A Toolkit for Martingale Convergence Problems
Martingale convergence problems often require more than one estimate. A maximal inequality controls the largest excursion over a finite time interval; an upcrossing count measures repeated passages between two levels; a moment bound can establish uniform integrability; and square-summable increments can force convergence in L2. The key is to match each tool to the quantity it actually controls.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \((\mathcal{F}_n)_{n\ge0}\) be a filtration, and let \((M_n)_{n\ge0}\) be an adapted process. It is a martingale if each \(M_n\) is integrable and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely. It is square-integrable if \(\mathbb{E}[M_n^2]<\infty\) for every \(n\). For a family of integrable random variables \(\{Z_\alpha\}\), uniform integrability means \(\sup_\alpha \mathbb{E}[|Z_\alpha|\mathbf{1}_{\{|Z_\alpha|>K\}}]\to0\) as \(K\to\infty\).
The factor \(4\) concerns the expected square of the maximum. It should not be confused with a claim that the terminal second moment itself bounds that expected square. For convergence, a different question is whether the increments have a finite total squared contribution. The next result gives a useful criterion and its proof.
Proof. Write \(D_n=M_n-M_{n-1}\) for \(n\ge1\). For \(i<j\), \(D_i\) is \(\mathcal{F}_{j-1}\)-measurable and \(\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]=0\). Therefore \(\mathbb{E}[D_iD_j]=\mathbb{E}[D_i\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]]=0\). Expanding a square and using this orthogonality gives, for \(m>n\), \(\mathbb{E}[(M_m-M_n)^2]=\sum_{j=n+1}^{m}\mathbb{E}[D_j^2]\). The assumed series has tails tending to zero, so \((M_n)\) is Cauchy in L2. Completeness of L2 gives an \(M_\infty\in L2\) with \(M_n\to M_\infty\) in L2.
It remains to obtain almost-sure convergence. Let \(v_n=\sum_{j=n+1}^{\infty}\mathbb{E}[D_j^2]\), which tends to zero. Choose increasing integers \(n_r\) such that \(v_{n_r}\le2^{-4r}\). For each finite \(m>n_r\), Doob’s L2 maximal inequality applied to the martingale \(M_k-M_{n_r}\), \(n_r\le k\le m\), gives \(\mathbb{P}(\max_{n_r\le k\le m}|M_k-M_{n_r}|>2^{-r})\le4v_{n_r}/2^{-2r}\le4\cdot2^{-2r}\). Letting \(m\) increase, continuity of probability for increasing events gives the same bound for \(\sup_{k\ge n_r}|M_k-M_{n_r}|>2^{-r}\). These bounds sum over \(r\), so the Borel–Cantelli lemma implies that almost surely, for all sufficiently large \(r\), \(\sup_{k\ge n_r}|M_k-M_{n_r}|\le2^{-r}\). For any \(p,q\ge n_r\), \(|M_p-M_q|\le|M_p-M_{n_r}|+|M_q-M_{n_r}|\le2^{1-r}\). Thus the sequence is Cauchy almost surely and converges almost surely. Its almost-sure limit equals its L2 limit almost surely, since L2 convergence implies convergence in probability and a sequence cannot have distinct almost-sure and in-probability limits. \(\square\)
Worked Problems
Worked Example: Bounding a Martingale Maximum
Let \(\varepsilon_1,\ldots,\varepsilon_4\) be independent random variables, each equal to \(1\) or \(-1\) with probability \(\tfrac12\), and define \(M_0=0\) and \(M_k=\tfrac12\sum_{j=1}^k\varepsilon_j\). Use Doob’s inequality to bound \(\mathbb{P}(\max_{0\le k\le4}|M_k|\ge2)\), then compute this probability exactly.
Use the filtration \(\mathcal{F}_k=\sigma(\varepsilon_1,\ldots,\varepsilon_k)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Independence and the zero mean of each increment give \(\mathbb{E}[M_{k+1}\mid\mathcal{F}_k]=M_k\). Also, the cross-products of distinct increments have expectation zero, so \(\mathbb{E}[M_4^2]=4\cdot\tfrac14=1\). The maximal inequality and then Markov’s inequality give \(\mathbb{P}(\max_{k\le4}|M_k|\ge2)\le\mathbb{E}[\max_{k\le4}|M_k|^2]/4\le4\cdot1/4=1\). This bound is valid but not informative.
There are \(2^4=16\) equally likely sign sequences. Since each increment has magnitude \(\tfrac12\), reaching absolute value \(2\) by time \(4\) requires all four signs to be positive or all four to be negative. There are two such sequences, so the exact probability is \(2/16=\tfrac18\). This example illustrates that a valid maximal bound may be loose; its role is to control a path maximum using a terminal moment, not necessarily to give the exact probability.
Worked Example: Counting Upcrossings of an Interval
For a real sequence \(x_0,\ldots,x_N\) and levels \(a<b\), an upcrossing is a completed passage from a value at most \(a\) to a later value at least \(b\). Count passages successively: after choosing an index with value at most \(a\), choose the first later index with value at least \(b\), then repeat after that index. Denote the number by \(U_N[a,b]\). Consider the two-step fair random walk \(M_0=0\), \(M_1=\varepsilon_1\), \(M_2=\varepsilon_1+\varepsilon_2\), and count upcrossings of \([0,1]\).
For any finite path, each completed upcrossing uses a time interval on which the total absolute increment is at least \(b-a\): if its starting value is \(x_s\le a\) and ending value is \(x_t\ge b\), then \(\sum_{j=s+1}^{t}|x_j-x_{j-1}|\ge|x_t-x_s|\ge b-a\). The intervals for successively counted upcrossings do not overlap in their increments. Summing these inequalities proves the finite-path bound \((b-a)U_N[a,b]\le\sum_{j=1}^{N}|x_j-x_{j-1}|\). This argument applies to every sequence, not only martingales.
Here \(a=0\), \(b=1\), and each of the four sign pairs has probability \(\tfrac14\). The paths are \((0,1,2)\), \((0,1,0)\), \((0,-1,0)\), and \((0,-1,-2)\). Their upcrossing counts are respectively \(1,1,0,0\), so \(\mathbb{E}[U_2[0,1]]=(1+1+0+0)/4=\tfrac12\). Each path has total absolute increment \(2\), and the pathwise bound reads \(1\cdot U_2[0,1]\le2\); taking expectations verifies \(\tfrac12\le2\). This simple bound is not a substitute for sharper upcrossing inequalities used in convergence theorems, but it is a direct check on what a completed crossing costs in path variation.
Worked Example: L2 Convergence and Uniform Integrability
Let \(\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}\), with each outcome having probability \(\tfrac14\), and let \(Y=\mathbf{1}_{\{\omega_1,\omega_2\}}\). Take \(\mathcal{F}_0\) to be trivial, \(\mathcal{F}_1\) generated by the partition \(\{\{\omega_1,\omega_2\},\{\omega_3,\omega_4\}\}\), and \(\mathcal{F}_2=\mathcal{F}\); set \(\mathcal{F}_n=\mathcal{F}\) for \(n\ge2\). Define \(M_n=\mathbb{E}[Y\mid\mathcal{F}_n]\).
Direct conditional expectations give \(M_0=\tfrac12\), \(M_1=1\) on \(\{\omega_1,\omega_2\}\) and \(0\) on \(\{\omega_3,\omega_4\}\), and \(M_n=Y\) for \(n\ge2\). These values verify the martingale property: the conditional average of \(M_1\) given the trivial \(\mathcal{F}_0\) is \(\tfrac12=M_0\), the conditional average of \(M_2=Y\) on each atom of \(\mathcal{F}_1\) equals the stated value of \(M_1\), and the process stays fixed thereafter.
The increments are \(M_1-M_0\), which equals \(\tfrac12\) or \(-\tfrac12\) with equal probabilities, and \(M_2-M_1=0\). Thus \(\mathbb{E}[(M_1-M_0)^2]=\tfrac14\), while every later increment has second moment zero. The sum of increment second moments is \(\tfrac14\); the theorem therefore gives convergence almost surely and in L2. Directly, \(M_n=Y\) for every \(n\ge2\), confirming both conclusions.
For a uniform integrability check, \(|M_n|\le1\) for all \(n\), so \(\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]=0\) for every \(K\ge1\). Hence the family is uniformly integrable. Its L1 convergence is also directly verified: for \(n\ge2\), \(\mathbb{E}|M_n-Y|=0\). This small example shows how a pathwise bound, an increment calculation, and a tail test answer distinct questions.
Moment Bounds as a Uniform Integrability Test
Proof. For \(K>0\), on the event \(\{|Z_\alpha|>K\}\) we have \(|Z_\alpha|\le |Z_\alpha|^p/K^{p-1}\), since \(|Z_\alpha|^{p-1}>K^{p-1}\). Therefore \(\mathbb{E}[|Z_\alpha|\mathbf{1}_{\{|Z_\alpha|>K\}}]\le \mathbb{E}|Z_\alpha|^p/K^{p-1}\le C/K^{p-1}\). The right-hand side is independent of \(\alpha\) and tends to zero as \(K\to\infty\). Taking the supremum over \(\alpha\) proves uniform integrability. \(\square\)
For a martingale, a uniform L2 bound is therefore one practical route to uniform integrability. The martingale convergence theorem under uniform integrability, established in the earlier material on martingale convergence, says that a uniformly integrable martingale converges almost surely and in L1. The hypotheses matter: a bound on first moments alone does not establish uniform integrability, and almost-sure convergence alone does not ensure L1 convergence. When L2 convergence is needed, the square-summable-increments theorem above provides a direct test; a uniform L2 bound also gives L2 convergence for a square-integrable martingale by the earlier L2 martingale convergence result.
A Common Mistake: Confusing the Maximum with the Endpoint
For a concrete check, take a two-step fair random walk \(M_0=0\), \(M_1=\varepsilon_1\), \(M_2=\varepsilon_1+\varepsilon_2\). The incorrect calculation \(\mathbb{E}[\max_{k\le2}|M_k|^2]\le\mathbb{E}[|M_2|^2]=2\) treats the endpoint as if it were the maximum. In fact, the four paths are \((0,1,2)\), \((0,1,0)\), \((0,-1,0)\), and \((0,-1,-2)\); their squared maxima are \(4,1,1,4\), so the expected squared maximum is \((4+1+1+4)/4=\tfrac52\), which exceeds \(2\). Doob’s inequality gives the correct upper bound \(4\mathbb{E}[|M_2|^2]=8\), and \(\tfrac52\le8\). Always apply the inequality to the maximum itself before using Markov’s inequality.
Check Your Understanding
For each question, identify the quantity being controlled and state the relevant hypothesis.
- For a square-integrable martingale with \(\mathbb{E}[M_N^2]=9\), what bound does Doob’s L2 inequality give for \(\mathbb{P}(\max_{k\le N}|M_k|\ge6)\)?
- Explain why disjoint completed upcrossings of \([a,b]\) require total absolute path variation of at least \((b-a)U_N[a,b]\).
- If \(\sup_n\mathbb{E}|Z_n|^3\le8\), give a uniform bound for \(\mathbb{E}[|Z_n|\mathbf{1}_{\{|Z_n|>K\}}]\).
- A square-integrable martingale has \(\sum_{n\ge1}\mathbb{E}[(M_n-M_{n-1})^2]=0\). What does the increment criterion imply about its paths?
- Why does a uniform bound on \(\mathbb{E}|Z_n|\) not by itself prove uniform integrability?