Lévy's Zero–one Law

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. 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 $X_1, X_2, X_3, \dots$ be a supermartingale. Suppose that the supermartingale is bounded in the sense that

: \sup_ \operatorname _t^-< \infty

where $X_t^-$ is the negative part of $X_t$, 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 $\pm 1$ 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 $X_t$. 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 $, X_n, \leq M$ always holds. In the event that the sequence $X_1,X_2,\dots$ does not converge, then $\liminf X_n$ and $\limsup X_n$ 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 ,b/math> 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 $X_t$. On the one hand, it can be shown from the definition of a supermartingale that for any $N \in \mathbf$ 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 ,b/math> 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 $u_N$ is the number of upcrossings in the sequence by time $N$, then the profit at time $N$ is at least $(b-a)u_N - 2M$: 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 $a \leq M$ and the current price is $-M$. But any strategy has expected profit at most $0$, so necessarily

: \operatorname \big _N\big\leq \frac.

By the monotone convergence theorem for expectations, this means that

: \operatorname \big lim_ u_N \bigleq \frac ,

so the expected number of upcrossings in the whole sequence is finite. It follows that the infinite-crossing event for interval ,b/math> 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 $a, b \in \mathbf$ there are finitely many upcrossings of interval ,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 $(X_n)_$ converges in mean (i.e. that \lim_ \operatorname stopping_time