�
n f (xi), then
i=1
�
2 max Ef,
1
n
n
�
�
f (xi) = 2Ef.
i=1
If Ef ≤ 1
n
�
n f (xi),
i=1
and ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
F with V C(F) = V , where d(f, g) = d1(f, g) = n
|f (xi) − g(xi)|.
e log 7
8
ε
ε
n
i=1
�V
�
Note that if f, g : X �→ [0, 1], then
d2(f, g) =
�
1
n
n
�
i=1
�
�1/2 �
�1/2
(f (xi) − g(xi))2
≤
|f (xi) − g(xi)|
.
1
n
n
�
i=1
Hence, ε < d2(f, g) ≤ d1(f, g) implies
D(F, ε, d2) ≤ D(F, ε2, d1) ≤
�
8e
ε2... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤ 2 log
1
x
1
x
x ≤ 1/e
1 ≤ log
Now, check for x ≥ 1/e.
� �
x
1
log dε =
ε
� 1 �
e
� �
x
1
log dε +
ε
1
log dε
ε
0
2
e
+
≤
�
x
1
e
1dx
= + x − = x +
≤ 2x
2
e
1
e
1
e
1
e
0
Using the above result, we get
�
P ∀f ∈ F, Ef −
n1 �
n
i=1
f (xi) ≤ K
�
α
1
Ef log Ef
n
�
�
tEf
n
+ K
≥... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
F = conv H =
�
T
�
λihi, hi ∈ H,
T
�
λi ≤ 1, λi ≥ 0, T ≥ 1
�
.
Then sign(f (x)) is the prediction of the label y. Let
i=1
i=1
Fd = convd H =
�
d
�
λihi, hi ∈ H,
T
�
λi ≤ 1, λi ≥ 0
�
.
Theorem 17.1. For any x = (x1, . . . , xn), if
i=1
i=1
then
log D(H, ε, dx) ≤ KV log 2/ε
log D(convd H, ε, dx) ≤ KV... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Hence, we can approximate any f ∈ Fd by f � ∈ FD,d within ε.
Now, let f = �d
i=1 λihi ∈ FD,d and consider the following construction. We will choose Y1(x), . . . , Yk(x)
i=1
i=1
from h1, . . . , hd according to λ1, . . . , λd:
P (Yj (x) = hi(x)) = λi and P (Yj (x) = 0) = 1 −
d
�
λi.
Note that with this construc... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
�
Yj − f
�
�
�
x
⎛
1
= Edx k
⎝
k
�
⎞
2
Yj , f ⎠ ≤ ε2 .
j=1
So, there exists a deterministic combination 1 �k Yj such that dx( 1 �k Yj , f ) ≤ ε.
Define
j=1
j=1
k
k
D,d F � =
⎧
⎨ 1
⎩
k
⎫
⎬
Yj : k = 4/ε2 , Yj ∈ {h1, . . . , hd} ⊆ {h1, . . . , hD}
⎭
k
�
j=1
Hence, we can approximate any f =
λihi ∈ F... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
number of such strings is
�
�
k+d−1 . Hence,
k
card F �
D,d ≤
≤
=
=
�
×
� � �
k + d
k
D
d
DD−dDd
dd(D − d)D−d
D(k + d) �d �
�
(k + d)k+d
kkdd
D �D−d �
d2
D − d
�
D(k + d) �d �
d2
1 +
D − d
k + d �k
k
d �D−d �
1 +
d �k
k
using inequality 1 + x ≤ e x
�
≤
�d
D(k + d)e2
d2
where k = 4/ε2 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
integral.
Theorem 18.1. Let (X , A, µ) be a measurable space, F ⊂ {f |f : X → R} be a class of measurable func
tions with measurable square integrable envelope F (i.e., ∀x ∈ X , ∀f ∈ F, |f (x)| < F (x), and �F �2 =
�
� �V
( F 2dµ)1/2 < ∞), and the �-net of F satisfies N (F, ��F �2, � · �) ≤ C 1
�
a constant K that... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
upper bounded (supk Ck ∨ Dk < ∞), such that
(18.1)
log N (convFn·kq , CkL · n−W , � · �2) ≤ Dk · n
for n, k ≥ 1, and q ≥ 3 + V . This implies the theorem, since if we let k → ∞, we have log N (convF, C∞L ·
n−W
, �·�2) ≤
�
��F �2, n = C
D
∞
∞
n. Let � = C∞C 1/V n−W , and K = D∞C V +2 C V +2 , we get C∞L n−W = C∞C ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
trivially. For general n, fix m = n/d for large enough
1 m− 1 �F � = Lm− 1
d > 1. For any f ∈ Fn, there exists a projection πmf ∈ Fm such that �f − πmf � ≤ C
0
V
V
V
by definition of Fm. Since
f ∈Fn
λf · f =
f ∈Fm
µf · f +
f ∈Fn
λf · (f − πmf ), we have convFn ⊂
�
�
�
convFm + convGn, and the number of ele... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
. It follows that EYi =
Y1, · · · , Yk be i.i.d. random variables such that P (Yi = fj ) =
�
λj fj for all i = 1,
λj for all j = 1,
· · ·
⎛
1
E ⎝
k
n
k
� �
Yi −
⎞
λj fj ⎠ ≤
i=1
j=1
⎛
1
E ⎝Y1 −
k
n
�
j=1
· · ·
, k, and
⎞
λj fj ⎠ ≤
1
k
(diamF)2 .
Thus at least one realization of 1 �k Yi has a dis... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(convGn
, �diamGn, � · �2) ≤
·
e + en
�
1
16
2 m 2/V n−2W
·
C1
�32 C1
·
−2 m 2/V n 2·W
�
=
e
e + C1
16
2d−2/V
�32 C−2d2/V n
1
·
By definition of Fm and and induction assumption, log N (convFm, C1L · m−W , � · �2) ≤ D1 · m. In other
words, the C1L m−W -net of convFm contains at most eD1m elements. This define... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
k > 1, construct Gn,k such that convFnkq ⊂ convFn(k−1)q + convGn,k in a similar way as before.
Gn,k contains at most nkq elements, and each has a norm smaller than L (n (k − 1)q)−1/V . To bound the
cardinality of a Lk−2n−W -net, we set � 2L (n (k − 1)q)−1/V = Lk−2n−W , get � = 1 n−1/2 (k − 1)q/V k−2 ,
46
·
2
Lectur... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
��er y = sign(f (x)) with minimum testing error. For any x, the term y f (x) is called margin can
·
be considered as the confidence of the prediction made by sign(f (x)). Classifiers like SVM and AdaBoost
are all maximal margin classifiers. Maximizing margin means, penalizing small margin, controling the
complexity of al... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
nition of dz
�1/2
�1/2
n
1 �
n
i=1
�
1
δ
n
1 �
n
i=1
≤
=
(yif (xi) − yi · g(xi))2
,Lipschetz condition
1
δ
dx(f (x), g(x)) ,definition of dx,
and the packing numbers for φδ(yF) and F satisfies inequality D(φδ(yF), �, dz) ≤ D(F, � δ, dx).
Recall that for a VC-subgraph class H, the packing number satisfies D(H, ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
H. The packing number satisfies
�V
� log k
�
D(H, �, dx) ≤ k
number: D(H, �, � · �1) ≤ � k
�V .
�
Since conv(H) satisfies the uniform entroy condition (Lecture 16) and f ∈ [−1, 1]X , with a probability
. D Haussler (1995) also proved the following two inequalities related to the packing
, and D(H, �, dx) ≤ K � 1
�... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
K · (Enφδ + n− 2 V +1 δ− V +1 +
1 V +2
V
u
)
n
for some constant K. We proceed to bound Eφδ for δ ∈ {δk = exp(−k) : k ∈ N}. Let exp(−uk) =
�
�2
1
k+1
e−u
, it follows that uk = u + 2
�
k∈N exp(−uk) = 1 −
�
1 −
�
k∈N k+1
1
�2
1
log(log δ
·
k
e−u = 1 − π2
6 ·
e−u < 1 − 2 e−u ,
·
·
log(k + 1) = u +... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
2 e−u ,
�
n
1
i=1 φδ(yi · f (xi)) ≤ n
�
n
·
·
·
1
·
P(y · f (x)) ≤ 0) ≤ K · inf
δ
Pn(y · f (x) ≤ δ) + n− 2(V +1) δ− V +1 +
V
V +2
�
2 log(log 1 + 1)
n
δ
�
.
u
n
+
49
Lecture 20
Bounds on the generalization error of voting classifiers.
18.465
As in the previous lecture, let H = {h : X �→ [−1, 1]}... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
the [0, 1] interval,
parametrized by a and b.
Then V C(H) = 2. Let F = conv H. First, rescale the functions: f = �T λihi = 2 �T λi
− 1 =
2f � −1 where f � = �T λih�
i, h� = hi+1 . We can generate any non-decreasing function f � such that f �(0) = 0
and f �(1) = 1. Similarly, we can generate any non-increasing f � ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
x)) −
1
n
�
I(yif (xi) ≤ δ) + Eϕδ (yf (x)) −
n
�
i=1
�
�
ϕδ (yif (xi))
1
n
n
�
i=1
ϕδ (yif (xi))
By going from 1
n
�
n
i=1 I(yif (xi) ≤ 0) to 1
n
�
n
i=1 I(yif (xi) ≤ δ), we are penalizing small confidence predic
tions. The margin yf (x) is a measure of the confidence of the prediction.
For the sake of... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
on the generalization error of voting classifiers.
18.465
we have
dx,y (ϕδ (yf (x)) , ϕδ (yg(x))) =
≤
�
�
(ϕδ (yif (xi)) − ϕδ (yig(xi)))2
�1/2
(yif (xi) − yig(xi))2
�1/2
n1 �
n
i=1
n
1 1 �
δ2 n
i=1
�
=
1
δ
n
1 �
n
i=1
�1/2
(f (xi) − g(xi))2
where f, g ∈ Fd.
Choose ε δ-packing of Fd so that
·
= dx(f, ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
error of voting classifiers.
18.465
We continue to prove the lemma from Lecture 20:
Lemma 21.1. Let Fd = convd H = {
λihi, hi ∈ H} and fix δ ∈ (0, 1]. Then
�d
i=1
�
ϕ¯δ
Eϕ
−
√δ
E
ϕδ
≤ K
��
dV log n
n
δ +
� ��
t
n
≥ 1 − e−t .
P ∀f ∈ Fd,
Proof. We showed that
log D(ϕδ (yFd) , ε/2, dx,y ) ≤ KdV log
2
. ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= x, ε = 2x . Without loss of generality, assume Eϕδ ≥ 1/n.
εδ
E
Otherwise, we’re doing better than in Lemma: √
E
≤
n ⇒ E ≤ log
n
� δ
log
n
Eϕδ
k � √
√
n 0
log1/2 D(ϕδ (yFd(x)) , ε)dε ≤ K
�
�
n . Hence,
√
log
dV Eϕδ
n
Eϕδ log
n
dV
δ
n
δ
2
n
So, with probability at least 1 − e−t ,
≤ K
Eϕδ (yf (x)) −... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
δ
∀f ∈ Fd,
. . . +
≥ 1 −
�
δ,d
e−t 6δ
d2π2
= 1 − e−t .
53
� �
�
�⎞
⎠
tδ,d
n
Lecture 21
Bounds on the generalization error of voting classifiers.
18.465
Since tδ,d = t + log d2π2
6δ ,
∀f ∈ Fd,
Eϕδ − ¯ϕδ
Eϕδ
√
≤ K
≤ K
≤ K �
�
⎛
⎝
��
�
dV log n
δ
n
+
t + log d2π2
6δ
n
⎞
⎠
dV log n... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
2
Eϕδ ≤ + �
2
2
+
n
1 �
n
i=1
ϕδ
Eϕδ ≤ 2
� ε �2
2
n
1 �
+ 2
n
ϕδ.
i=1
The bound becomes
⎛
⎜ n
⎜ 1 �
P (yf (x) ≤ 0) ≤ K ⎜
⎝ n
i=1
⎞
I(yif (xi) ≤ δ) +
⎟
n
t
dV
⎟
log + ⎟
n
n ⎠
δ
� �� �
(*)
where K is a rough constant.
(*) not satisfactory because in boosting the bound should get better when t... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x) ≤ δ) +
�
��
inc. with δ
�
V min(T, (log n)/δ2) log n
n
��
dec. with δ
⎞
t ⎟
δ + ⎟
⎠ .
n
�
Proof. Let f = �
T
=1 λihi, g = k
i
1 �
k
=1 Yj , where
j
P (Yj = hi) = λi and P (Yj = 0) = 1 −
T
�
λi
as in Lecture 17. Then EYj (x) = f (x).
i=1
P (yf (x) ≤ 0) = P (yf (x) ≤ 0, yg(x) ≤ δ) + P (yf (x) ≤... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
+
�
δ
�
2 � EYj
� ≤
⎞
1
⎠
2
≤ (by Hoeffding’s ineq.) Exe−kD(EY1
δ
�+ 2 ,EY1
�)
because D(p, q) ≥ 2(p − q)2 (KL-divergence for binomial variables, Homework 1) and, hence,
≤ Exe−kδ2/2 = e−kδ2/2
�
D EY1
� +
δ
, EY1
2
�
� ≥ 2
� �2
δ
2
= δ2/2.
We therefore obtain
(22.1)
P (yf (x) ≤ 0) ≤ P (yg(x) ≤ δ) + e... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x))
� �
t
n
+
V k log n
δ
n
≤ K
= ε/2.
y is increasing with x and decreasing with y.
Note that Φ(x, y) = x√−
x
By inequalities (22.1) and (22.2),
and
Eϕδ (yg(x)) ≥ P (yg(x) ≤ δ) ≥ P (yf (x) ≤ 0) −
1
n
1
n
n
�
ϕδ (yig(xi)) ≤ Pn (yg(x) ≤ 2δ) ≤ Pn (yf (x) ≤ 3δ) +
i=1
By decreasing x and increasing y in Φ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
18.465
Let f = �T λihi, where λ1 ≥ λ2 ≥ . . . ≥ λT ≥ 0. Rewrite f as
i=1
d
�
λihi +
T
�
f =
λihi =
d
�
λihi + γ(d)
T
�
λ�
ihi
i=1
i=d+1
i=1
i=d+1
where γ(d) = �
T
=d+1 λi and λ�
i
i = λi/γ(d).
Consider the following random approximation of f ,
where, as in the previous lectures,
i=1
d
�
g =
λihi +... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
2
� = yYj +1 ∈ [0, 1] and applying Hoeffding’s inequality, we get
⎛
⎛
1
PY ⎝γ(d)y ⎝
k
Hence,
⎞
⎛
�
1
Yj (x) − EY ⎠ ≥ δ � (x, y)⎠ = PY ⎝
k
⎞
k
�
j=1
k
�
Yj
�(x) − EY1
� ≥
j=1
≤ e−
2kδ
2γ(d)2 .
P (yf (x) ≤ 0) ≤ P (yg(x) ≤ δ) + e−
kδ
2
2γ2(d) .
If we set e−
2
kδ
2γ(d
)2 = n
1 , then k = 2γ
δ
2
(d) lo... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
define γ(d, f ) = �T
i=d+1
λi. Then with probability at least 1 − e−t ,
P (yf (x) ≤ 0) ≤
� �
ε + Pn (yf (x) ≤ δ) + ε2
�2
inf
δ∈(0,1)
where
��
·
V e(f, δ)
n
n
log +
δ
� �
t
n
ε = K
Example 23.1. Consider the zero-error case. Define
δ∗ = sup{δ > 0, Pn (yf (x) ≤ δ) = 0}.
Hence, Pn (yf (x) ≤ δ∗) = 0 for c... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
α log n
δ2d2α−1
= 0
d = Kα ·
log1/(2α−1) n
δ2/(2α−1)
≤ K
log n
.
δ2/(2α−1)
59
Lecture 23
Bounds in terms of sparsity.
18.465
Hence,
Plugging in,
e(f, δ) ≤ K
log n
δ2/(2α−1)
P (yf (x) ≤ 0) ≤ K
�
V log n
n(δ∗)2/(2α−1)
n
log +
δ∗
�
.
t
n
As α → ∞, the bound behaves like
V log n
n
log
n
.
δ∗
60... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= var(Zk) =
k=1 Yk
g =
c − EYk
k=1 Zk. Then EZk = Eg = f . We define σc
c)2 . (If we define {Yk}k=1,
h∈c λc
var(Yk
∀h ∈ H, P(Yk = h) = λh, and define g = 1 �N Yk, we might get much larger var(Yk)).
�N
1
c = N
c�2 = �
1
c αc N
c) = �Yk
c . Then EYk
c = h) = λh
h (h − EYh
c αcYk
h h. Let Zk =
c αc
,N be random v... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
x)|yf (x) ≤ 0, σc
2 < r
�
�
�
1
N
N
�
k=1
�
N1 �
N
�
exp −
k=1
≤ P
(yZk − EyZk) > δ|yf (x) ≤ 0, σ2 < r
c
,Bernstein’s inequality
�
N 2δ2
2N σ2 + 2 N δ 2
3
·
��
N 2δ2 N 2δ2
,
4N σ2
N δ
c
8
3
≤
≤
≤
c
�
�
�
exp − min
�
exp −
N δ2
4r
, for r small enough
(24.2)
set 1
≤
n
.
As... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
n
δ
with probability at least 1 − e−u . Using a technique developed earlier in this course, and taking the union
bound over all m, δ, we get, with probability at least 1 − Ke−u ,
P(yg ≤ δ)
≤
K inf Pn(yg ≤ 2δ) +
�
m,δ
·
V Nm
n
n
log +
δ
�
.
u
n
(Since EPn(yg ≤ 2δ) ≤ EPn(yf (x) ≤ 3δ) + EPn(σc
the same ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1) , Zk
(2) ≤ 1 ,and Zk
(1) − Zk
�
�
�2
(2) ≤ 4
�
≤ E Zk
(1) − Zk
(2)
�2
= 2σ2
c
var (1,2) σ2
N ≤
Y1,
···
,N
2 .
·
2 σc
62
sϕ (s)δ12δδ Lecture 24
Bounds in terms of sparsity (example).
18.465
We start with
PY1,
···
,N
(σ2
c ≥ 4r) ≤ P (1,2)
Y1,
···
,N
(σ2
N ≥ 3r) + P (1,2)
(σ2
σ2
c ≥ 4r| N... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
) ≤ KV log 2 by the assumption of our problem, we have log D(convNm {hi ·
hj : hi, hj ∈ H}, �) ≤ KV · Nm · log 2
� by the VC inequality, and
�
�
Eφr(σ2
N ) − En
� �
φr(σ2 ) / Eφr(σ2 )
N
N ≤
��
K
n
V Nm log /n + u/n
r
�
·
�
with probability at least 1 − e−u . Using a technique developed earlier in this co... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
25.1)
|Z(x1, . . . , xn) − Z(x1, . . . , xi−1, xi
� , xi+1, . . . , xn)| ≤ ci.
Decompose Z − EZ as follows
Z(x1, . . . , xn) − Ex� Z(x1
� , . . . , x�
n) = (Z(x1, . . . , xn) − Ex� Z(x1
� , x2, . . . , xn))
+ (Ex� Z(x�
1, x2, . . . , xn) − Ex� Z(x�
1, x2
� , x3, . . . , xn))
. . .
�
+ Ex� Z(x�
�
1, . . . , xn
−1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
−
2
e−λ =
e−λ
eλ +
2
+ s
e−λ
eλ −
2
≤ e λ2 /2 + s · sh(x)
using Taylor expansion. Now use Zi
ci
= s, where, by assumption, −1 ≤ Zi
ci
≤ 1. Then
e λZi = e λci · Zi
ci ≤ e λ2 c 2
i /2 +
Zi
ci
sh(λci).
Since Exi Zi = 0,
Exi e λZi ≤ e λ2 c 2
i /2 .
We now prove McDiarmid’s inequality
�
64
Lecture 2... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
� e λ2 (c 2
λ2 P n
≤ e
i=1 c 2
i /2
Hence,
P (Z − EZ > t) ≤ e−λt+λ2 P
n
i=1 ci /2
2
and we minimize over λ > 0 to get the result of the theorem.
�
Example 25.1. Let F be a class of functions: X �→ [a, b]. Define the empirical process
Then, for any i,
�
�
Ef −
�
Z(x1, . . . , xn) = sup
�
�
f ∈F
1
n
n
�
f ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= b−a for all i,
�
�
n
P (Z − EZ > t) ≤ exp −
t2
�
2
n
i=1 n
(b−a)2
2
= e−
2nt
2(b−a)2
.
By setting t =
�
2u (b − a), we get
n
�
P Z − EZ >
�
(b − a) ≤ e−u .
�
2u
n
Let ε1, . . . , εn be i.i.d. such that P (ε = ±1) = 1 . Define
2
�
�
�
Z((ε1, x1), . . . , (εn, xn)) = sup
�
�
f ∈F
1
n
Then, for ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
e−u .
�
8u
n
�
8u
n
66
Lecture 26
Comparison inequality for Rademacher processes.
18.465
Define the following processes:
�
Z(x) = sup Ef −
f ∈F
1
n
�
n
�
i=1
f (xi)
and
R(x) = sup
1
f ∈F n
n
�
εif (xi).
i=1
Assume a ≤ f (x) ≤ b for all f, x. In the last lecture we proved Z is concentrated arou... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
if −M ≤ f (x) ≤ M for all f, x, then
with probability at least 1 − e−t ,
Hence, with high probability,
Theorem 26.1. If −1 ≤ f ≤ 1, then
�
ER ≤ R + M
�
2t
n
.
Z(x) ≤ 2R(x) + 4M
�
2t
.
n
P Z(x) ≤ 2ER(x) + 2
� �
2t
n
≥ 1 − e−t .
67
Lecture 26
Comparison inequality for Rademacher processes.
18.465
If ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�(t2) ≤ EεG sup t1 + εt2
�
�
,
t∈T
t∈T
i.e. enough to show that we can erase contraction for 1 coordinate while fixing all others.
Since P (ε = ±1) = 1/2, we need to prove
�
�
�
G sup t1 + ϕ(t2) + G sup
t∈T
t∈T
1
2
1
2
�
t1 − ϕ(t2) ≤
�
�
G sup t1 + t2 + G sup
t∈T
1
2
�
t∈T
1
2
�
.
t1 − t2
As... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�(s2) ≤ −s2. Hence
�
�
�
Σ ≤ G(t1 + t2) + G(s1 − s2) ≤ G sup t1 + t2 + G sup t1 − t2
t∈T
t∈T
�
.
�
.
68
Lecture 26
Comparison inequality for Rademacher processes.
18.465
Case 3: t2 ≥ 0, s2 ≥ 0
Case 3a: s2 ≤ t2
It is enough to prove
G(t1 + ϕ(t2)) + G(s1 − ϕ(s2)) ≤ G(t1 + t2) + G(s1 − s2).
Note that s2 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
s1 − ϕ(s2) − G s1 − s2 = G (s1 − s2) + (s2 − ϕ(s2)) − G s1 − s2
�
�
≤ G t1 + t2 − G t1 + ϕ(t2) .
�
�
Case 3a: t2 ≤ s2
Again, it’s enough to show
Σ ≤ G(s1 + s2) + G(t1 − t2)
G(t1 + ϕ(t2)) − G(t1 − t2) ≤ G(s1 + s2) − G(s1 − ϕ(s2))
We have
t1 − t2 ≤ t1 − ϕ(t2) ≤ s1 − ϕ(s2)
since s1, s2 achieves maximum and sin... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
way as Case 3.
We now apply the theorem with G(s) = (s)+ .
Lemma 26.1.
Proof. Note that
�
n
�
�
E sup
�
�
�
t∈T
i=1
�
�
n
�
�
�
εiϕi(ti) ≤ 2E sup
�
�
�
�
�
�
t∈T
i=1
�
�
�
εiti
�
�
x
|
|
−
= ( )+ + ( ) = ( )+ + (
x
x
x
−
)+
x .
We apply the Contraction Inequality for Rademacher processes with G(s) = (s)... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
of [0, 1] valued functions, Z = supf
supf
any f ∈ F unknown and to be estimated, the empirical error Z can be probabilistically bounded by R in the
following way. Using the fact that Z ≤ 2R and by Martingale inequality, P Z ≤ EZ +
n
=1 f (xi) , and R =
i
�n are Rademarcher random variables. For
, xn where �1,
· ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤
≤
≤
����
with probability 1−e−u
Eφδ(yf (x))
En
n1 �
φδ(yf (x)) + sup (Eφδ(yf (x)) −
n
f ∈F
��
�
Z
i=1
φδ (yif (xi)))
�
Enφδ(yf (x)) + 2 · E sup(
f ∈F
�
n1 �
n
i=1
�
2u
n
�iφδ(yif (xi))) +
��
R
�
�
�iyif (xi) +
≤
����
contraction
Enφδ(yf (x)) +
2
δ
E sup
f ∈F
=
≤
Enφδ(yf (x)) + E sup ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= P sup
n
1 �
i=1
n
1 �
h n
i=1
�ih(xi) ≤ K
�ih(xi) ≤ K
� � 1
1
n 0
� � 1 �
1
n 0
log1/2 D(H, �, dx)d� +
� ��
u
n
1
V log d� +
�
� ��
u
n
≥ 1 − e−u .
71
Lecture 27 Application of Martingale inequalities. Generalized Martingale inequalities.
18.465
Thus E sup n
1
�
�ih(xi) ≤ K
��
�
� �
1 ≤... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, · · · , xn) −
��
�
···
d2 (x2 ,
,xn)
with the assumptions that Exi di = 0, and �di�∞ ≤ ci.
We will give a generalized martingale inequality below.
�
n di = Z − EZ where di = di(xi,
i=1
· · ·
, xn),
maxi �di�∞ ≤ C, σi
2 = σi
2(xi+1,
· · ·
, xn) = var(di), and Edi = 0. Take � > 0,
n
� �
di − �
n
P(
2 ≥ t) ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
exp(λdn) exp(λ�σ2 ) ≤ 1. Iterate over
·
n
�
n
� �
di − �
n
σi
P
�
2 ≥ t ≤ exp (−λ · t)
i=1
i=1
72
Lecture 27 Application of Martingale inequalities. Generalized Martingale inequalities.
18.465
. Take t = u/λ, we get
To minimize the sum �
�
�
P
n
� �
di ≥ �
n
i=1
i=1
2 +
σi
u
�
2
�
(1 + 2�C) ≤ ex... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(yi − h(xi))2
i=1
over Hk, where k is the number of layers.
Define L(y, h(x)) = (y − h(x))2, 0 ≤ L(y, h(x)) ≤ 4. We want to bound EL(y, h(x)).
From the previous lectures,
�
�
EL(y, h(x)) −
�
sup
�
�
1
n
with probability at least 1 − e−t .
Define
n
�
i=1
�
�
L(yi, h(xi)) ≤ 2E sup
�
�
�
�
�
1
�
�
�
n
n
�
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�→ R is a contraction because largest derivative of s on
2
[−2, 2] is 4. Hence,
E
�
�
1
�
sup
�
�
n
h∈Hk (A1,...,Ak )
n
�
i=1
�
�
εi(yi − h(xi))2
�
�
�
= EEε
n
�
�
�
1
�
sup
�
�
n
h∈Hk (A1,...,Ak )
�
�
1
�
sup
�
�
n
h∈Hk (A1,...,Ak )
i=1
�
n1
�
�
�
sup
�
�
n
h∈Hk (A1,...,Ak )
i=1
n
�
εi
i=1
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1
� �
α�
j
n
�
εihj (xi)
εihj (xi)
��
�
�
�
�
�
��
�
�
�
�
�
�
�
1
�
sup
�
n
�
�
h∈Hk (A1,...,Ak )
j
�
n1
�
�
sup
�
�
n
h∈Hk−1(A1 ,...,Ak−1)
�
i=1
i=1
�
�
�
εihj (xi)
�
�
≤ 2LAkEε
= 2LAkEε
�
The last equality holds because sup
|
λj sj = maxj sj , i.e. max is attained at one of the vertices.
|
|... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
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