Two Roles in a Convergence Theorem
A martingale can fluctuate for a long time, so convergence is not a consequence of the martingale identity at any single step. The proof of the convergence theorem uses two different controls. Upcrossing bounds prevent a path from repeatedly moving between two separated levels. Uniform integrability controls the tails of the variables and, in particular, prevents a finite limiting argument from losing expectation into increasingly rare, increasingly large values.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\). Let \((M_n)_{n\ge0}\) be a real-valued process, with state space \(S=\mathbb{R}\), such that \(M_n\) 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 \(n\). These conditions say that \((M_n)\) is a martingale. Uniform integrability is the following tail condition:
Uniform integrability implies a uniform first-moment bound. Indeed, for any \(K>0\), split according to whether \(|M_n|\le K\) to obtain \(\mathbb{E}|M_n|\le K+\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]\). Choose one \(K\) for which the supremum of the tail expectations is finite. This gives a finite constant \(C\) with \(\sup_n\mathbb{E}|M_n|\le C\).
Upcrossings Give Almost-Sure Convergence
For real numbers \(a<b\), an upcrossing of \([a,b]\) is a completed passage from a value at most \(a\) to a later value at least \(b\). Let \(U_n[a,b]\) be the maximum number of non-overlapping upcrossings that can be completed using \(M_0,\ldots,M_n\). The following bound is the key pathwise-to-probability step for a martingale.
Proof. Starting at time \(0\), use the following rule: buy one unit after observing a value at most \(a\), hold it until a value at least \(b\) is observed, sell it there, and repeat. Let \(H_{i-1}\) be the number of units held over the step from time \(i-1\) to time \(i\). The rule uses only observations through time \(i-1\), so \(H_{i-1}\) is \(\mathcal{F}_{i-1}\)-measurable and takes values in \(\{0,1\}\). The total gain through time \(n\) is \(G_n=\sum_{i=1}^n H_{i-1}(M_i-M_{i-1})\). Each summand is integrable, and the martingale property gives \(\mathbb{E}[H_{i-1}(M_i-M_{i-1})]=0\); thus \(\mathbb{E}G_n=0\). Every completed purchase-and-sale earns at least \(b-a\). If a unit is still held at time \(n\), it was bought at a value at most \(a\), so its unsold gain is at least \(M_n-a\), which is at least \(-(a-M_n)^+\). Therefore \(G_n\ge (b-a)U_n[a,b]-(a-M_n)^+\). Taking expectations and using \(\mathbb{E}G_n=0\) proves the inequality. \(\square\)
To apply the lemma, take the finite constant \(C=\sup_n\mathbb{E}|M_n|\) obtained from uniform integrability. Since \((a-M_n)^+\le |a|+|M_n|\), for every \(n\), $$ \mathbb{E}U_n[a,b]\le \frac{|a|+C}{b-a}. $$ The counts \(U_n[a,b]\) do not decrease as \(n\) increases. By monotone convergence, the total number \(U_\infty[a,b]\) of upcrossings satisfies \(\mathbb{E}U_\infty[a,b]\le (|a|+C)/(b-a)<\infty\). A nonnegative random variable with finite expectation is finite almost surely, so there are only finitely many upcrossings of each fixed interval, almost surely. Applying this to the countable collection of rational \(a<b\) gives one event of probability one on which every rational interval has only finitely many upcrossings.
On that event, the extended-real lower and upper limits of \(M_n\) must agree. If they did not, one could choose rational \(a<b\) strictly between them; the sequence would then visit values at most \(a\) and at least \(b\) infinitely often, producing infinitely many upcrossings. Thus \(M_n\) has an extended-real limit almost surely. It remains to rule out an infinite limit. For any \(K>0\), on the event \(\{M_n\to+\infty\}\), the indicators \(\mathbf{1}_{\{|M_n|>K\}}\) are eventually one. Fatou's lemma and Markov's inequality therefore give $$ \mathbb{P}(M_n\to+\infty) \le \liminf_{n\to\infty}\mathbb{P}(|M_n|>K) \le \frac{C}{K}. $$ Letting \(K\to\infty\) makes this probability zero. The same argument applies to \(M_n\to-\infty\). Consequently the limit is finite almost surely; call it \(M_\infty\). Fatou's lemma also gives \(\mathbb{E}|M_\infty|\le\liminf_n\mathbb{E}|M_n|\le C\), so the limit is integrable.
Worked Example: A Bounded Martingale with Two Possible Limits
Let \(\Omega=\{-1,1\}\), with each outcome having probability \(\tfrac12\), and let \(Y(\omega)=\omega\). Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_n=\sigma(Y)\) for \(n\ge1\), and state space \(S=\mathbb{R}\). Define \(M_0=0\) and \(M_n=Y\) for \(n\ge1\).
At the first step, \(\mathbb{E}[M_1\mid\mathcal{F}_0]=\mathbb{E}Y=\tfrac12(1)+\tfrac12(-1)=0=M_0\). For \(n\ge1\), \(M_{n+1}=M_n=Y\), so the conditional expectation identity holds as well. The possible values of every \(M_n\) lie in \([-1,1]\). For \(K\ge1\), the strict event \(\{|M_n|>K\}\) is empty, so all the tail expectations are zero. The family is uniformly integrable.
On the outcome \(Y=1\), the sequence is \(0,1,1,\ldots\), and on \(Y=-1\), it is \(0,-1,-1,\ldots\). Thus \(M_\infty=Y\). The check in \(L^1\) is \(\mathbb{E}|M_n-M_\infty|=\tfrac12|0-1|+\tfrac12|0-(-1)|=1\) for \(n=0\), and \(0\) for every \(n\ge1\). In particular, the error tends to zero, and \(\mathbb{E}M_\infty=\tfrac12-\tfrac12=0=\mathbb{E}M_0\).
Uniform Integrability Upgrades the Limit to \(L^1\)
Almost-sure convergence alone does not allow expectations of absolute differences to be passed to the limit. Here the missing control comes from uniform integrability. The short argument below establishes the upgrade directly, including tail control for the limit.
Proof of the \(L^1\) conclusion. For \(K>0\), define the clipped value \(T_K(x)=\max(-K,\min(x,K))\). This function is bounded and continuous. Since \(M_n\to M_\infty\) almost surely, bounded convergence gives \(\mathbb{E}|T_K(M_n)-T_K(M_\infty)|\to0\) for fixed \(K\). Write \(\eta(K)=\sup_n\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]\), which tends to zero as \(K\to\infty\). On \(\{|M_\infty|>2K\}\), convergence implies that eventually \(|M_n|>K\) and \(|M_n|\to|M_\infty|\). Fatou's lemma then gives $$ \mathbb{E}\bigl[|M_\infty|\mathbf{1}_{\{|M_\infty|>2K\}}\bigr] \le \liminf_{n\to\infty}\mathbb{E}\bigl[|M_n|\mathbf{1}_{\{|M_n|>K\}}\bigr] \le \eta(K). $$ For any real \(x\), \(|x-T_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}\). The triangle inequality therefore yields $$ \mathbb{E}|M_n-M_\infty| \le \mathbb{E}|T_K(M_n)-T_K(M_\infty)| +\eta(K) +\mathbb{E}\bigl[|M_\infty|\mathbf{1}_{\{|M_\infty|>K\}}\bigr]. $$ The last term is at most \(\eta(K/2)\), by the preceding tail bound with \(K/2\). For fixed \(K\), take the upper limit as \(n\to\infty\); the clipped term tends to zero. Then let \(K\to\infty\). Both \(\eta(K)\) and \(\eta(K/2)\) tend to zero, proving \(\mathbb{E}|M_n-M_\infty|\to0\). Finally, \(|\mathbb{E}M_n-\mathbb{E}M_\infty|\le\mathbb{E}|M_n-M_\infty|\), and \(\mathbb{E}M_n=\mathbb{E}M_0\) for every \(n\) by iterated conditional expectation. Hence \(\mathbb{E}M_\infty=\mathbb{E}M_0\). \(\square\)
Worked Example: A Bounded Rademacher-Series Martingale
Let \((\xi_k)_{k\ge1}\) be independent random variables with \(\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=\tfrac12\), on their product probability space. Let \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\), and take \(S=\mathbb{R}\). Set \(M_0=0\) and \(M_n=\sum_{k=1}^n2^{-k}\xi_k\).
Independence and the equal probabilities give \(\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=0\). Hence \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n+2^{-(n+1)}\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=M_n\). Also, $$ |M_n|\le\sum_{k=1}^n2^{-k}=1-2^{-n}<1. $$ Thus, for \(K\ge1\), each strict tail event \(\{|M_n|>K\}\) is empty, which verifies uniform integrability.
The infinite series \(M_\infty=\sum_{k=1}^{\infty}2^{-k}\xi_k\) converges absolutely for every outcome, since \(\sum_{k=1}^{\infty}2^{-k}=1\). Its remainder satisfies $$ |M_\infty-M_n|\le\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}. $$ This proves almost-sure convergence and gives the explicit \(L^1\) check \(\mathbb{E}|M_\infty-M_n|\le2^{-n}\to0\). In addition, independence gives \(\mathbb{E}M_\infty=\sum_{k=1}^{\infty}2^{-k}\mathbb{E}\xi_k=0=\mathbb{E}M_0\).
Why Bounded First Moments Are Not Enough
The upcrossing argument only needs a uniform bound on \(\mathbb{E}|M_n|\) to obtain almost-sure convergence of a martingale to a finite limit. Uniform integrability has a stronger role: it controls tail expectations uniformly and supplies the \(L^1\) conclusion. The distinction matters because a sequence can keep a fixed amount of expectation in values that become larger and rarer.
Worked Example: A Martingale with Bounded First Moments but No \(L^1\) Convergence
Let \(\Omega=\{-1,1\}^{\mathbb{N}}\) have independent fair coordinates \(\xi_1,\xi_2,\ldots\), and let \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\) with the usual trivial \(\mathcal{F}_0\). For \(n\ge0\), define \(M_n=2^n\mathbf{1}_{\{\xi_1=\cdots=\xi_n=1\}}\), interpreting the event at \(n=0\) as \(\Omega\). The state space is \(S=\mathbb{R}\).
On the event that the first \(n\) coordinates are all \(1\), the next value is \(2^{n+1}\) with conditional probability \(\tfrac12\) and \(0\) with conditional probability \(\tfrac12\); its conditional expectation is \(2^n\). If one of the first \(n\) coordinates is \(-1\), both \(M_n\) and \(M_{n+1}\) are zero. This verifies the martingale identity in both cases. Since the all-ones event has probability \(2^{-n}\), \(\mathbb{E}|M_n|=2^n2^{-n}=1\) for every \(n\).
The probability that the first \(n\) coordinates are all \(1\) is \(2^{-n}\), which tends to zero. Almost surely, some coordinate is therefore \(-1\), after which all later \(M_n\) are zero. Thus \(M_n\to0\) almost surely, but \(\mathbb{E}|M_n-0|=1\) for every \(n\), so there is no \(L^1\) convergence to this limit. Uniform integrability fails: for any \(K>0\), choose \(n\) with \(2^n>K\); then $$ \mathbb{E}\bigl[|M_n|\mathbf{1}_{\{|M_n|>K\}}\bigr] =2^n\mathbb{P}(\xi_1=\cdots=\xi_n=1) =2^n2^{-n}=1. $$ Consequently the supremum of the tail expectations at this same threshold \(K\) is at least \(1\), for every \(K\).
Common mistake. In the last example, the calculation \(\mathbb{E}|M_n|=1\) for every \(n\) does not show uniform integrability. It controls the whole first moment, not the tails. The required test fixes \(K\), then takes the supremum over \(n\) of \(\mathbb{E}[|M_n|\mathbf{1}_{\{|M_n|>K\}}]\). For any fixed \(K\), choosing \(n\) with \(2^n>K\) makes that tail expectation exactly \(1\), so it does not tend to zero as \(K\) increases.
Check Your Understanding
Use the upcrossing and tail-control arguments to justify your answers.
- In the expected upcrossing bound, where is the martingale property used, and why must the holding decision be measurable before the next increment?
- Why does finiteness of the upcrossing count for every rational interval imply that the extended-real limit exists?
- Which step in the proof rules out \(M_n\to+\infty\), and what uniform bound does it require?
- For the Rademacher-series example, compute the exact bound on \(|M_\infty-M_n|\) when \(n=3\).
- For the final example, at a fixed threshold \(K=5\), give an integer \(n\) for which the tail expectation is \(1\), and verify it.