text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
�
n
n
�
i=1
εih(xi) +
�
�
�
�
�
8
√
n
�
�
EL(y, h(x)) −
�
sup
�
�
h∈Hk (A1,...,Ak )
1
n
n
�
�
�
�
L(yi, h(xi))
�
�
�
t
n
+ 8
i=1
�
�
8
�
+ √
εih(xi)
�
�
n
n
�
i=1
Z (Hk(A1, . . . , Ak)) :=
≤ 8
k
�
j=1
�
�
1
E sup
�
·
(2LAj )
�
�
n
h∈H
with probability at least 1 − e−t .
Assume H is ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
t
1
n n
Since −1 ≤ h ≤ 1 for all h ∈ H,
n
�
h2(xi)
�⎞
⎠
i=1
≥ 1 − e−t ,
n
�
h2(xi)
�⎞
⎠
.
i=1
≥ 1 − 2e−t ,
�
�
�
1
�
sup
�
�
n
h∈H
Pε
n
�
i=1
�
�
�
εih(xi)
�
�
≤ √
K
n
� 1
0
log1/2 D(H, ε, dx)dε + K
� �
t
n
≥ 1 − 2e−t
,
Since H is a VC-subgraph class with V C(H) = V ,
log D(H, ε, dx) ≤ KV l... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
have
�
∞
�
n
�
�
� 1 �
Eε sup �
h∈H � n
i=1
�
�
�
� ≤ K
εih(xi)
�
2e−
nu
2
K2 du
V
n
V
n
+
0
�
+
�
�
V
n
= K
≤ K
∞ K
√
n
0
2
e−x dx
�
K
+ √
n
≤ K
V
n
2
for V ≥ 2. We made a change of variable so that x = K2 . Constants K change their values from line to
line.
2
nu
We obtain,
Z (Hk(A1,... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
wj = �j if �j =�
j
0 and wj = 1 if
With this t,
Z (Hk(�1, . . . , �k)) ≤ K
k
�
j=1
(2L · 2−�j ) ·
�
�
V
n
8
+ √
n
+ 8
t + 2 �k
j=1
n
log |wj |
77
Lecture 29
Generalization bounds for neural networks.
18.465
with probability at least
1 − e−t−2
j=1 log |wj | = 1 −
Pk
k
� 1
|wj |2
j=1
e−t .
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
j and |�j | ≤ | log Aj | + 1. Hence, |wj | ≤ | log Aj | + 1.
Therefore, with probability at least 1 − e−u ,
∀(A1, . . . , Ak), Z (Hk(A1, . . . , Ak)) ≤ K
(4L · Aj ) ·
k
�
j=1
�
V
n
8
+ √
n
�
2 �k
j=1
+8
log (| log Aj | + 1) + k log 5 + u
n
.
Notice that log (| log Aj | + 1) is large when Aj is very lar... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1
Function φ is called feature map.
Family of classifiers:
FH = {(w, z)H : �w�H ≤ 1}.
F = {(w, φ(x))H : �w�H ≤ 1} � f : X �→ R.
Algorithms:
(1) SVMs
f (x) =
n
�
i=1
n
�
αiK(xi, x) = (
i=1
�
αiφ(xi), φ(x))H
�
��
w
Here, instead of taking any w, we only take w as a linear combination of images of data po... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤ E sup Eϕδ (yf (x)) −
f ∈F
ϕδ (yif (xi)) +
� �
2t
n
1
n
n
�
i=1
79
Lecture 30
Generalization bounds for kernel methods.
18.465
Using the symmetrization technique,
�
E sup
f ∈F
E(ϕδ (yf (x)) − 1) −
Since δ (ϕδ − 1) is a contraction,
·
�
n1
n
i=1
�
�
1
(ϕδ (yif (xi)) − 1) ≤ 2E sup
�
�
�
n
f ∈F... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
4
δn
�
=
E
εiφ(xi),
εiφ(xi)
=
n
�
i=1
E
4
δn
��
εiεj (φ(xi), φ(xi))
H
i,j
εiεj K(xi, xj ) ≤
εiεj K(xi, xj )
H
�
�
E
i,j
4
δn
i=1
�
�
1
E sup
�
�
�
n
f ∈F
4
δ
i=1
�
�
�
sup
(w,
�
�
�w�≤1
�
n
�
4
δn
E
�
�
�
�
E
4
δn
i=1
4 ��
δn
E
i,j
=
=
=
=
=
n
�
�
�
� �
i=1
4
δn
EK... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= supx∈X |f (x) − g(x)|.
The following theorem appears in Cucker & Smale:
Theorem 31.1. ∀h ≥ d,
where Ch is a constant.
Note that for any x1, . . . , xn,
�
dx(f, g) =
Hence,
� 2d
h
Ch
ε
log N (F, ε, d) ≤
�
�1/2
n1 �
n
i=1
(f (xi) − g(xi))2
≤ d(f, g) = sup
|f (x) − g(x)| ≤ ε.
x
Assume the loss function ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
} - random collection of sets. Assume that C(x1, . . . , xn) satisfies:
(1) C(x1, . . . , xn) ⊆ C(x1, . . . , xn, xn+1)
(2) C(π(x1, . . . , xn)) = C(x1, . . . , xn) for any permutation π.
Let
and
�C (x1, . . . , xn) = card {C ∩ {x1, . . . , xn}; C ∈ C}
G(n) = E�C(x1,...,xn)(x1, . . . , xn).
81
Lecture 31
Optimi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
∈ Cx) + n
⎛
⎜
⎜
·
n) exp ⎜
⎝
−
�
�
2
t2
√
1 (I(x�
n
P
n
i=1(I(xi
1
n
2
i∈Cx )−I(xi∈Cx ))
� ∈Cx)+I(xi∈Cx))
⎞
⎟
⎟
�2 ⎟
⎠
≤ 4E�C(x1,...,xn,x�
1,...,x� )(x1, . . . , xn, x�
n
1, . . . , xn
� ) e− nt
·
4
2
= 4G(2n)e− nt
4
2
83
Lecture 32
Applications of random VC inequality to voting algorithms a... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
classifiers): algorithm outputs f = �
T
=1 λihi. Take random approxi
i
mation g(x) = k
Fix δ > 0.
1 �
k
j
=1 Yj (x), where Y1, . . . , Yk i.i.d with P (Yj = hi) = λi, EYj (x) = f (x).
P (yf (x) ≤ 0) = P (yf (x) ≤ 0, yg(x) ≤ δ) + P (yf (x) ≤ 0, yg(x) > δ)
⎛
≤ P (yg(x) ≤ δ) + Ex,yPY ⎝y
1
k
k
�
⎞
Yj (x) > δ, yE... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, 1]}
Fk =
hj (x) : hj ∈ H
⎧
⎨
⎩ k
k
1 �
j=1
⎫
⎬
⎭
Note that H(x1, . . . , xn) ⊆ H(x1, . . . , xn, xn+1) and H(π(x1, . . . , xn)) = H(x1, . . . , xn).
84
Lecture 32
Applications of random VC inequality to voting algorithms and SVM.
18.465
In the last lecture, we proved
�
Px,y
sup
C∈C
where
P (C) − n... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Setting the above bound to e−u and solving for t, we get
�
2
n
(u + k log N (2n) + log(8n + 4))
t =
So, with probability at least 1 − e−u, for all C
�
P (C) − n
1
�
n
=1 I(xi ∈ C)
i
P (C)
�2
≤
2
n
(u + k log N (2n) + log(8n + 4)) .
In particular,
�
�
n
P (yg(x) ≤ δ)
I
−
i=1
P
δ
(yg(x) ≤
1
n
�2
yig(x... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
) + e−kδ2/2 .
i=1
Choose k such that e−kδ2
/2
1
= n , i.e. k =
2 log n
δ2
b)2
. Plug (32.2) and (32.3) into (32.1) (look at (a−
a
). Hence,
�
P (yf (x) ≤ 0) − 2
�
1
n − n
n I(yif (xi) ≤ 2δ)
i=1
P (yf (x) ≤ 0) −
�2
2
n
with probability at least 1 − e−u .
Recall that for SVM, N (n) = card {±K(xi, x)} ≤ 2n.... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
/2 ≥ r
1
4 − log(2 − r) ≤ 0 ⇔ r ≤ 2 − e
Case a): r ≥ e−1/2 , λ = 1 + 2 log r
1
4 . e−1/2 ≤ 2 − e
4 .
1
Let
(log r)2 − log r − 2(log r)2 − log(2 − r) ≤ 0
f (r) = log(2 − r) + log r + (log r)2 .
Is f (r) ≥ 0? Enough to prove f �(r) ≤ 0. Is
Enough to show (rf �(r))� ≥ 0:
f �(r) = −
1
2 − r
+
1
r
+ 2 log r ·... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, x) as follows:
(1) V (A, x) = {(s1, . . . , sn) : si ∈ {0, 1}, ∃y ∈ A s.t. if si = 0 then xi = yi}
x = ( x1, x2,
. . . , xn)
= =�
. . . =
y = ( y1,
y2,
. . . , yn)
s = (
0,
1,
. . . ,
0)
Note that it can happen that xi = yi but si = 0
.
(2) U (A, x) = conv V (A, x) =
(3) d(A, x) = minu∈U (A,x) |u|2 = mi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
≤
1
P (A)
1
e
4
≤
1 + P (A)
P (A)
.
Let x = (x1, . . . , xn, xn+1) = (z, xn+1). Define
A(xn+1) = {(y1, . . . , yn) : (y1, . . . , yn, xn+1) ∈ A}
and
B = {(y1, . . . , yn) : ∃yn+1, (y1, . . . , yn, yn+1) ∈ A}
One can verify that
and
Take 0 ≤ λ ≤ 1. Then
s ∈ U (A(xn+1, z)) ⇒ (s, 0) ∈ U (A, (z, xn+1))
t ∈ U (... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
,(z,xn+1))dP n(z)dP (xn+1).
e
Then inner integral is
�
X n
1
4 d(A,(z,xn+1))dP n(z) ≤
e
�
4 (λd(A(xn+1),z)+(1−λ)d(B,z)+(1−λ)2)dP n(z)
1
e
X n
= e 4 (1−λ)2
1
�
1 d(A(xn+1),z))λ+( 1
e( 4
4 d(B,z))(1−λ)dP n(z)
We now use H ¨older’s inequality:
�
��
f gdP ≤
�1/p ��
f pdP
�1/q
gq dP
1
where + = 1
p
1
q... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
P n(B)
1 �
≤
P n(B)
2 −
P n(A(xn+1)) �
.
P n(B)
90
Lecture 33
Talagrand’s convex-hull distance inequality.
18.465
Now, integrate over the last coordinate. When averaging over xn+1, we get measure of A.
�
1
4 d(A,x)dP n+1(x) =
e
� �
1
4 d(A,(z,xn+1))dP n(z)dP (xn+1)
e
�
2 −
P n(A(xn+1))
P n(B)
�
)
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1,
. . . ,
0)
Build conv V (A, x) = U (A, x). Finally, d(A, x) = min{|x − u|2; u ∈ conv A}
Theorem 34.1. Consider a convex and Lipschitz f : Rn �→ R, |f (x) − f (y)| ≤ L|x − y|, ∀x, y ∈ Rn . Then
and
�
f (x1, . . . , xn) ≥ M + L
P
√
�
t
≤ 2e−t/4
�
f (x1, . . . , xn) ≤ M − L
P
√
�
t
≤ 2e−t/4
where M is m... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, x) ≤
√
t.
92
xconv Ad Lecture 34
Consequences of Talagrand’s convex-hull distance inequality.
18.465
So, f (x) ≤ f (u0) + L
√
t.
What is f (u0)? We know that u0 ∈ conv A, so u0 =
�
λiai, ai ∈ A, and
�
λi ≥ 0,
λi = 1. Since f is convex,
f (u0) = f
��
�
λiai ≤
�
λif (ai) ≤
�
λia = a.
This implies f (... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
n be a bounded set. Let
�
n
�
�
�
f (x1, . . . , xn) = sup
�
�
h∈H
i=1
�
�
�
hixi .
�
�
Let’s check:
(1) convexity:
�
n
�
�
�
f (λx + (1 − λ)y) = sup
�
�
h∈H
i=1
�
n
�
�
�
λ
= sup
�
�
h∈H
�
�
�
hi(λxi + (1 − λ)yi)
�
�
n
�
hixi + (1 − λ)
�
�
�
hiyi
�
�
�
n
�
�
�
+ (1 − λ) sup
�
�
h∈H
i=1
i=1
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
hi(xi − yi)
�
�
≤ (by Cauchy-Schwartz)
�� ��
2
hi
(xi − yi)2
sup
h∈H
= |x − y|
��
h2
i
�
sup
h∈H
�
��
L=Lipschitz constant
Theorem 34.2. If M is the median of f (x1, . . . , xn), and x1, . . . , xn are i.i.d with P (xi = 1) = p and
P (xi = 0) = 1 − p, then
and
hixi ≥ M + sup
h∈H
��
h2
i ·
√
�
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
x� = (x�
1, . . . , x�
n), an independent copy of x. Let
V (x) = Ex� sup
f ∈F
(f (xi) − f (x� ))2
i
n
�
i=1
be ”random uniform variance” (unofficial name)
Theorem 35.1.
�
�
P Z(x) ≥ EZ(x) + 2 V (x)t ≤ 4e e−t/4
�
P Z(x) ≤ EZ(x) − 2 V (x)t ≤ 4e e−t/4
�
�
·
·
�
Recall the Symmetrization lemma:
Lemma 35.1. ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
4 ,
i
n
�
�
� �
f ∈F i=1
⎞
i.e.
⎛
P ⎝sup
n
�
f ∈F i=1
n
�
f ∈F i=1
n
�
�
� �
f ∈F i=1
f (xi) ≥ sup
f (x� ) + 2 �t sup
i
(f (xi) − f (x� ))2⎠ ≤ 4e−t/4 .
i
⎞
95
Lecture 35
Talagrand’s concentration inequality for empirical processes.
18.465
If we switch xi ↔ x� , nothing changes, so we can switch ra... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
f ∈F
(f (x� ) + εi(f (xi) − f (x� ))) ≥ sup
f ∈F
i
i
(f (xi) − εi(f (xi) − f (x� )))
i
n
�
i=1
�
�
� �
+2�t sup
f ∈F i=1
n
⎞
(f (xi) − f (x� ))2⎠
i
�
= Ex,x� Pε
sup . . . ≥ sup . . . + 2
f ∈F
f ∈F
�
. . . for fixed x, x�
√
Φ1(ε) = sup
f ∈F
(f (x� ) + εi(f (xi) − f (x� )))
i
i
Φ2(ε) = sup
f ∈F
(f (xi)... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
t ≤ Φ2 + 2L
√
t.
Thus,
and
�
Φ1 ≥ Φ2 + 2L
√
�
t
Pε
≤ 4e−t/4
�
Px,x�,ε Φ1 ≥ Φ2 + 2L
√
�
t ≤ 4e−t/4 .
96
Lecture 35
Talagrand’s concentration inequality for empirical processes.
18.465
The ”random uniform variance” is
For example, if F = {f }, then
V (x) = Ex� sup
(f (xi) − f (x� ))2 .
i
n
�
f ∈F i=... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
x)
=
�
2d(A1,A2,x)dP n(x) ≤
1
P n(A1)P n(A2)
P (d(A1, A2, x) ≥ t) ≤
1
P n(A1)P n(A2)
· 2−t
We first prove the following lemma:
Lemma 36.1. Let 0 ≤ g1, g2 ≤ 1, gi : X �→ [0, 1]. Then
�
min 2,
dP (x) ·
�
�
�
1
,
g1(x)
1
g2(x)
�
g1(x)dP (x) ·
g2(x)dP (x) ≤ 1
Proof. Notice that log x ≤ x − 1.
So enough to sh... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1 otherwise
�
�
2d(A1 ,A2 ,x)dP (x) = min 2,
�
1
�
1
,
I(x ∈ A1) I(x ∈ A2)
dP (x)
≤ �
1
I(x ∈ A1)dP (x) �
·
I(x ∈ A2)dP (x)
=
1
P (A1)P (A2)
�
98
Lecture 36
Talagrand’s two-point inequality.
18.465
n
→
n + 1 :
Let x ∈ X n+1 , A1, A2 ⊆ X n+1 . Denote x = (x1, . . . , xn, xn+1) = (z, xn+1).
Define
a... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(z)
�
2d(B1,B2,z)dP n(z)
= 2
Moreover, by induction,
≤ 2 ·
1
P n(B1)P n(B2)
�
�
I(xn+1) ≤
I(xn+1) ≤
2d(A1(xn+1),B2,z)dP n(z) ≤
1
P n(A1(xn+1))P n(B2)
2d(B1,A2(xn+1),z)dP n(z) ≤
1
P n(B1)P n(A2(xn+1))
and
Hence,
I(xn+1) ≤ min
�
2
,
P n(B1)P n(B2)
⎛
1
,
P n(A1(xn+1))P n(B2)
1
P n(B1)P n(A2(xn+1)... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
+1(A2)/P n(B2)
·
because P n(A1(xn+1))dP (xn+1) = P n+1(A1).
�
=
1
P n+1(A1)P n+1(A2)
�
100
Lecture 37
Talagrand’s concentration inequality for empirical processes.
18.465
Lemma 37.1. Let
and a ≤ f ≤ b for all f ∈ F. Then
V (x) = Ex� sup
(f (xi) − f (x� ))2
i
n
�
f ∈F i=1
�
P V ≤ 4EV + (b − a)2t ≥ 1 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1 , xi = yi
yi
2},
Then we can decompose V as follows
I3 = {i ≤ n : xi =�
1 , xi =�
yi
2}
yi
V (x) = Ex� sup
(f (xi) − f (x�
i))2
n
�
f ∈F
i=1
�
= Ex� sup
f ∈F
�
i∈I1
�
f ∈F i∈I1
n
�
(f (xi) − f (x�
i))2 +
�
(f (xi) − f (x�
i))2 +
�
(f (xi) − f (x�
i))2
�
i∈I2
i∈I3
≤ Ex� sup
(f (xi) − f (x� ))... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Talagrand’s concentration inequality for empirical processes.
18.465
Now, let Z(x) = supf ∈F | n f (xi)|. Then
�
EZ + 2 V (x)t
Z(x)
i=1
�
≤
����
with prob. ≥1−(4e)e−t/4
√
√
√
Using inequality
c + d ≤
c +
d,
≤
����
with prob. ≥1−4 2−t
·
�
EZ + 2 (4EV + (b − a)2t)t.
Z(x) ≤ EZ + 4
√
EV t + 2(b − a)t
with ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
f ∈F
f ∈F
E sup
f ∈F
(f (xi) − f (x� ))2 = E sup
i
f ∈F
n
�
i=1
� �
n
(f (xi) − f (x� ))2 − 2nVar(f ) + 2nVar(f )
i
�
i=1
n
�
�
(f (xi) − f (x�
≤ E sup
f ∈F
i=1
(by symmetrization)
i))2 − E(f (xi) − f (x�
�
i))2 + 2n sup Var(f )
f ∈F
n
�
i=1
≤ 2E sup
f ∈F
�
≤ 2E sup
f ∈F
n
�
i=1
εi(f (xi) − f (... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
| + 2nσ2
i
n
�
≤ 4(b − a) · 2E sup
f ∈F
i=1
= 8(b − a)EZ + 2nσ2
εi|f (xi)| + 2nσ2
We have proved the following
Lemma 37.2.
where σ2 = supf ∈F Var(f ).
EV ≤ 8(b − a)EZ + 2nσ2 ,
Corollary 37.1. Assume a ≤ f ≤ b for all f ∈ F. Let Z = supf ∈F | n f (xi)| and σ2 = supf ∈F Var(f ).
Then
i=1
�
�
�
P Z ≤ EZ + 4... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
2nσ2
t + (b − a)
t
3
�
�
�
�
�
�
with probability at least ≥ 1 − e−t . Here, a ≤ f ≤ b for all f ∈ F and σ2 = supf ∈F Var(f ).
Now divide by n to get
�
�
1
�
sup
�
�
n
f ∈F
n
�
i=1
�
�
f (xi) − Ef
�
�
�
≤ E sup |. . .| +
f ∈F
4(b − a)E sup |. . .| + 2σ2
f ∈F
�
t
n
+ (b − a)
t
3n
Compare this result... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
�
�
�
�
�
�
�
�
�
1
n
�
Ef −
What if we knew that Ef0 ≤ ε and the family Fε
�
n f (xi)
at supf ∈F
�
i=1
Pin down location of f .0
�
�
�Ef�
�
�
�
�
f0(xi)
�
�
�
�
Ef −
�
�
�
≤ sup
f ∈Fε
n
�
1
n
1
n
is too conservative.
−0
i=1
n
�
i=1
�
�
�
f (xi)
�
�
Pretend we know Ef0 ≤ ε, f0 ∈ Fε.
Then with p... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
.
t
n
+
t
3n
104
Lecture 38
Applications of Talagrand’s concentration inequality.
18.465
Take ε = 2−k , k = 0, 1, 2, . . .. Change t
→
t + 2 log(k + 2). Then, for a fixed k, with probability at least
1 − e−t
1
(k+2)2 ,
�
�
Ef0 −
�
�
�
1
n
n
�
i=1
�
�
�
�
�
�
f0(xi) ≤ ϕ(ε) +
(4ϕ(ε) + 2ε)
t + 2 log(k ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
+ 2 log(log2 E
f0
+ 2)
n
+
1
t + 2 log(log2 E
f0
+ 2)
3n
= Φ(Ef0)
Hence, Ef0 ≤ 1
n
�
n f0(xi) + Φ(Ef0). Denote x = Ef0. Then x ≤
i=1
f ¯ + Φ(x).
Theorem 38.1. Let 0 ≤ f ≤ 1 for all f ∈ F. Define Fε = {f ∈ F, Ef ≤ ε} and ϕ(ε) = E supf ∈Fε
Then, with probability at least 1 − e−t, for any f0 ∈ F, Ef0 ≤ x∗, where... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
V (A, x) = {(I(x1 =�
y1), . . . , I(xn =�
yn)) : y = (y1, . . . , yn) ∈ A},
and
U (A, x) = conv V (A, x)
d(A, x) = min{|s 2 =
|
n
�
2 , s ∈ U (A, x)}
si
i=1
In the previous lectures, we proved
Theorem 39.1.
Today, we prove
P (d(A, x) ≥ t) ≤ P (A)
e−t/4 .
1
Theorem 39.2. The following are equivalent:
(1) d... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
αiI(xi =�
yi) ≤
n
�
�
�
� �
α2
i · t
i=1
i=1
Note that
�
0 is constant on L because s0 is perpendicular to the face.
αis
i
�
��
�
αisi
0 ≤
αiI(xi =�
yi) ≤
Hence,
�
0)2
(s
i
≤
��
(s0)2t and
i
��
(s0)2 ≤
i
√
t.
Therefore, d(A, x) ≤
0)2
i ≤ t.
2t
αi
�
(s
We now turn to an application of the... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
B − 1 ≤
�
Theorem 39.3.
�
P B(x1, . . . , xn) ≤ M + 2
i t + 1 ≥ 1 − 2e−t/4 .
x2
·
��
�
Proof. Let A = {y : B(y1, . . . , yn) ≤ M }, where P (B ≥ M ) ≥ 1/2, P (B ≤ M ) ≥ 1/2. We proved that
P (d(A, x) ≥ t) ≤ P (A)
e−t/4 .
1
Take x such that d(A, x) ≤ t. Take α = (x1, . . . , xn). Since d(A, x) ≤ t, there exi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
−t .
Hence,
B(x1, . . . , xn) � M + 2 nEx2
1
�
t·
�
108
Lecture 40
Entropy tensorization inequality. Tensorization of Laplace transform.
18.465
In this lecture, we expose the technique of deriving concentration inequalities with the entropy tensorization
inequality. The entropy tensorization inequality enable... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
we define x pointsizely by ∂
∂x
�
get x = udP > 0.
�
(u (log u − log x) − (u − x)) dP = 0, and
·
��
(ug) dP − λ
For the second formulation, the Laplacian corresponding to sup
�
is L(g, λ) =
exp(g)dP − 1 . It is linear in λ and concave in g, thus supg inf λ≥0 L = inf λ≥0 supg L. Define
�
λ dP − udP + λ. We s... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
( n EntPi (u)) dPn .
i=1
Proof. Proof by induction. When n = 1, the above inequality is trivially true. Suppose
�
u log udPn ≤
�
�
udPn log udPn +
� n
�
EntPi (u)dPn .
i=1
109
Lecture 40
Entropy tensorization inequality. Tensorization of Laplace transform.
18.465
Integrate with regard to Pn+1,
�
�
≤
u ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
exity of entropy
udPn+1
log udPn+1 + EntPn+1 (u)dPn+1 +
EntPi (u)dPn+1
� �
·
� �
� �
� � n+1
�
udPn+1 ·
log udPn+1 +
EntPi (u)dPn+1 .
=
����
=
����
≤
����
�
�
=
≤
� n
�
i=1
� n
�
i=1
i=1
By definition of entropy, EntPn+1 (u) ≤
� �
n+1 EntPi (u)dPn+1 .
i=1
�
The tensorization of entropy lemma can ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x1,
, xi−1, x�
, xn),
�
n
=1 φ(−λ(Z − Z i)). We will use the ten
i
sorization of entropy technique to prove the following Hoeffding-type inequality. This theorem is Theorem 9 of
i, xi+1,
· · ·
· · ·
n
Pascal Massart. About the constants in Talagrand’s concentration inequalities for empiri
cal processes. The An... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
and
2
2
Ee λZ λZ − Ee λZ log Ee λZ ≤ Ee λZ
�
�
�
φ −λ(Z − Z i)
i
≤
≤
Center Z, and we get
1 Ee λZ �
2
i
λ2(Z − Z i)2
1
2
Lλ2Ee λZ .
Ee λ(Z−EZ)λ(Z − EZ) − log Ee λ(Z−EZ) ≤
1
Lλ2Ee λ(Z−EZ).
2
112
�
Lecture 41
Application of the entropy tensorization technique.
18.465
Let F (λ) = Eeλ(Z−EZ). It follo... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
get P(Z ≥ EZ+ Lt/2) ≤
e−t . As a result, The above inequality improves the constant of Hoeffding’s inequality.
�
The following Bennett type concentration inequality is Theorem 10 of
Pascal Massart. About the constants in Talagrand’s concentration inequalities for empiri
cal processes. The Annals of Probability, 200... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
property Ψ ≤ Ψ0:
·
1
eλ −1
λ
(e −1)(1−e−λ)
� �
e λ − 1
�
1 − e−λ
Substract 41.2 from 41.1, and let Ψ1=Ψ−Ψ0
(1 − e−λ)Ψ�
1 − Ψ1
=1−e−λ ,and (e −1)(1−e−λ)
λ
�
�
�
1
eλ − 1
Ψ1
(eλ
eλ
−1)2 =1
�
eλ
� −
(eλ − 1)2 Ψ1
��
”
�
e −1 λ
Ψ1
λ
�
“
≤ 0
≤ 0
Ψ1(λ)
eλ − 1
≤
Ψ1(λ)
lim
λ→0 eλ − 1
= 0.
It follows... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
+
�
fi, Efi = 0, supf ∈F var(f ) = supf ∈F
�
3 ≤ e−x .
(1 + u)EZ + nσ2x + x
�
�
def.
n f 2 = σ2
i=1 i
, ∀i ∈ {1, · · ·
, n}, fi ≤
Proof. Let
Z =
Zk =
n
� �
def.
◦
fi
fi =
n
i=1
sup
f ∈F
sup
f ∈F
i=1
�
i=k
fi
� = fk such that Zk = sup
Zk
�
f ∈F i=� k
fi.
It follows that Zk
� ≤ Z − Zk ≤ u. Let ψ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Thus
≤ ασ2Ee λZ .
�
�
E λZeλZ − Ee λZ log Ee λZ ≤ Ee λZ
�
ψ(λ(Z − Zk))
k
Let Z0 = Z − E, and center Z, we get
≤ f (λ)EZeλZ + f (λ)αnσ2Ee λZ .
�
�
E λZ0e λZ0 − Ee λZ0 log Ee λZ0 ≤ f (λ)EZ0e λZ0 + f (λ) αnσ2 + EZ Ee λZ0 .
�
�
Let F (λ) = EeλZ0 , and Ψ(λ) = log F (λ), we get
(λ − f (λ)) F �(λ) − F (λ) log ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, X �) be an exchangeable pair on X (i.e., dP(X, X �) = dP(X �, X). Let F (x, x�) =
−F (x, x�) be antisymmetric, f (x) = E(F (x, x�)|x). Then Ef (x) = 0. If further
then P(f (x) ≥ t) ≤ exp −
2
t
2(C+Bt)
�
�
1
2
�
.
Δ(x) =
E |(f (x) − f (x�)) F (x, x�)| x ≤ Bf (x) + C,
�
�
�
Proof.
by definition of f (X)
E (... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
X
�
�
�
�
⎞
�
⎟
⎟
⎟
⎟
⎠
�
=
=
117
Lecture 42
Stein’s method for concentration inequalities.
18.465
Since m(λ) is a convex function in λ, and m�(0) = 0, m�(λ) always has the same sign as λ. In the interval
0 ≤ λ < 1/B, the above inequality can be expressed as
m�(λ) ≤ λ · (B · m�(λ) + C · m(λ))
(log m(λ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
| probabilistically. To apply the above theorem, take Xi
· · ·
, n} be a random variable uniformly distributed
an independent copy of Xi for i = 1,
, n, I ∼ unif{1,
Let X =
· · ·
�
n
i
i
over 1,
· · ·
, n, and X � =
i=I Xi + XI . Define F (X, X �), f (X), and Δ(X) as the following,
�
F (X, X �) = n (X − X �... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
Xi − E Xi ≥ t ≤ exp − �
�
�
�� � �
� � �
P − Xi − E − Xi ≥ t ≤ exp − �
�
�
+ σ2
i
t2
2
i ci
�
t2
2
2 + σi
i ci
���
�
�
�
P � Xi − E Xi� ≥ t ≤ 2 exp − �
union bound
�
�
�
�
t2
2
i c2
i + σi
.
118
�
Lecture 42
Stein’s method for concentration inequalities.
18.465
Example 42.3. Let (a... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
ai,π(i) −
ai,π�(i)
�
i=1
i=1
n 1 �
n
2
·
J
aJ,π(J) −
n 1 �
n2
2
·
I,J
aI,J −
n 1 �
n2
2
·
I,J
aI,J
�
aI,π(I) + aJ,π(J) − aI,π(J) − aJ,π(I)|π
=
f (X) def. = E (F (X, X �)|X)
n �
2
n 1 �
2
n
�
=
·
I
=
ai,π(i) −
aI,π(I) +
1 �
n
ai,j
Δ(X)
def. =
·
i,j
=
i
X − EX
�
�
n
1
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
⎝
β �
n
i<j
σiσj + β h
·
⎞
σi⎠ .
n
�
i=1
1
We are interested in the concentration of m(σ) = n
e x−e−x
ex+e−x
to the conditional distribution of σi on {σj : j = i} (Gibbs sampling):
. Given any σ, we can pick I uniformly and independently from {1,
i σi around tanh (β m(σ) + β h) where tanh(x) =
, n}, and ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.080 / 6.089 Great Ideas in Theoretical Computer Science
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.080/6.089 GITCS
Feb 07, 2008
Lecturer: Scott Aaronson
Scribe: Mergen Nachin
Lecture 2
Administrative... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
about Euclid’s GCD algorithm, which was one of the first non-trivial algorithms known to
humankind.
Here is some digression. Area of a circle is A = πr2 . It’s obvious that the area of a circle should
go like the r2; the question is why the constant of proportionality (π) should be the same one that
relates circumfe... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
men are mortal, Socrates is man, therefore Socrates is a mortal.
This is a syllogism. In more modern language, we call it transitivity of implications. In general,
a syllogism is
If A ⇒ B is valid and B ⇒ C is valid, then A ⇒ C is valid.
Remark: What do we mean by ”⇒”? ”A ⇒ B” is valid if A is false or B is true or... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
rule ”If you are under 21 the you are not drinking”. Who do you
have to check to test this rule: someone who is drinking, someone who isn’t drinking, someone
who’s over 21, someone who’s under 21?
And then, of course, almost everyone gets it right. Even though this problem is logically
equivalent in every way to th... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
As
we all see in our everyday life, everyone draws inferences all the time. However, what he was (as
far as I know) was the first person in the historical record to (as it were) draw a box around the
inference rule, to say that this is a general law of thought.This is crucial because it allows us to
reason about the... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
what today we would see as the right picture of what such a machine
would be. To him, it’s not that you would take some lifeless clump of clay, and utter some mystical
incantation that would magically imbue it the power of speech – like in the legend of Pinocchio, or
the legend of the Golem, or even in a lot of scie... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
can’t be analyzed into thousands
of baby steps? Maybe they can! Many things seem magical until you know the mechanism. So
why not logical reasoning itself?
To me, that’s really the motivation for studying logic: to discover ”The Laws of Thought.” But
to go further, we need to roll up our sleeves, and talk about som... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
all four sentences?
No. By applying the rules, we can reach a contradiction! You agree that if we reach a logical
contradiction by applying the rules, then the sentences can’t all be valid?
Suppose a set of sentences is inconsistent (i.e., there’s no way of setting the variables so that
all of them are satisfied). C... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
that’s true, you can get by cranking the rules.
In the next lecture: a bit of first-order logic!
2-4 | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
Introduction to Simulation - Lecture 10
Modified Newton Methods
Jacob White
Thanks to Deepak Ramaswamy, Jaime Peraire, Michal
Rewienski, and Karen Veroy
Outline
• Damped Newton Schemes
– Globally Convergent if Jacobian is Nonsingular
– Difficulty with Singular Jacobians
• Introduce Continuation Schemes
– Problem with... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
converge regardless of initial guess!
SMA-HPC ©2003 MIT
Non-converging
Case
f(x)
1-D Picture
1x
X
0x
Limiting the changes in X might improve convergence
SMA-HPC ©2003 MIT
Newton Method
with Limiting
Newton Algorithm
Newton Algorithm for Solving
( )
F x =
0
0
x
=
Initial Guess,
k =
0
Repeat {
Compute
k
(
F x
)... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
)
b
)
J
−
F
J
F
Then
1
k
x
(
( )
x
)
−
≤
β
(
Inverse is bounded
)
J
F
(
y
)
≤
(cid:65)
x
−
y
(
Derivative is Lipschitz Cont
)
There exists a set of
(
)
F x
(
F x
=
1
+
k
k sα ∈
'
(
k
k
x
+ ∆
α
0,1 such that
(
F x
γ
<
1
+
k
]
)
k
)
with <1
γ
Every Step reduces F-- Global Convergence!
SMA-HPC ©2003 MIT
Newton Method
w... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
2 0
v
−
)
+
10
−
16
(
e
0.025
− =
1) 0
Newton Method
with Limiting
Damped Newton
Example cont.
(
f v
2
)
=
)
1
(
−
v
2
10
(
2 0
v
−
)
+
10
−
16
(
e
0.025
− =
1) 0
Newton Method
with Limiting
0
x
Repeat {
Initial Guess,
=
k =
Damped Newton
Nested Iteration
0
Compute
Solve
FJ
Find
k
1
+
k
α
k
=
x
x
k
k
k
k
F
,
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
F
(
k
x
)
+
J
F
(
)
k
k
x α
⎡
⎢
⎣
1
−
J
F
(
k
x
)
k
(
F x
)
≤⎤
⎥
⎦
(cid:65)
2
k
α
J
F
1
−
(
k
x
)
F
(
k
x
)
2
SMA-HPC ©2003 MIT
Newton Method
with Limiting
Damped Newton
Theorem Proof-Cont
From the previous slide
k
(
F x
)
1
+ −
F
(
k
x
)
+
J
F
(
)
k
k
x α
⎡
⎢
⎣
1
−
J
F
(
k
x
)
k
(
F x
)
≤⎤
⎥
⎦
(cid:65)
2
k
α
J... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
F x
1
⎢
⎣
(
k
+
α α
(
F x
2
β
2
≤
−
)
)
(cid:65)
1
+
2
k
k
)
⎤
⎥
⎦
k
(
F x
)
Two Cases:
(cid:65)
1)
2
β
2
(
kF x
)
<
1
2
Pick
k
α
=
1 (Standard Newton)
⎛
⇒ −
⎜
⎝
1
k
(
k
+
α α
)
2
(cid:65)
2
β
2
(
kF x
)
<
⎞
⎟
⎠
1
2
2)
(cid:65)
2
β
2
k
(
F x
)
>
1
2
Pick
k
α
=
1
2
β
(cid:65)
k
(
F x
)
⎛
⇒ −
⎜
⎝
1
k
(
k
α α
+
)
2
(c... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
β
(cid:65)
0
(
F x
)
≤
0
γ
First – Show that the iterates do not increase
Second – Use the non-increasing fact to prove convergence
SMA-HPC ©2003 MIT
Newton Method
with Limiting
Initial Guess,
k =
0
=
x
Repeat {
Damped Newton
Nested Iteration
0
Compute
Solve
FJ
Find
1
k
+
k
α
k
=
x
x
k
k
k
k
F
,
1
+
x
x
J
)... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
solve
0 where:
Starts the continuation
( )
F x
Ends the continuation
a)
b)
=
(
(cid:4)
F x
(
(cid:4)
F x
)
( )
0 , 0
)
( )
1 ,1
=
c)
(
)
x λ
is sufficiently smooth
Hard to insure!
(
)
x λ
0
SMA-HPC ©2003 MIT
Dissallowed
λ
1
Continuation Schemes
Basic Concepts
Template Algorithm
x
( )
0
)
(
λ =
prev
x
(
(cid:4)
F x
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
:4)
F x
)
(
λ λ λ
=
)
,
)
(
(
)
F x
λ
Observations
(
1
+ −
)
(
)
x
λ λ
=
λ
=
)
(
(cid:4)
( )
0 , 0
0 F x
(
(cid:4)
( )
F x
0 , 0
∂
x
∂
x
)
( )
0
=
0
=
I
Problem is easy to solve and
Jacobian definitely nonsingular.
λ
=1
(cid:4)
(
F x
(
(cid:4)
F x
∂
=
)
)
( )
1 ,1
( )
0 , 0
x
∂
SMA-HPC ©2003 MIT
(
F x
=
)
( )
1
(
F x... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
λ δλ λ δλ
+
+
(
(
)
,
≈
Update Improvement
(cid:4)
0
)
(
,
F x λ λ +
(
(
(cid:4)
F x
∂
)
)
(
)
,
λ λ
x
∂
)
(cid:4)
(
F x λ λ
∂
∂
λ
(
)
,
δλ
(
x
(
)
λ δλ
+
−
x
(
)
λ
)
+
⇒
(
(cid:4)
F x
∂
)
(
)
,
λ λ
x
∂
(
0
x
(
λ
+
)
δλ
−
x
(
)
λ
)
= −
(cid:4)
F
∂
(
x
)
(
)
,
λ λ
∂
λ
δλ
Have From last
step’s Newton
Better Guess
for n... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
uation Schemes
(
)
x λ
Jacobian Altering Scheme
Still can have problems
Must switch back to
increasing lambda
Arc-length
steps
0
lambda steps
Must switch from
increasing to
decreasing lambda
λ
1
SMA-HPC ©2003 MIT
Continuation Schemes
(
)
x λ
Jacobian Altering Scheme
Arc-length Steps?
Arc-length
steps
0
arc-length... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
2003 MIT
Continuation Schemes
)x λ
(
0
Jacobian Altering Scheme
Arc-length Turning point
What happens here?
λ
1
Upper left-hand
Block is singular
SMA-HPC ©2003 MIT
⎡
⎢
⎢
⎢
⎢
⎣
(
(cid:4)
F x
∂
k
k
λ
,
)
(
(cid:4)
F x
∂
k
k
λ
,
)
x
∂
(
λ
x
prev
)
T
)
2
2
(
k
x
−
∂
λ
k
(
−
λ λ
prev
)
⎤
⎥
⎥
⎥
⎥
⎦
Summary
• Damped Newton ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
6.867 Machine learning, lecture 8 (Jaakkola)
1
Lecture topics:
• Support vector machine and kernels
• Kernel optimization, selection
Support vector machine revisited
Our task here is to first turn the support vector machine into its dual form where the exam
ples only appear in inner products. To this end, assume ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
��rst that
J(θ, θ0) really is equivalent to the original problem. Suppose we set θ and θ0 such that at
least one of the constraints, say the one corresponding to (xi, yi), is violated. In that case
�
�
− αi yi(θT φ(xi) + θ0) − 1 > 0
(4)
for any αi > 0. We can then set αi = ∞ to obtain J(θ, θ0) = ∞. You can think... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
the right hand side by first obtaining θ and θ0
as a function of the Lagrange multipliers (and the data). To this end
d
dθ0
d
dθ
J(θ, θ0; α) = −
αtyt = 0
J(θ, θ0; α) = θ −
αtytφ(xt) = 0
n
�
t=1
n
�
t=1
(7)
(8)
So, again the solution for θ is in the span of the feature vectors corresponding to the training... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10)
6.867 Machine learning, lecture 8 (Jaakkola)
3
the input vectors does not appear explicitly as part of the optimization problem. It is
formulated solely on the basis of the Gram matrix:
⎡
K = ⎣
φ(x1)T φ(x1)
· · ·
φ(xn)T φ(... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
. Moreover, the identity of the support vectors will depend on the
feature mapping or the kernel function.
But what is θˆ
solving for ˆαt by looking at the support vectors. Indeed, for all i ∈ SV we should have
0? It appeared to drop out of the optimization problem. We can set θ0 after
yi(θˆT φ(xi) + θˆ
0) = yi
� ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
the best kernel function. Unfortunately
this won’t work without some care. For example, if we multiply all the feature vectors by
2, then the resulting geometric margin will also be twice as large (we just expanded the
space; the relations between the points remain the same). It is necessary to perform some
normali... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
yiyj [φ(xi)T φ(xj )],
t=1
i=1 j=1
subject to 0 ≤ αt ≤ C,
n
�
αtyt = 0
(21)
(22)
t=1
The resulting discriminant function has the same form except that the ˆαt values can be
different. What about θˆ
0 we need to identify classification constraints that
are satisfied with equality. These are no longer simply the o... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
flexible as finding the best
convex combination of basic (fixed) kernels. Key to such an approach is the measure we
would optimize. Ideally, this measure would be the generalization error but we obviously
have to settle for a surrogate measure. The surrogate measure could be cross-validation or
an alternative criterio... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.969 Topics in Geometry: Mirror Symmetry
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
MIRROR SYMMETRY: LECTURE 9
DENIS AUROUX
1. The Quintic (contd.)
To recall where we were, we had
(1)
Xψ = {(x0 : · · ·... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
)5n
(n!)
T0
In terms of z = (5ψ)−5, the period is proportional to
(5)
φ0(z) =
∞ (5n)! n
�
z
(n!)5
n=0
�
�
cnzn) =
ncnzn, we obtained the Picard-Fuchs equation
d
Setting Θ = z dz : Θ(
(6)
θ4φ0 = 5z(5Θ + 1)(5Θ + 2)(5Θ + 3)(5Θ + 4)φ0
Proposition 1. All periods Ωˇ
ψ satisfy this equation.
�
Note that all ... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
it has a residue
C).
on X which is ideally a 3-form on X, but is at least a class in H 3(X,
�
1
πi S1 φ(z)dz.
Recall from complex analysis, if φ(z) has a pole at 0, res0(φ) = 2
Now, let’s say that we have a 3-cycle C in X: we can associate a “tube” 4-cycle
in P4 which is the preimage of C in the boundary of a tub... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
0 ∧ · · · ∧ �
dxi ∧ · · · ∧ �
dxj ∧ · · · ∧ dx4
with deg (g0 · · · g4) = 5� − 4, then
(13)
dφ =
�
�
1 � ∂fψ � ∂gj
gj ∂xj
f �+1
∂xj
ψ
− fψ
j
j
�
Ω
MIRROR SYMMETRY: LECTURE 9
3
In particular, if we have something of the form (
�
gj
∂fψ
∂xj
)
Ω
+1 (the Jacobian ideal
f �
ψ
∂fψ
is the span of { ∂xi
})... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
⎞
⎟
⎟
⎟
⎟
⎠
, w(z) =
⎛
⎜
⎜
⎝
f (z)
Θf (z)
. . .
Θs−1f (z)
⎞
⎟
⎟
⎠
1
−Bs−1(z)
.
.
.
0
· · ·
The fundamental theorem of these differential equations states that there exists
a constant s × s matrix R and an s × s matrix of holomorphic functions S(z) s.t.
(16)
Φ(z) = S(z) exp((log z)R) = S(z)(id + (log z)R +
log2 z
2... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
0 0 0
⎞
⎟
⎟
⎠
4
DENIS AUROUX
is nilpotent, and our assumption holds. The corresponding monodromy is
(19)
T = e
⎛
⎜
2πiR ⎜ 0
=
⎝
0
0
1 2πi
1
0
0
(2πi)2
2
2πi
1
0
⎞
(2πi)3
6
(2πi)2 ⎟
⎟
2
2πi
⎠
1
�
If ω(z) = β Ωˇ
ψ is a period, then it is a solution of the Picard-Fuchs equation, and
thus a linear combi... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
) exp(R log z), φ1(z) = φ0(z) log z + φ˜(z),
with φ˜(z) holomorphic. Now
n=0 (n!)5
�
(20)
Θj (f (z) log z) = (Θj f ) log z + j(Θj−1f )
If we write F (x) = x4 − 5z �
4
=1(5x + j), then
j
(21)
Dφ1(z) = F (Θ)(φ0(z) log z + φ˜(z))
= (F (Θ)φ0) log z + F �(Θ)φ0 + F (Θ)φ˜
Since 0 = Dφ0 = Dφ1, we find Dφ˜(z) = −F �(Θ)φ0... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
Substitution of Power Series
We can find the power series of e−t by starting with the power series for ex
and making the substitution x = −t2 .
2
e x = 1 + x +
2
x
2!
+
e−t2
= 1 + (−t2) +
+
· · ·
(R = ∞)
+
(−t2)3
3!
+
· · ·
3
x
3!
(−t2)2
2!
t6
3!
= 1 − t2 +
t4
2!
−
+ · · ·
The signs of the terms a... | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/33f76c6f2abf4190beb0a0cf29065926_MIT18_01SCF10_Ses100d.pdf |
L2: Combinational Logic Design
L2: Combinational Logic Design
(Construction and Boolean Algebra)
(Construction and Boolean Algebra)
Acknowledgements:
Materials in this lecture are courtesy of the following sources and are used with permission.
Prof. Randy Katz (Unified Microelectronics Corporation Distinguished Prof... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
n+
p
inversion layer�
channel
gate oxide
p+
n
D
ID
-4
x 10
6
5
4
)
A
(
3
I
D
2
1
0
0
VGS= 2.5 V
Resistive
Saturation
VGS= 2.0 V
VGS= 1.5 V
VGS= 1.0 V
0.5
1
1.5
2
2.5
VDS(V)
VT = 0.5V
(cid:190) MOS is a very non-linear.
(cid:190) Switch-resistor model
sufficient for first order analysis.
G
+
VGS
-
S
L2: 6.111 S... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
Vin = 2.5
Vin = 1
Vin = 0.5
Vin = 0
Vout
CMOS gates have:
(cid:131) Rail-to-rail swing (0V to VDD)
(cid:131) Large noise margins
(cid:131) “zero” static power dissipation
2.5
2
1.5
)
V
(
t
u
o
V
1
0.5
0
0
0.5
1
1.5
2
2.5
V
(V)
in
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
7
Possible Function of Two... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
X
0
0
1
1
X
0
0
1
1
X
0
0
1
1
Y
0
1
0
1
Y
0
1
0
1
Y
0
1
0
1
Y
0
1
0
1
Z
1
1
1
0
Z
0
0
0
1
Z
1
0
0
0
Z
0
1
1
1
Z = X • Y
Z = X • Y
Z = X + Y
Z = X + Y
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
9
Exclusive (N)OR Gate
Exclusive (N)OR Gate
XOR
(X ⊕ Y)
XNOR
(X ⊕ Y)
X
Y
X
Y
Z
Z
X
0
0
1
1
X
0
0
1
1
Y
0
1
... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
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