L2-Bounded Martingales Converge Almost Surely

From Bounded Second Moments to Almost-Sure Convergence

A martingale may have no deterministic trend and still fluctuate from one time to the next. A uniform bound on its second moments rules out fluctuations that remain large indefinitely: the total squared size of its future increments must become small. To turn that L2 control into a statement about almost every sample path, we use a maximal inequality to control all times in a tail at once.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\), meaning an increasing sequence of sigma-fields contained in \(\mathcal{F}\). A real-valued process \((M_n)_{n\ge0}\) is a martingale with respect to this filtration if \(M_n\) is \(\mathcal{F}_n\)-measurable, \(\mathbb{E}|M_n|<\infty\), and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely for every \(n\). In this tutorial, assume also that \(\mathbb{E}[M_n^2]<\infty\) for every \(n\).

Almost-sure convergence means that the set of sample points \(\omega\) for which the numerical sequence \(M_n(\omega)\) converges has probability one. The theorem does not assert here that convergence also holds in L2. We prove the almost-sure claim using Doob’s L2 maximal inequality, a result established in the earlier material on martingale inequalities. In the form needed, if \((Y_j)_{0\le j\le N}\) is a square-integrable martingale with \(Y_0=0\), then for every \(a>0\),

The constant \(4\) comes from the L2 maximal inequality. We will apply it to the martingale formed by subtracting the value at the start of a tail, so its final second moment measures the size of the increments in that tail—not the size of the martingale from time zero.

Proof of the Convergence Theorem

Proof. Assume \(C=\sup_{n\ge0}\mathbb{E}[M_n^2]<\infty\). Define the increment \(\Delta_{n+1}=M_{n+1}-M_n\). The martingale property gives \(\mathbb{E}[\Delta_{n+1}\mid\mathcal{F}_n]=0\). Since \(M_n\) is \(\mathcal{F}_n\)-measurable,

Expanding the square therefore gives \(\mathbb{E}[M_{n+1}^2]=\mathbb{E}[M_n^2]+\mathbb{E}[\Delta_{n+1}^2]\). In particular, the sequence \(\mathbb{E}[M_n^2]\) is nondecreasing and bounded above by \(C\), so it has a finite limit \(L\). Summing the same identity from \(n\) to \(N-1\), for \(N\ge n\), yields

Here the first equality follows by applying the expansion to the martingale started at \(M_n\); distinct increments have zero cross-expectation by the same conditional-expectation calculation. Since \(\mathbb{E}[M_n^2]\to L\), the last bound tends to zero as \(n\to\infty\), uniformly over all \(N\ge n\).

Choose increasing integers \(n_r\) such that \(L-\mathbb{E}[M_{n_r}^2]\le 2^{-4r}\) for each integer \(r\ge1\). For fixed \(n=n_r\), the process \(Y_k=M_k-M_n\), \(n\le k\le N\), is a martingale starting at zero at time \(n\). Applying Doob’s inequality over this finite time interval gives, for every \(a>0\),

Let \(N\) increase. The finite-horizon events increase to the event that \(|M_k-M_n|>a\) for some \(k\ge n\), so continuity of probability for increasing events gives

The sum over \(r\ge1\) of these upper bounds is finite: \(\sum_{r\ge1}4\cdot2^{-2r}=4\sum_{r\ge1}4^{-r}=4/3\). Consequently, the probability that infinitely many of the displayed tail events occur is zero. Indeed, for any \(R\), the probability of at least one such event with index \(r\ge R\) is at most \(\sum_{r\ge R}4\cdot2^{-2r}\), which tends to zero as \(R\to\infty\).

Thus, with probability one, for all sufficiently large \(r\) and every \(k\ge n_r\), \(|M_k-M_{n_r}|\le2^{-r}\). For any \(p,q\ge n_r\), the triangle inequality then gives \(|M_p-M_q|\le |M_p-M_{n_r}|+|M_q-M_{n_r}|\le2^{1-r}\). This proves that, on an event of probability one, \((M_n)\) is a Cauchy sequence of real numbers. Completeness of the real numbers gives a finite limit on that event. Defining \(M_\infty=0\) on the remaining null set gives the claimed almost-sure convergence. \(\square\)

Worked Examples

Tail Control and a Common Mistake

The essential quantity in the proof is the second moment of a tail difference, \(\mathbb{E}[(M_N-M_n)^2]\). Bounded second moments alone would not make the maximal inequality useful if applied to \(M_k\) itself: the resulting bound would concern the whole process, not the fluctuations after time \(n\). The variance decomposition identifies the tail difference with \(\mathbb{E}[M_N^2]-\mathbb{E}[M_n^2]\), which tends uniformly to zero over \(N\ge n\). Doob’s inequality then controls the maximum over the entire tail.

Common mistake. In the random-series example, a mistaken tail calculation might use \(\mathbb{E}[M_k^2]\le\tfrac13\) to claim that the maximal probability after time \(n\) is at most \(\tfrac{4}{3a^2}\). This bound does not decrease with \(n\), so it does not establish convergence. The correction is to center the process at \(M_n\). For \(k\ge n\), the tail variance is \(\mathbb{E}[(M_k-M_n)^2]=\sum_{j=n+1}^k4^{-j}\le4^{-n}/3\). Doob’s inequality consequently gives the bound \(4\cdot4^{-n}/(3a^2)\), which tends to zero for each fixed \(a>0\). The distinction between the whole process and its increments after time \(n\) is the key to the argument.

This proof yields almost-sure convergence by showing that almost every path is Cauchy. It does not need to identify the distribution of the limit, nor does the conclusion alone provide an L2 convergence claim. Those are separate questions: here the result is precisely that a martingale with uniformly bounded second moments has a finite almost-sure limit.