Why Bounded First Moments Control Oscillation
A submartingale can move upward on average while still fluctuating along individual sample paths. To prove that its paths settle down, it is useful to count a specific kind of fluctuation: an upcrossing of an interval \([a,b]\), where \(a<b\). Each upcrossing requires the process to visit at or below \(a\), and later at or above \(b\). A bound on the expected number of these crossings will rule out persistent oscillation.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with filtration \((\mathcal{F}_n)_{n\ge0}\). A real-valued process \((X_n)_{n\ge0}\) is a submartingale if each \(X_n\) is \(\mathcal{F}_n\)-measurable and integrable, and \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]\ge X_n\) almost surely for every \(n\). We assume throughout that the process is \(L^1\)-bounded, meaning that there is a finite constant \(C\) such that \(\mathbb{E}|X_n|\le C\) for every \(n\).
The Expected-Upcrossing Estimate
The estimate comes from a bounded predictable strategy. The strategy holds one unit of the process after observing a value at or below \(a\), and stops holding after observing a value at or above \(b\). A submartingale has nonnegative expected gain for this strategy. Applying the same fact to the complementary strategy lets us bound that gain from above as well.
Proof. For each increment \(X_k-X_{k-1}\), with \(1\le k\le n\), let \(H_k\) be \(1\) if the strategy is holding after observing \(X_{k-1}\), and \(0\) otherwise. It begins not holding, starts holding when it observes \(X_{k-1}\le a\), and stops holding when it observes \(X_{k-1}\ge b\). Thus \(H_k\) is \(\mathcal{F}_{k-1}\)-measurable. Set \(G_n=\sum_{k=1}^n H_k(X_k-X_{k-1})\). Since \(0\le H_k\le1\), integrability and the submartingale property give \(\mathbb{E}[H_k(X_k-X_{k-1})]=\mathbb{E}[H_k\mathbb{E}[X_k-X_{k-1}\mid\mathcal{F}_{k-1}]]\ge0\). Hence \(\mathbb{E}G_n\ge0\). Applying the same argument to \(1-H_k\) shows \(\mathbb{E}\sum_{k=1}^n(1-H_k)(X_k-X_{k-1})\ge0\). The two sums add to \(X_n-X_0\), so \(\mathbb{E}G_n\le\mathbb{E}[X_n-X_0]\).
Path by path, every completed holding period buys at a value at most \(a\) and sells at a value at least \(b\), gaining at least \(b-a\). If the strategy is still holding at time \(n\), its last, unfinished gain is \(X_n-X_s\), where the purchase value satisfies \(X_s\le a\). Therefore that gain is at least \(X_n-a\), and hence at least \(-(a-X_n)^+\). There are \(U_n(a,b)\) completed holding periods, so $$ G_n\ge (b-a)U_n(a,b)-(a-X_n)^+. $$ Taking expectations and using \(\mathbb{E}G_n\le\mathbb{E}[X_n-X_0]\) proves the first inequality. For the uniform bound, \(\mathbb{E}X_n\le C\), \(-\mathbb{E}X_0\le C\), and \((a-X_n)^+\le |a|+|X_n|\), so the right-hand side is at most \(3C+|a|\). This proves the second inequality. \(\square\)
The estimate is uniform in \(n\) when \(a\) and \(b\) are fixed. That uniformity is essential: because \(U_n(a,b)\) is nondecreasing, the limiting count \(U_\infty(a,b)=\lim_{n\to\infty}U_n(a,b)\) satisfies $$ \mathbb{E}U_\infty(a,b)=\lim_{n\to\infty}\mathbb{E}U_n(a,b)\le \frac{3C+|a|}{b-a}. $$ Here the equality follows from monotone convergence. In particular, \(U_\infty(a,b)\) is finite almost surely: a nonnegative random variable with finite expectation cannot be infinite on an event of positive probability.
Worked Example: A Deterministic Submartingale
Take a one-point probability space, with its trivial filtration, and define \(X_n=1-2^{-n}\) for \(n\ge0\). The process is adapted and integrable. Since \(X_{n+1}-X_n=2^{-(n+1)}>0\), it is a submartingale. Also \(0\le X_n<1\), so \(\mathbb{E}|X_n|=X_n\le1\) for every \(n\), and we can take \(C=1\).
Consider the interval \([a,b]=[\tfrac14,\tfrac34]\). The values at times \(0,1,2\) are \(X_0=0\), \(X_1=\tfrac12\), and \(X_2=\tfrac34\). Thus the process begins below \(a\) and first reaches or exceeds \(b\) at time \(2\). Since it is increasing, it cannot complete another upcrossing of this interval. Therefore \(U_n(a,b)=0\) for \(n=0,1\), and \(U_n(a,b)=1\) for every \(n\ge2\).
Check the bound: For \(n\ge2\), its right-hand side is $$ \frac{3C+|a|}{b-a} =\frac{3+\tfrac14}{\tfrac34-\tfrac14} =\frac{13}{2}. $$ The actual expected count is \(1\), which is no greater than \(\tfrac{13}{2}\). Directly, the sharper right-hand side of the first inequality at \(n=2\) is \(X_2-X_0+(a-X_2)^+=\tfrac34+0=\tfrac34\), while \((b-a)\mathbb{E}U_2=\tfrac12\). The inequality holds.
Worked Example: A Random One-Step Martingale
Let the probability space have two outcomes, with probabilities \(\tfrac13\) and \(\tfrac23\). Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_n\) for \(n\ge1\) contain all events, and set \(X_0=0\). On the first outcome set \(X_n=2\) for all \(n\ge1\); on the second set \(X_n=-1\) for all \(n\ge1\). Then \(\mathbb{E}X_1=\tfrac13(2)+\tfrac23(-1)=0=X_0\), and after time \(1\) the process is constant. It is therefore a martingale, and in particular a submartingale.
For every \(n\ge1\), \(\mathbb{E}|X_n|=\tfrac13(2)+\tfrac23(1)=\tfrac43\); at time \(0\) the expected absolute value is \(0\). Thus \(C=\tfrac43\) works. On the first outcome the sequence makes one upcrossing of \([0,1]\), since \(X_0=0\) and \(X_1=2\). On the second outcome it makes none. For \(n\ge1\), this gives \(\mathbb{E}U_n(0,1)=\tfrac13\).
Check the estimate: For \(n\ge1\), \(\mathbb{E}[X_n-X_0]=0\) and \(\mathbb{E}[(0-X_n)^+]=\tfrac23\). Thus the first bound gives \(\mathbb{E}U_n(0,1)\le\tfrac23\), consistent with the computed value \(\tfrac13\). The bound is not generally an equality.
Almost-Sure Convergence to a Finite Limit
Proof. Let \(C=\sup_n\mathbb{E}|X_n|\). For each rational pair \(a<b\), the expected-upcrossing estimate gives \(\sup_n\mathbb{E}U_n(a,b)\le(3C+|a|)/(b-a)<\infty\). By monotone convergence, \(U_\infty(a,b)=\lim_n U_n(a,b)\) has finite expectation and is finite almost surely. There are only countably many rational pairs, so with probability one every such pair has only finitely many upcrossings.
On any path with this property, the extended-real sequence \((X_n)\) must have an extended limit. Indeed, if its lower limit were strictly less than its upper limit, there would be rational numbers \(a<b\) strictly between them. The path would then have values at or below \(a\) arbitrarily late, and values at or above \(b\) arbitrarily late. Those two kinds of values must occur alternately in the order needed to make infinitely many upcrossings, contradicting finiteness of \(U_\infty(a,b)\). Thus the lower and upper limits agree, allowing at this stage a limit of \(+\infty\) or \(-\infty\).
It remains to rule out an infinite limit. For any \(K>0\), Markov's inequality gives \(\mathbb{P}(|X_n|>K)\le C/K\) for every \(n\). If \(|X_n|\to\infty\), then \(|X_n|>K\) eventually, so this event is contained in \(\liminf_n\{|X_n|>K\}\). Fatou's lemma applied to the indicators yields $$ \mathbb{P}\bigl(\liminf_n\{|X_n|>K\}\bigr) \le \liminf_{n\to\infty}\mathbb{P}(|X_n|>K) \le \frac{C}{K}. $$ Consequently, the probability that \(|X_n|\to\infty\) is at most \(C/K\) for every \(K>0\), and hence is zero as \(K\to\infty\). Together with the existence of an extended limit, this proves that the limit is finite almost surely. \(\square\)
Worked Example: Why the First-Moment Bound Matters
On a one-point probability space with the trivial filtration, let \(X_n=n\). This is an integrable submartingale, since \(X_{n+1}=n+1\ge n=X_n\). Its expected absolute value is \(\mathbb{E}|X_n|=n\), so there is no finite uniform \(L^1\) bound. The sequence tends to \(+\infty\), not to a finite real limit.
For the interval \([0,1]\), its successive values \(0,1,2,\ldots\) make exactly one upcrossing: the first pair is \(X_0=0\) and \(X_1=1\), and the process never returns to a value at or below \(0\). Thus \(U_n(0,1)=1\) for every \(n\ge1\). The upcrossing count alone permits convergence to an infinite limit; the uniform first-moment bound is also needed to ensure that the limit is finite.
Check: The process is deterministic, so \(\mathbb{E}|X_n|=n\) exactly. In the estimate, the right-hand side for \(a=0,b=1\) is \(\mathbb{E}[X_n-X_0]+\mathbb{E}[(-X_n)^+]=n+0=n\), which bounds the actual expected count \(1\) for \(n\ge1\). The bound is not uniform in \(n\).
What the Result Does—and Does Not—Say
The theorem gives almost-sure convergence, a statement about the sample paths outside an event of probability zero. It does not assert convergence in \(L^1\), which would require \(\mathbb{E}|X_n-X_\infty|\to0\). A uniform bound on \(\mathbb{E}|X_n|\) is enough for the pathwise conclusion proved here, but is not by itself a guarantee of that stronger mode of convergence.
Common mistake: using nonnegative expected strategy gain as an upper bound on upcrossings. The holding strategy above has \(\mathbb{E}G_n\ge0\), but that does not imply \((b-a)\mathbb{E}U_n(a,b)\le\mathbb{E}G_n\). For a concrete check, use the two-outcome martingale in the previous example with \(a=0\), \(b=1\), and \(n=1\). Its expected strategy gain is \(\mathbb{E}G_1=0\), while \(\mathbb{E}U_1(0,1)=\tfrac13\). The proposed inequality would say \(\tfrac13\le0\), which is false. The unfinished holding period can have a negative gain. The valid pathwise bound includes the correction \(-(a-X_n)^+\); accounting for that term is what makes the upcrossing estimate correct.
Check Your Understanding
For each question, distinguish the pathwise upcrossing count from its expectation, and state which assumptions are being used.
- For \(X_n=1-2^{-n}\), how many upcrossings of \([\tfrac14,\tfrac34]\) have occurred by time \(n=1\), and by time \(n=2\)?
- In the two-outcome martingale example, what is \(\mathbb{E}|X_n|\) for \(n\ge1\), and what is \(\mathbb{E}U_n(0,1)\)?
- Why does finite expected \(U_\infty(a,b)\) imply that \(U_\infty(a,b)\) is finite almost surely?
- In the convergence proof, what does the existence of only finitely many upcrossings for every rational interval rule out about the lower and upper limits?
- For the deterministic process \(X_n=n\), which hypothesis of the convergence theorem fails, and what does this show about the role of that hypothesis?