Doob's martingale convergence theorems explained

In mathematicsspecifically, in the theory of stochastic processesDoob's martingale convergence theorems are a collection of results on the limits of supermartingales, named after the American mathematician Joseph L. Doob.[1] Informally, the martingale convergence theorem typically refers to the result that any supermartingale satisfying a certain boundedness condition must converge. One may think of supermartingales as the random variable analogues of non-increasing sequences; from this perspective, the martingale convergence theorem is a random variable analogue of the monotone convergence theorem, which states that any bounded monotone sequence converges. There are symmetric results for submartingales, which are analogous to non-decreasing sequences.

Statement for discrete-time martingales

A common formulation of the martingale convergence theorem for discrete-time martingales is the following. Let

X1,X2,X3,...

be a supermartingale. Suppose that the supermartingale is bounded in the sense that

\supt

-]
\operatorname{E}[X
t

<infty

where

-
X
t
is the negative part of

Xt

, defined by X_t^- = -\min(X_t, 0) . Then the sequence converges almost surely to a random variable

X

with finite expectation.

There is a symmetric statement for submartingales with bounded expectation of the positive part. A supermartingale is a stochastic analogue of a non-increasing sequence, and the condition of the theorem is analogous to the condition in the monotone convergence theorem that the sequence be bounded from below. The condition that the martingale is bounded is essential; for example, an unbiased

\pm1

random walk is a martingale but does not converge.

As intuition, there are two reasons why a sequence may fail to converge. It may go off to infinity, or it may oscillate. The boundedness condition prevents the former from happening. The latter is impossible by a "gambling" argument. Specifically, consider a stock market game in which at time

t

, the stock has price

Xt

. There is no strategy for buying and selling the stock over time, always holding a non-negative amount of stock, which has positive expected profit in this game. The reason is that at each time the expected change in stock price, given all past information, is at most zero (by definition of a supermartingale). But if the prices were to oscillate without converging, then there would be a strategy with positive expected profit: loosely, buy low and sell high. This argument can be made rigorous to prove the result.

Proof sketch

The proof is simplified by making the (stronger) assumption that the supermartingale is uniformly bounded; that is, there is a constant

M

such that

|Xn|\leqM

always holds. In the event that the sequence

X1,X2,...

does not converge, then

\liminfXn

and

\limsupXn

differ. If also the sequence is bounded, then there are some real numbers

a

and

b

such that

a<b

and the sequence crosses the interval

[a,b]

infinitely often. That is, the sequence is eventually less than

a

, and at a later time exceeds

b

, and at an even later time is less than

a

, and so forth ad infinitum. These periods where the sequence starts below

a

and later exceeds

b

are called "upcrossings".

Consider a stock market game in which at time

t

, one may buy or sell shares of the stock at price

Xt

. On the one hand, it can be shown from the definition of a supermartingale that for any

N\inN

there is no strategy which maintains a non-negative amount of stock and has positive expected profit after playing this game for

N

steps. On the other hand, if the prices cross a fixed interval

[a,b]

very often, then the following strategy seems to do well: buy the stock when the price drops below

a

, and sell it when the price exceeds

b

. Indeed, if

uN

is the number of upcrossings in the sequence by time

N

, then the profit at time

N

is at least

(b-a)uN-2M

: each upcrossing provides at least

b-a

profit, and if the last action was a "buy", then in the worst case the buying price was

a\leqM

and the current price is

-M

. But any strategy has expected profit at most

0

, so necessarily

\operatorname{E}[uN]\leq

2M
b-a

.

By the monotone convergence theorem for expectations, this means that

\operatorname{E}[\limNuN]\leq

2M
b-a

,

so the expected number of upcrossings in the whole sequence is finite. It follows that the infinite-crossing event for interval

[a,b]

occurs with probability

0

. By a union bound over all rational

a

and

b

, with probability

1

, no interval exists which is crossed infinitely often. If for all

a,b\inQ

there are finitely many upcrossings of interval

[a,b]

, then the limit inferior and limit superior of the sequence must agree, so the sequence must converge. This shows that the martingale converges with probability

1

.

Failure of convergence in mean

Under the conditions of the martingale convergence theorem given above, it is not necessarily true that the supermartingale

(Xn)n

converges in mean (i.e. that

\limn\operatorname{E}[|Xn-X|]=0

).

As an example,[2] let

(Xn)n

be a

\pm1

random walk with

X0=1

. Let

N

be the first time when

Xn=0

, and let

(Yn)n

be the stochastic process defined by

Yn:=Xmin(N,

. Then

N

is a stopping time with respect to the martingale

(Xn)n

, so

(Yn)n

is also a martingale, referred to as a stopped martingale. In particular,

(Yn)n

is a supermartingale which is bounded below, so by the martingale convergence theorem it converges pointwise almost surely to a random variable

Y

. But if

Yn>0

then

Yn+1=Yn\pm1

, so

Y

is almost surely zero.

This means that

\operatorname{E}[Y]=0

. However,

\operatorname{E}[Yn]=1

for every

n\geq1

, since

(Yn)n

is a random walk which starts at

1

and subsequently makes mean-zero moves (alternately, note that

\operatorname{E}[Yn]=\operatorname{E}[Y0]=1

since

(Yn)n

is a martingale). Therefore

(Yn)n

cannot converge to

Y

in mean. Moreover, if

(Yn)n

were to converge in mean to any random variable

R

, then some subsequence converges to

R

almost surely. So by the above argument

R=0

almost surely, which contradicts convergence in mean.

Statements for the general case

In the following,

(\Omega,F,F*,P)

will be a filtered probability space where

F*=(Ft)t

, and

N:[0,infty) x \Omega\toR

will be a right-continuous supermartingale with respect to the filtration

F*

; in other words, for all

0\leqs\leqt<+infty

,

Ns\geq\operatorname{E}[Nt\midFs].

Doob's first martingale convergence theorem

Doob's first martingale convergence theorem provides a sufficient condition for the random variables

Nt

to have a limit as

t\to+infty

in a pointwise sense, i.e. for each

\omega

in the sample space

\Omega

individually.

For

t\geq0

, let
-
N
t

=max(-Nt,0)

and suppose that

\supt\operatorname{E}[

-
N
t

]<+infty.

Then the pointwise limit

N(\omega)=\limtNt(\omega)

exists and is finite for

P

-almost all

\omega\in\Omega

.[3]

Doob's second martingale convergence theorem

It is important to note that the convergence in Doob's first martingale convergence theorem is pointwise, not uniform, and is unrelated to convergence in mean square, or indeed in any Lp space. In order to obtain convergence in L1 (i.e., convergence in mean), one requires uniform integrability of the random variables

Nt

. By Chebyshev's inequality, convergence in L1 implies convergence in probability and convergence in distribution.

The following are equivalent:

(Nt)t>0

is uniformly integrable, i.e.

\limC\supt

\int
\{\omega\in\Omega\mid|Nt(\omega)|>C\
} \left| N_t (\omega) \right| \, \mathrm \mathbf (\omega) = 0;

N\inL1(\Omega,P;R)

such that

Nt\toN

as

t\toinfty

both

P

-almost surely and in

L1(\Omega,P;R)

, i.e.

\operatorname{E}\left[\left|Nt-N\right|\right]=\int\Omega\left|Nt(\omega)-N(\omega)\right|dP(\omega)\to0ast\to+infty.

Doob's upcrossing inequality

The following result, called Doob's upcrossing inequality or, sometimes, Doob's upcrossing lemma, is used in proving Doob's martingale convergence theorems.[3] A "gambling" argument shows that for uniformly bounded supermartingales, the number of upcrossings is bounded; the upcrossing lemma generalizes this argument to supermartingales with bounded expectation of their negative parts.

Let

N

be a natural number. Let

(Xn)n

be a supermartingale with respect to a filtration

(l{F}n)n

. Let

a

,

b

be two real numbers with

a<b

. Define the random variables

(Un)n

so that

Un

is the maximum number of disjoint intervals
[n
i1

,

n
i2

]

with
n
i2

\leqn

, such that
X
n
i1

<a<b<

X
n
i2

. These are called upcrossings with respect to interval

[a,b]

. Then

(b-a)\operatorname{E}[Un]\le\operatorname{E}[(Xn-a)-].

where

X-

is the negative part of

X

, defined by X^- = -\min(X, 0) .[4] [5]

Applications

Convergence in Lp

Let

M:[0,infty) x \Omega\toR

be a continuous martingale such that

\supt\operatorname{E}[|Mt|p]<+infty

for some

p>1

. Then there exists a random variable

M\inLp(\Omega,P;R)

such that

Mt\toM

as

t\to+infty

both

P

-almost surely and in

Lp(\Omega,P;R)

.

The statement for discrete-time martingales is essentially identical, with the obvious difference that the continuity assumption is no longer necessary.

Lévy's zero–one law

Doob's martingale convergence theorems imply that conditional expectations also have a convergence property.

Let

(\Omega,F,P)

be a probability space and let

X

be a random variable in

L1

. Let

F*=(Fk)k

be any filtration of

F

, and define

Finfty

to be the minimal σ-algebra generated by

(Fk)k

. Then

\operatorname{E}[X\midFk]\to\operatorname{E}[X\midFinfty]ask\toinfty

both

P

-almost surely and in

L1

.

This result is usually called Lévy's zero–one law or Levy's upwards theorem. The reason for the name is that if

A

is an event in

Finfty

, then the theorem says that

P[A\midFk]\to1A

almost surely, i.e., the limit of the probabilities is 0 or 1. In plain language, if we are learning gradually all the information that determines the outcome of an event, then we will become gradually certain what the outcome will be. This sounds almost like a tautology, but the result is still non-trivial. For instance, it easily implies Kolmogorov's zero–one law, since it says that for any tail event A, we must have

P[A]=1A

almost surely, hence

P[A]\in\{0,1\}

.

Similarly we have the Levy's downwards theorem :

Let

(\Omega,F,P)

be a probability space and let

X

be a random variable in

L1

. Let

(Fk)k

be any decreasing sequence of sub-sigma algebras of

F

, and define

Finfty

to be the intersection. Then

\operatorname{E}[X\midFk]\to\operatorname{E}[X\midFinfty]ask\toinfty

both

P

-almost surely and in

L1

.

See also

References

. Bernt Øksendal. Stochastic Differential Equations: An Introduction with Applications . Sixth. Springer. Berlin . 2003 . 3-540-04758-1. (See Appendix C)

Notes and References

  1. Book: Doob, J. L.. Stochastic Processes. Wiley. New York. 1953.
  2. Book: Durrett , Rick . Rick Durrett

    . Rick Durrett. Probability: theory and examples. Second. Duxbury Press. 1996 . 978-0-534-24318-0.

    Book: 4th edition. 2010. 9781139491136. Durrett. Rick. Cambridge University Press .
  3. Web site: Martingale Convergence Theorem. Massachusetts Institute of Tecnnology, 6.265/15.070J Lecture 11-Additional Material, Advanced Stochastic Processes, Fall 2013, 10/9/2013.
  4. Book: Functional Analysis for Probability and Stochastic Processes: An Introduction. Bobrowski, Adam. 2005. Cambridge University Press. 113–114. 9781139443883.
  5. Gushchin, A. A.. On pathwise counterparts of Doob's maximal inequalities. Proceedings of the Steklov Institute of Mathematics. 287. 287. 118–121. 1410.8264. 2014. 10.1134/S0081543814080070. 119150374 .
  6. Book: Doob, Joseph L.. Measure theory. Graduate Texts in Mathematics, Vol. 143. Springer. 1994. 197. 9781461208778.