Lecture 25
Martingale-difference inequalities.
18.465
Let Z(x1 , . . . , xn ) : X n �→ R. We would like to bound Z − EZ. We will be able to answer this question if for
any x1 , . . . , xn , x�1 , . . . , x�n ,
|Z(x1 , . . . , xn ) − Z(x1 , . . . , xi−1 , x�i , xi+1 , . . . , xn )| ≤ ci .
(25.1)
Decompose Z − EZ as follows
Z(x1 , . . . , xn ) − Ex� Z(x�1 , . . . , x�n ) = (Z(x1 , . . . , xn ) − Ex� Z(x�1 , x2 , . . . , xn ))
+ (Ex� Z(x�1 , x2 , . . . , xn ) − Ex� Z(x�1 , x�2 , x3 , . . . , xn ))
...
�
�
+ Ex� Z(x�1 , . . . , x�n−1 , xn ) − Ex� Z(x�1 , . . . , x�n )
= Z1 + Z2 + . . . + Zn
where
Zi = Ex� Z(x�1 , . . . , x�i−1 , xi , . . . , xn ) − Ex� Z(x�1 , . . . , x�i , xi+1 , . . . , xn ).
Assume
(1) |Zi | ≤ ci
(2) EXi Zi = 0
(3) Zi = Zi (xi , . . . , xn )
Lemma 25.1. For any λ ∈ R,
2 2
ci /2
Exi eλZi ≤ eλ
.
Proof. Take any −1 ≤ s ≤ 1. With respect to λ, function eλs is convex and
eλs = eλ(
Then 0 ≤
1+s 1−s
2 , 2
≤ 1 and
eλs ≤
1+s
2
+
1−s
2
1+s
2
)+(−λ)( 1−s
2 )
.
= 1 and therefore
2
1 + s λ 1 − s −λ
eλ + e−λ
eλ − e−λ
e +
e =
+s
≤ eλ /2 + s · sh(x)
2
2
2
2
using Taylor expansion. Now use
Zi
ci
= s, where, by assumption, −1 ≤
Z
λci · c i
eλZi = e
i
2 2
ci /2
≤ eλ
+
Zi
ci
≤ 1. Then
Zi
sh(λci ).
ci
Since Exi Zi = 0,
2 2
ci /2
Exi eλZi ≤ eλ
.
�
We now prove McDiarmid’s inequality
64
Lecture 25
Martingale-difference inequalities.
18.465
Theorem 25.1. If condition (25.1) is satisfied,
−
P (Z − EZ > t) ≤ e
2
t2
Pn
c2
i=1 i
.
Proof. For any λ > 0
�
� Eeλ(Z−EZ)
P (Z − EZ > t) = P eλ(Z−EZ) > eλt ≤
.
eλt
Furthermore,
Eeλ(Z−EZ) = Eeλ(Z1 +...+Zn )
= EEx1 eλ(Z1 +...+Zn )
�
�
= E eλ(Z2 +...+Zn ) Ex1 eλZ1
�
�
2 2
≤ E eλ(Z2 +...+Zn ) eλ c1 /2
�
�
2 2
= eλ c1 /2 EEx2 eλ(Z2 +...+Zn )
�
�
2 2
= eλ c1 /2 E eλ(Z3 +...+Zn ) Ex2 eλZ2
2
(c21 +c22 )/2
2
Pn
≤ eλ
≤ eλ
i=1
Eeλ(Z3 +...+Zn )
c2i /2
Hence,
2
P (Z − EZ > t) ≤ e−λt+λ
Pn
i=1
ci2 /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
�
�
n
�
�
1�
�
�
f (xi )� .
Z(x1 , . . . , xn ) = sup �Ef −
�
n
f ∈F �
i=1
Then, for any i,
|Z(x1 , . . . , x�i , . . . , xn ) − Z(x1 , . . . , xi , . . . , xn )|
�
�
�
�
�
�
1
�
= �sup ��Ef − (f (x1 ) + . . . + f (x�i ) + . . . + f (xn ))��
� f
n
�
���
�
��
1
− sup ��Ef − (f (x1 ) + . . . + f (xi ) + . . . + f (xn ))���
�
n
f
≤ sup
f ∈F
1
b−a
|f (xi ) − f (x�i )| ≤
= ci
n
n
because
sup f (t) − sup g(t) ≤ sup(f (t) − g(t))
t
t
t
65
Lecture 25
Martingale-difference inequalities.
18.465
and
|c| − |d| ≤ |c − d|.
Thus, if a ≤ f (x) ≤ b for all f and x, then, setting ci =
�
b−a
n
for all i,
�
t2
P (Z − EZ > t) ≤ exp − �n (b−a)2
2 i=1 n2
By setting t =
�
2u
n (b
2
nt
− 2(b−a)
2
=e
.
− a), we get
�
�
P Z − EZ >
�
2u
(b − a) ≤ e−u .
n
Let ε1 , . . . , εn be i.i.d. such that P (ε = ±1) = 12 . Define
� n
�
�
�1 �
�
�
εi f (xi )� .
Z((ε1 , x1 ), . . . , (εn , xn )) = sup �
�
�
n
f ∈F
i=1
Then, for any i,
|Z((ε1 , x1 ), . . . , (ε�i , x�i ), . . . , (εn , xn )) − Z((ε1 , x1 ), . . . , (εi , xi ), . . . , (εn , xn ))|
�
�
�1 �
� 2M
�
�
≤ sup � (εi f (xi ) − εi f (xi ))�� ≤
= ci
n
f ∈F n
where −M ≤ f (x) ≤ M for all f and x.
Hence,
�
P (Z − EZ > t) ≤ exp − �n
2 i=1
By setting t =
�
8u
n M,
�
t2
(2M )2
n2
nt2
= e− 8M 2 .
we get
�
�
P Z − EZ >
8u
M
n
�
8u
M
n
�
≤ e−u .
Similarly,
�
P EZ − Z >
�
≤ e−u .
66