Martingales - Almost Sure Convergence and Doob's … · Martingales Almost Sure Convergence and Doob’s Inequality In^es B. Almeida Probability Theory Course December 12th, 2016

MartingalesAlmost Sure Convergence and Doob’s Inequality

Ines B. Almeida

Probability Theory Course

December 12th, 2016

Ines B. Almeida (Probability Theory Course) Martingales December 12th, 2016 1 / 24

Outline

1. Introduction

2. Predictable Sequences and Stopping Times

3. Upcrossing Inequality

4. Almost Sure Convergence

5. Doob’s Decomposition

6. Doob’s Inequality


Filtrations

Definition (Filtration)

A filtration is an increasing sequence of σ-fields: Fn ⊆ Fn+1 ∀n.

A sequence Xn is adapted to a filtration Fn if Xn ∈ Fn ∀n.

The σ-field Fn = σ(X1, ...,Xn) can be thought of as the informationknown at time n.


Martingales

A martingale is the fortune of a player betting on a fair game.

Definition (Martingale)

A sequence Xn,Fn is said to be a martingale iif

1. E|Xn| <∞,

2. Xn ∈ Fn ∀n,

3. E(Xn+1|Fn) = Xn ∀n.

Property (3) may be an inequality:

I E(Xn+1|Fn) ≤ Xn ∀n iif Xn is a supermartingale,

I E(Xn+1|Fn) ≥ Xn ∀n iif Xn is a submartingale.


Predictable Sequences

Definition (Predictable Sequence)

Let Fn be a filtration. Hn is said to be a predictable sequence ifHn ∈ Fn−1 ∀n

A predictable sequence at time n can be predicted with certainty from theinformation available at time n − 1.



Let’s play a game...

I flip an even coin over and over,

I win (lose) bet if a flip results in head (tails),

I bet one euro each time.

Let Xm be the player’s fortune at time m. The winnings at time n are

(H · X )n =n∑

m=1

Hm(Xm − Xm−1).

One popular strategy is to double the bet when we lose. Is this strategyany good?



Theorem (5.2.5)

Let Xnn≥0 be a supermartingale. If Hn ≥ 0 is a predictable sequenceand each Hn is bounded, then (H · X )n is a supermartingale.

Proof:

E((H · X )n+1|Fn) = E((H · X )n|Fn) + E(Hn+1(Xn+1 − Xn)|Fn) =

= (H · X )n + Hn+1E(Xn+1 − Xn|Fn) =

= (H · X )n + Hn+1 [E(Xn+1|Fn)− Xn] ≤≤ (H · X )n

Similar results can be proved for submartingales and martingales.


Stopping Time

Definition (Stopping Time)

A r.v. N taking values in 1, 2, 3, . . . ∪ ∞ is said to be a stopping timeif N = n ∈ Fn ∀n.

In words: if N is the time a player stops gambling, the decision to stop attime n has to be measurable w.r.t. the information available at that time.


Stopping Time

Suppose Hn = 1N≥n. Since Hn is predictable (why?), we have that(H · X )n = XN∧n − X0 is a supermartingale. X0 is also a supermartingale,and it follows that

Theorem (5.2.6)

If N is a stopping time and Xn is a supermartingale, then XN∧n is also asupermartingale.


Upcrossing Inequality

Let Xn be a submartingale, a < b, N0 = −1, and

N2k−1 = inf m > N2k−2 : Xm ≤ a ,N2k = inf m > N2k−1 : Xm ≥ b .

Notice Nj are stopping times. Also, define the predictable sequence

Hm =

1 if N2k−1 < m ≤ N2k for some k0 otherwise

From N2k−1 to N2k , Xm crosses from below a to above b. Hm profits fromthese upcrossings.



Let Un =∑k : N2k ≤ n be the number of upcrossings completed up to

time n.

Theorem (5.2.7 Upcrossing Inequality)

If Xmm≥0 is a submartingale then

(b − a)E(Un) ≤ E((Xn − a)+)− E((X0 − a)+).

In words: you cannot lose money by betting on a favorable game.

This result is useful to prove the Martingale Convergence Theorem (next).



Proof: Notice that, for φ ≥ 0 convex and increasing:

E(φ(Xn+1)|Fn) ≥ φ(E(Xn+1|Fn)) ≥ φ(Xn)

where we used Jensen’s inequality and the fact that Xm is a submartingaleto show that φ(Xn) is also a submartingale. It follows that

Ym := a + (Xm − a)+

is a submartingale.



On one hand

(H · Y )n =n∑

m=1

Hm(Ym − Ym−1) =

=Un∑k≥1

N2k∑m>N2k−1

(Ym − Ym−1) +n∑

m≥n∧N2Un+1

(Ym − Ym−1) =

=Un∑k≥1

(YN2k− YN2k−1

) + (Yn − Yn∧N2Un+1) ≥

≥ Un(b − a) + (Yn − Yn∧N2Un+1) ≥ Un(b − a)

since (Yn − Yn∧N2Un+1) ≥ 0. It follows that

E((H · Y )n) ≥ (b − a)E(Un).



On the other hand, let Km = 1− Hm. It is clear that

Yn − Y0 =n∑

m=1

(Ym − Ym−1) = (H · Y )n + (K · Y )n

Since Hm, Km are predictable and Ym is a submartingale, (K · Y )n is alsoa submartingale. Thus,

E((K · Y )n) ≥ E((K · Y )0) = 0

It follows that

E((H · Y )n) = E(Yn − Y0)− E((K · Y )n) ≤≤ E(Yn − Y0) =

= E((Xn − a)+)− E((X0 − a)+).



We thus arrive to the desired result,

(b − a)E(Un) ≤ E((H · Y )n) ≤ E((Xn − a)+)− E((X0 − a)+).


Almost Sure Convergence

Theorem (5.2.8 Martingale Convergence Theorem)

If Xnn≥0 is a submartingale with E(X+n ) <∞, then Xn −→ X a.s., with

|X | <∞.

Theorem (5.2.9)

If Xn ≥ 0 is a supermartingale then Xn −→ X a.s. and E(X ) ≤ E(X0).

The above theorems do not guarantee convergence in Lp.



Theorem (5.2.8 Martingale Convergence Theorem)

If Xnn≥0 is a submartingale with E(X+n ) <∞, then Xn −→ X a.s., with

|X | <∞.

Proof: We begin by showing the number of upcrossings is finite. Since(X − a)+ ≤ X+ + |a|, and using the Upcrossing Inequality,

(b − a)E(Un) ≤ E((Xn − a)+)− E((X0 − a)+) ≤≤ E(X+

n ) + |a| − E((X0 − a)+) ≤≤ E(X+

n ) + |a| <∞

Because Un is discrete, we have Un <∞.



Un <∞ for any a < b. It follows that

P(∪a,b∈Qlim Xn < a < b < lim Xn

)= 0

As having non-zero probability would imply Un →∞. By making a, barbitrarily close, we conclude limXn exists a.e.Using Fatou’s Lemma, we have that

E(X+) = E(lim X+n ) ≤ lim E(X+

n ) <∞

and, because Xn is a submartingale,

E(X−) ≤ lim E(X−n ) = lim E(X+n )− E(Xn) ≤

≤ lim E(X+n )− E(Xn) ≤ lim E(X+

n )− E(X0) <∞

from which follows −∞ < E(X ) <∞.



Theorem (5.2.9)

If Xn ≥ 0 is a supermartingale then Xn −→ X a.s. and E(X ) ≤ E(X0).

Proof: Let Yn = −Xn ≤ 0 be a submartingale. Since Y +n = 0, limYn

exists and E|Y | <∞. by the previous theorem. It follows that limXn alsoexists E|X | <∞. Finally, notice that by Fatou’s Lemma,

E(X ) ≤ lim E(Xn) ≤ lim E(X0) = E(X0)

and we have completed the proof.

The above theorems do not guarantee convergence in Lp.


Doob’s Decomposition

Theorem (5.2.10 Doob’s Decomposition)

Any submartingale Xnn≥0 can be uniquely written as Xn = Mn + An,where Mn is a martingale and An is a predictable increasing sequence withA0 = 0.

Doob’s Decomposition Theorem allows one to reduce questions aboutsubmartingales to questions about martingales.


Doob’s Decomposition

Proof: From

E(Xn|Fn−1) = E(Mn|Fn−1) + E(An|Fn−1) =

= Mn−1 + An = Xn−1 − An−1 + An

we conclude that

a) An − An−1 = E(Xn|Fn)− Xn−1

b) Mn = Xn − An

Setting A0 = 0 uniquely defines both An and Mn. From a) follows thatAn ≥ An−1 and An ∈ Fn−1. It is easy to verify that Mn ∈ Fn. Finally,

E(Mn|Fn−1) = E(Xn|Fn)− E(An|Fn) =

= E(Xn|Fn)− An =

= Xn−1 − An−1 = Mn−1.


Doob’s Inequality

Theorem (5.4.1)

If Xn is a submartingale and N is a stopping time with P(N ≤ k) = 1 then

E(X0) ≤ E(XN) ≤ E(Xk)

Proof: Notice XN∧n is a submartingale. It follows that

E(X0) = E(XN∧0) ≤ E(XN∧k) = E(XN)

(since N ≤ k w.p. 1), thus proving the first inequality.Now let K = 1N≤n−1; Kn is predictable and (K · X )n is a submartingale.Then

E(Xk)− E(XN) = E((K · X )k) ≥ E((K · X )0) = 0

thus proving the second inequality.


Doob’s Inequality

Theorem (5.4.2 Doob’s Inequality)

Let Xm be a submartingale and define, for λ > 0,

A =

max

0≤m≤nX+m ≥ λ

.

ThenλP(A) ≤ E(Xn1A) ≤ E(X+

n ).

It means we can prove results on n positive r.v.s by looking at theexpected value of the last one of them.


Doob’s Inequality

Proof: Let N = infm : Xm ≥ λorm = n. Then

λP(A) =

∫Aλ dP ≤

∫AXn dP =

∫ΩXn1A dP = E(Xn1A).

By the previous theorem, E(XN1A) ≤ E(Xn1A). Thus,

E(Xn1A) =

∫AXn dP ≤

∫AX+n dP ≤

∫ΩX+n dP = E(X+

n )

and we proved the result.


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Martingales - Almost Sure Convergence and Doob's … · Martingales Almost Sure Convergence and Doob’s Inequality In^es B. Almeida Probability Theory Course December 12th, 2016