Doob's Upcrossing Inequality and the Almost-Sure Martingale Convergence Theorem

Intuition Beginner

A martingale is the running fortune in a fair game: given everything seen so far, your best forecast for tomorrow's fortune is exactly today's. The previous unit asked what happens when you stop such a game. This unit asks a different question: what happens if you never stop, and just let the game run forever? Does the fortune settle down to some final number, or can it wander up and down without end?

The surprising answer is that a fair game whose fortune stays bounded on average must settle down. It cannot oscillate forever. To see why, count how many times the fortune crosses from below some low level up past some high level — an "upcrossing" of the band between them. Every full upcrossing means the fortune climbed through the whole band. Doob's insight is that a fair game cannot afford to make many such climbs: each completed climb is, in effect, a guaranteed profit of the band's width, and a fair game forbids guaranteed profit. So the expected number of upcrossings of any fixed band is finite.

Now picture a sequence that fails to settle: it must dip below the low level and rise above the high level again and again, infinitely often, for some band. That would require infinitely many upcrossings of that band. Since every band can only be upcrossed finitely many times on average, no band is crossed infinitely often, and the fortune is trapped into converging to a single limiting value.

The takeaway: bounded fair games converge. But there is a catch that the rest of the unit is about. The fortune converges to a final value, yet that final value need not be the "true" target the fortune was tracking — mass can leak away to infinity. Spotting exactly when this leak happens, and when it does not, is the heart of the matter.

Visual Beginner

Picture a wandering fortune drawn as a jagged line over time, with two horizontal guide lines marking a low level $a$ and a high level $b$. An upcrossing is one complete journey of the line from at-or-below $a$ up to at-or-above $b$.

Each arrow is one completed upcrossing: a full trip from the lower line to the upper line. Doob's count says that, for a bounded fair game, the average number of such arrows over the whole infinite future is finite. A path that failed to settle would need infinitely many arrows for some choice of band, which the count forbids. So the path flattens to a limit on the right.

The shaded band can be any pair of levels you like. Because no band is crossed infinitely often, the high points and low points of the path eventually stop straddling any fixed band, which is exactly what it means for the path to home in on one value.

Worked example Beginner

We watch a concrete fair game and see both the convergence and the catch. Consider a fortune that starts at $1$ and, at each step, either doubles or drops to $0$. Specifically: while the fortune is positive, flip a fair-ish coin that says "survive" with probability $1/2$ and "die" with probability $1/2$; on "survive" the fortune doubles, on "die" it goes to $0$ and stays there forever.

Step 1. Check it is a fair game. If the current fortune is $x>0$, the next fortune is $2x$ with probability $1/2$ and $0$ with probability $1/2$. The average next fortune is $\frac{1}{2}(2x)+\frac{1}{2}(0)=x$, the current value. So the expected next fortune equals the current fortune: this is a martingale, a fair game. If the fortune is already $0$ it stays $0$, which is also fair.

Step 2. Find the limit. The fortune survives step after step only as long as every coin says "survive." The chance of surviving the first $n$ steps is $(1/2{)}^n$, which shrinks to $0$. So with probability $1$ the game eventually dies, and once dead the fortune is $0$ forever. The limiting fortune is therefore $0$, with probability $1$. The bounded-below fair game has converged, exactly as promised.

Step 3. Spot the catch. The starting fortune was $1$, and the average fortune at every single step is also $1$ (a fair game keeps its average). Yet the limit is $0$. The average of the limit is $0$, but the limit of the averages is $1$. These two numbers disagree.

Step 4. Where did the dollar go? On the rare surviving paths the fortune is enormous — after $n$ survivals it is $2^n$ — and these huge-but-rare values are exactly what keeps the running average at $1$. As $n$ grows, the dollar's worth of average is carried by an ever-rarer, ever-larger spike that escapes to infinity. In the limit the spike has vanished from view (probability $0$), taking the dollar with it.

What this tells us: a bounded fair game always converges to some final value, but the average can leak away to infinity along rare exploding paths. The final value tracks the typical outcome; it need not preserve the mean. The condition that plugs this leak — keeping the rare spikes from carrying away mass — is what the higher tiers call uniform integrability.

Check your understanding Beginner

Formal definition Intermediate+

Fix a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_n{)}_{n\ge 0},\mathbb{P})$. We use the martingale vocabulary of 37.04.01: an adapted integrable process $(X_n)$ is a supermartingale if $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]\le X_n$ a.s., a submartingale if the reverse inequality holds, and a martingale when equality holds. The martingale transform $C\bullet X$ of a predictable $(C_n)$ against $(X_n)$, the stopping-time apparatus, and the optional-stopping theorem are taken as verified there; conditional expectation and its tower/Jensen properties are taken as verified from the Radon-Nikodym theorem 02.07.08.

Definition (upcrossings of an interval). Let $(X_n)$ be a process and $a<b$ real. Define stopping times $S_1<T_1<S_2<T_2<\cdots$ by $$ S_1 = \inf{n \ge 0 : X_n \le a}, \qquad T_k = \inf{n \ge S_k : X_n \ge b}, \qquad S_{k+1} = \inf{n \ge T_k : X_n \le a}, $$ with $\inf \varnothing =\infty$. The interval $[S_k,T_k]$ is the $k$-th upcrossing of $[a,b]$ once $T_k<\infty$. The number of upcrossings completed by time $N$ is $$ U_N([a,b]) = \sup{k \ge 0 : T_k \le N}, $$ a nondecreasing-in-$N$, ${\mathcal{F}}_N$-measurable random variable; its monotone limit $U_{\infty }([a,b])={\lim }_{N\to \infty }{U}_N([a,b])$ counts all upcrossings over the infinite horizon.

Definition ($L^1$-boundedness). A process $(X_n)$ is $L^1$-bounded if ${\sup }_n\mathbb{E}|X_n|<\infty$. This is the standing hypothesis of the convergence theorem; it is strictly weaker than uniform integrability.

Definition (uniform integrability). A family $(X_n)$ is uniformly integrable (UI) if $$ \lim_{c \to \infty} \sup_n \mathbb{E}\big[|X_n|,\mathbf{1}_{{|X_n| > c}}\big] = 0. $$ UI implies $L^1$-boundedness; the converse fails. The role of UI in this unit is exactly to forbid the escape-of-mass-to-infinity seen in the Beginner worked example.

Definition (closed martingale). A martingale $(X_n)$ is closed (by $X_{\infty }$) if there exists $X_{\infty }\in L^1$ with $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ for all $n$. A closed martingale is the Radon-Nikodym density process $X_n=d(X_{\infty }d\mathbb{P})/d\mathbb{P}{|}_{{\mathcal{F}}_n}$ of 02.07.08; the closure question — whether the a.s. limit $X_{\infty }$ closes the martingale — is the central dichotomy below.

Counterexamples to common slips Intermediate+

• A.s. convergence does not give $\mathbb{E}[X_{\infty }]=\mathbb{E}[X_0]$. The Beginner game converges to $X_{\infty }=0$ while $\mathbb{E}[X_n]=1$ for all $n$. The mean is not preserved in the limit without uniform integrability; $L^1$-boundedness alone is too weak.

• $L^1$-bounded is not uniformly integrable. The same game is $L^1$-bounded ($\mathbb{E}|X_n|=1$) yet not UI: the mass at level $2^n$ on the surviving event of probability $2^{-n}$ refuses to be controlled uniformly in $n$.

• Convergence is to a finite limit, but only because of $L^1$-boundedness. A martingale that is not $L^1$-bounded, such as the symmetric simple random walk, oscillates and does not converge: $\mathrm{limsup}{X}_n=+\infty$ and $\mathrm{liminf}{X}_n=-\infty$ a.s. The upcrossing bound is vacuous there because $\mathbb{E}[(X_N-a{)}^{-}]$ is unbounded in $N$.

• Closure requires more than a limit existing. Even when $X_n\to X_{\infty }$ a.s. with $X_{\infty }\in L^1$, it can fail that $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$: in the Beginner game $\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]=0\ne X_n$ on the survival event. A.s. convergence and closure are different statements; UI is the bridge between them.

• The interval must be fixed before counting. $U_{\infty }([a,b])$ is finite for each fixed rational pair $a<b$; one cannot conclude convergence from finiteness for a single cleverly chosen $\omega$-dependent band. The argument quantifies over the countable family of rational bands.

Key theorem with proof Intermediate+

Theorem (Doob's upcrossing inequality and almost-sure martingale convergence). Let $(X_n)$ be a supermartingale and $a<b$.

(Upcrossing inequality.) For every $N$, $$ (b - a),\mathbb{E}[U_N([a,b])] \le \mathbb{E}[(X_N - a)^-]. $$

(Convergence.) If in addition $(X_n)$ is $L^1$-bounded, then $X_n\to X_{\infty }$ a.s. for some $X_{\infty }\in L^1$. In particular every $L^1$-bounded martingale converges a.s. to an integrable limit.

Proof. The upcrossing inequality. Build the predictable "buy at $a$, sell at $b$" strategy $C$: set $C_1=1_{\{X_0\le a\}}$ — hold the asset only while below $a$ awaiting a rise — and for $n\ge 2$, $$ C_n = \mathbf{1}{{C{n-1} = 1}}\mathbf{1}{{X{n-1} < b}} + \mathbf{1}{{C{n-1} = 0}}\mathbf{1}{{X{n-1} \le a}}. $$ The first term keeps you holding once bought until the price reaches $b$; the second buys afresh whenever the price is at or below $a$ and you are not holding. Each $C_n$ is a function of $X_0,\dots ,X_{n-1}$, hence ${\mathcal{F}}_{n-1}$-measurable: $C$ is predictable, and $C_n\in \{0,1\}$.

During the $k$-th upcrossing $[S_k,T_k]$ you hold the asset throughout, so the transform accumulates $X_{T_k}-X_{S_k}\ge b-a$ from that completed upcrossing. After the last completed upcrossing there may be one incomplete holding period in progress at time $N$, contributing $X_N-X_S$ where $X_S\le a$ at the buy; this contributes at least $-(X_N-a{)}^{-}=\min (X_N-a,0)$. Summing the gains, $$ (C \bullet X)N = \sum{k} (\text{completed upcrossing gains}) + (\text{current incomplete gain}) \ge (b - a),U_N([a,b]) - (X_N - a)^-. $$ Because $C$ is predictable, bounded, and nonnegative and $X$ is a supermartingale, the transform $C\bullet X$ is itself a supermartingale 37.04.01, so $\mathbb{E}[(C\bullet X{)}_N]\le (C\bullet X{)}_0=0$. Taking expectations of the displayed inequality, $$ 0 \ge \mathbb{E}[(C \bullet X)_N] \ge (b - a),\mathbb{E}[U_N([a,b])] - \mathbb{E}[(X_N - a)^-], $$ which rearranges to the claimed bound.

From the inequality to convergence. Suppose $(X_n)$ is $L^1$-bounded, $\Lambda :={\sup }_n\mathbb{E}|X_n|<\infty$. Fix rationals $a<b$. Since $(X_N-a{)}^{-}\le |X_N|+|a|$, the upcrossing bound gives, for every $N$, $$ \mathbb{E}[U_N([a,b])] \le \frac{\mathbb{E}[(X_N - a)^-]}{b - a} \le \frac{\Lambda + |a|}{b - a}. $$ By monotone convergence $\mathbb{E}[U_{\infty }([a,b])]={\lim }_N\mathbb{E}[U_N([a,b])]\le \frac{\Lambda +|a|}{b-a}<\infty$, so $U_{\infty }([a,b])<\infty$ a.s. Let $$ \Lambda_{a,b} = {\liminf_n X_n < a < b < \limsup_n X_n}. $$ On ${\Lambda }_{a,b}$ the process dips below $a$ and rises above $b$ infinitely often, forcing $U_{\infty }([a,b])=\infty$; hence $\mathbb{P}({\Lambda }_{a,b})=0$. The event that $(X_n)$ fails to converge in $[-\infty ,+\infty ]$ is exactly $$ {\liminf_n X_n < \limsup_n X_n} = \bigcup_{\substack{a < b \ a, b \in \mathbb{Q}}} \Lambda_{a,b}, $$ a countable union of null sets, hence null. So $X_n\to X_{\infty }$ a.s. in $[-\infty ,+\infty ]$. Fatou's lemma 02.07.05 applied to $|X_n|\ge 0$ gives $\mathbb{E}|X_{\infty }|\le {\mathrm{liminf}}_n\mathbb{E}|X_n|\le \Lambda <\infty$, so $X_{\infty }$ is a.s. finite and integrable. For a martingale, $L^1$-boundedness is the hypothesis and the same argument applies verbatim. $\square$

Bridge. The upcrossing inequality builds toward the entire convergence theory of martingales and appears again in the $L^p$ and uniform-integrability refinements that decide closure, and it generalises the elementary observation that a bounded monotone sequence converges: where monotonicity forbids any band-crossing outright, the martingale property only forbids too many on average, and the foundational reason is identical — a guaranteed gain of $b-a$ per crossing is incompatible with a fair (or unfavourable) game. This is exactly the discrete shadow of the fact that a bounded local martingale is a genuine martingale that converges, the central insight being that oscillation across a fixed band is a profit machine the supermartingale inequality cannot finance. Putting these together, $L^1$-boundedness controls $\mathbb{E}[(X_N-a{)}^{-}]$, the upcrossing bound converts that single integrability estimate into a.s. convergence simultaneously for every rational band, and the bridge is that almost-sure convergence is, at bottom, a finiteness statement about upcrossing counts. The limit $X_{\infty }$ produced here is the candidate that, under the extra hypothesis of uniform integrability, will close the martingale as the Radon-Nikodym density of 02.07.08; without that hypothesis the limit exists but is dual to a loss of mass, which is exactly the dichotomy the Master tier resolves.

Exercises Intermediate+

Advanced results Master

The upcrossing inequality and the a.s. convergence theorem package, together with uniform integrability, into a complete classification of the asymptotics of an $L^1$-bounded martingale. The convergence dichotomy reads: every $L^1$-bounded martingale $(X_n)$ converges a.s. to an integrable $X_{\infty }$, and the following are equivalent — (i) $(X_n)$ is uniformly integrable; (ii) $X_n\to X_{\infty }$ in $L^1$; (iii) $(X_n)$ is closed, $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$; (iv) there exists $Y\in L^1$ with $X_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for all $n$. When these fail, the a.s. limit still exists but $\mathbb{E}[X_{\infty }]<\mathbb{E}[X_0]$ for a nonnegative martingale, with the deficit $\mathbb{E}[X_0]-\mathbb{E}[X_{\infty }]$ measuring exactly the mass that escapes to infinity. The closed case is the Radon-Nikodym picture of 02.07.08 run to its limit: $X_{\infty }$ is the density $dQ/d\mathbb{P}$ where $dQ=X_{\infty }d\mathbb{P}$, and $X_n$ is its conditional projection onto ${\mathcal{F}}_n$, so closure is the statement that the consistent system of finite-horizon densities is the restriction of a single limit density on ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Lévy's upward theorem $\mathbb{E}[Y\mid {\mathcal{F}}_n]\to \mathbb{E}[Y\mid {\mathcal{F}}_{\infty }]$ a.s. and in $L^1$ is the closed-martingale instance, and Lévy's $0$-$1$ law $\mathbb{E}[1_A\mid {\mathcal{F}}_n]\to 1_A$ for $A\in {\mathcal{F}}_{\infty }$ is its indicator specialisation.

The Galton-Watson process is the canonical proving ground for the dichotomy. With offspring mean $m>1$ and $Z_0=1$, the normalised population $W_n=Z_n/m^n$ is a nonnegative martingale, hence $W_n\to W_{\infty }$ a.s. with $W_{\infty }\in L^1$. The Kesten-Stigum theorem (1966) sharpens this: $\mathbb{E}[W_{\infty }]=1$ (equivalently $(W_n)$ is UI, equivalently $W_{\infty }>0$ on the survival event) holds iff $\mathbb{E}[Z_1{\log }^{+}{Z}_1]<\infty$; otherwise $W_{\infty }=0$ a.s. and all the mean leaks to infinity. The $L^2$ case is computed directly: if ${\sigma }^2=\mathrm{Var}(Z_1)<\infty$ then $\mathbb{E}[W_n^2]=1+{\sigma }^2{m}^{-2}\frac{1-m^{-n}}{1-m^{-1}}$ is bounded, so $(W_n)$ is $L^2$-bounded, UI, and $W_{\infty }$ is a genuine nondegenerate limit with $\mathbb{E}[W_{\infty }]=1$ — the asymptotic growth rate $Z_n\sim W_{\infty }{m}^n$ with a random prefactor.

Polya's urn exhibits the closed case in its purest form. Starting from $r$ red and $b$ black, the red fraction $M_n$ is a bounded martingale, hence UI and convergent a.s. and in $L^1$ to $M_{\infty }\sim \mathrm{Beta}(r,b)$. The urn sequence is exchangeable, and de Finetti's theorem identifies $M_{\infty }$ as the directing random parameter: conditionally on $M_{\infty }=\theta$, the draws are i.i.d. Bernoulli$(\theta )$. The martingale convergence theorem is thus the analytic engine behind de Finetti's representation of exchangeable sequences, with $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$ the running posterior mean of the unknown bias.

These limit theorems also drive a clean proof of the strong law of large numbers via a backward (reversed) martingale: for i.i.d. integrable $({\xi }_k)$ with $S_n={\sum }_{k\le n}{\xi }_k$, the averages $S_n/n=\mathbb{E}[{\xi }_1\mid {\mathcal{G}}_n]$ where ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ is a decreasing filtration; a reversed martingale is automatically UI, converges a.s. and in $L^1$, and the Hewitt-Savage $0$-$1$ law identifies the limit as the constant $\mathbb{E}[{\xi }_1]$. The same backward-martingale convergence underlies the Lévy downward theorem and the ergodic decomposition.

Synthesis. The foundational reason a bounded fair game converges is that oscillation across any fixed band is a profit engine: each completed upcrossing banks $b-a$, and the supermartingale inequality caps the expected total bankable to $\mathbb{E}[(X_N-a{)}^{-}]/(b-a)$, so this is exactly the mechanism by which $L^1$-boundedness, through a single negative-part estimate, forbids infinitely many crossings of every rational band simultaneously and pins the path to a limit. Putting these together, a.s. convergence, $L^1$ convergence, and closure are one structure viewed through three integrability lenses: $L^1$-boundedness buys the a.s. limit via upcrossings and Fatou 02.07.05, uniform integrability upgrades it to $L^1$ convergence via Vitali and is dual to closure as the Radon-Nikodym density of 02.07.08, and $L^p$-boundedness for $p>1$ forces UI for free through Doob's maximal inequality. This is exactly the central insight that organises the asymptotic theory: a martingale is a consistent system of conditional-expectation densities, the convergence theorem is the time-asymptotic Radon-Nikodym identification of that system with a single limit density when no mass escapes, and the escape of mass — visible in the Galton-Watson $W_{\infty }=0$ regime and the doubling game — is precisely the failure of uniform integrability. The whole optional-stopping calculus of 37.04.01 generalises into this limit theory, and it is dual to the continuous-time martingale convergence that closes Brownian and Itô martingales downstream.

Full proof set Master

The upcrossing inequality and the a.s. convergence theorem are proved in full in the Key theorem section. The remaining Master claims are recorded here.

Proposition (the convergence dichotomy). Let $(X_n)$ be an $L^1$-bounded martingale with a.s. limit $X_{\infty }\in L^1$. The following are equivalent: (i) $(X_n)$ is uniformly integrable; (ii) $X_n\to X_{\infty }$ in $L^1$; (iii) $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ for all $n$; (iv) $X_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for some $Y\in L^1$.

Proof. (i)$\Rightarrow$(ii): UI plus a.s. convergence (which gives convergence in probability) is the hypothesis of Vitali's convergence theorem, yielding $\mathbb{E}|X_n-X_{\infty }|\to 0$. (ii)$\Rightarrow$(iii): for $m$ fixed and $A\in {\mathcal{F}}_m$, the martingale property gives ${\int }_A{X}_{n}d\mathbb{P}={\int }_A{X}_{m}d\mathbb{P}$ for all $n\ge m$; $L^1$ convergence lets $n\to \infty$ on the left to give ${\int }_A{X}_{\infty }d\mathbb{P}={\int }_A{X}_{m}d\mathbb{P}$, the defining identity $X_m=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_m]$. (iii)$\Rightarrow$(iv): take $Y=X_{\infty }$. (iv)$\Rightarrow$(i): for any fixed $Y\in L^1$, the family $\{\mathbb{E}[Y\mid \mathcal{G}]:\mathcal{G}\subseteq \mathcal{F}\text{ a sub-}\sigma \text{-algebra}\}$ is uniformly integrable. Indeed by conditional Jensen $|\mathbb{E}[Y\mid \mathcal{G}]|\le \mathbb{E}[|Y|\mid \mathcal{G}]$, and on $\{|\mathbb{E}[Y\mid \mathcal{G}]|>c\}=:A_c\in \mathcal{G}$, $$ \mathbb{E}\big[|\mathbb{E}[Y\mid\mathcal{G}]|,\mathbf{1}{A_c}\big] \le \mathbb{E}\big[\mathbb{E}[|Y|\mid\mathcal{G}]\mathbf{1}{A_c}\big] = \mathbb{E}[|Y|\mathbf{1}_{A_c}], $$ while $\mathbb{P}(A_c)\le c^{-1}\mathbb{E}|\mathbb{E}[Y\mid \mathcal{G}]|\le c^{-1}\mathbb{E}|Y|\to 0$ uniformly in $\mathcal{G}$; absolute continuity of the integral $Y\mapsto \mathbb{E}[|Y|1_A]$ then makes the bound uniformly small. $\square$

Proposition (Lévy's upward theorem). For $Y\in L^1$ and a filtration $({\mathcal{F}}_n)$ with ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$, $\mathbb{E}[Y\mid {\mathcal{F}}_n]\to \mathbb{E}[Y\mid {\mathcal{F}}_{\infty }]$ a.s. and in $L^1$.

Proof. The process $X_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ is a martingale (tower property), and by the previous proposition's (iv)$\Rightarrow$(i) it is uniformly integrable, hence $L^1$-bounded; the convergence theorem gives an a.s. and (by UI) $L^1$ limit $X_{\infty }$. It remains to identify $X_{\infty }=\mathbb{E}[Y\mid {\mathcal{F}}_{\infty }]$. For $A\in {\mathcal{F}}_m$, ${\int }_A{X}_{\infty }d\mathbb{P}={\lim }_n{\int }_A{X}_{n}d\mathbb{P}={\int }_{A}Yd\mathbb{P}$ (the last equality since ${\int }_A{X}_n={\int }_{A}Y$ for $n\ge m$ by definition of conditional expectation). The collection of $A$ with ${\int }_A{X}_{\infty }={\int }_{A}Y$ is a $\lambda$-system containing the $\pi$-system ${\bigcup }_m{\mathcal{F}}_m$, hence contains ${\mathcal{F}}_{\infty }$ by Dynkin's lemma. As $X_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, $X_{\infty }=\mathbb{E}[Y\mid {\mathcal{F}}_{\infty }]$ a.s. $\square$

Proposition (Galton-Watson $L^2$ regime). Let $(Z_n)$ be a Galton-Watson process with $Z_0=1$, offspring mean $m=\mathbb{E}[Z_1]\in (1,\infty )$ and variance ${\sigma }^2=\mathrm{Var}(Z_1)<\infty$. Then $W_n=Z_n/m^n$ is an $L^2$-bounded martingale; consequently $W_n\to W_{\infty }$ a.s. and in $L^2$ with $\mathbb{E}[W_{\infty }]=1$ and $\mathrm{Var}(W_{\infty })={\sigma }^2/(m^2-m)$.

Proof. Write $Z_{n+1}={\sum }_{i=1}^{Z_n}{\chi }_i^{(n)}$ with ${\chi }_i^{(n)}$ i.i.d. copies of $Z_1$ independent of ${\mathcal{F}}_n=\sigma (Z_0,\dots ,Z_n)$. Then $\mathbb{E}[Z_{n+1}\mid {\mathcal{F}}_n]=m{Z}_n$, so $\mathbb{E}[W_{n+1}\mid {\mathcal{F}}_n]=W_n$: a martingale, nonnegative with $\mathbb{E}[W_n]=1$. For the second moment, $\mathrm{Var}(Z_{n+1}\mid {\mathcal{F}}_n)=Z_n{\sigma }^2$ and $\mathbb{E}[Z_{n+1}^2\mid {\mathcal{F}}_n]=m^2{Z}_n^2+{\sigma }^2{Z}_n$, so dividing by $m^{2(n+1)}$, $$ \mathbb{E}[W_{n+1}^2] = \mathbb{E}[W_n^2] + \frac{\sigma^2}{m^{n+2}}\mathbb{E}[W_n], m^{n} \cdot m^{-n}!\cdot m^{n} = \mathbb{E}[W_n^2] + \sigma^2 m^{-(n+2)}, $$ using $\mathbb{E}[Z_n]=m^n$. Summing the geometric series from $\mathbb{E}[W_0^2]=1$ gives $\mathbb{E}[W_n^2]=1+{\sigma }^2{\sum }_{k=1}^n{m}^{-(k+1)}\to 1+\frac{{\sigma }^2}{m^2}\cdot \frac{1}{1-m^{-1}}=1+\frac{{\sigma }^2}{m^2-m}<\infty$. So $(W_n)$ is $L^2$-bounded, hence UI; the $L^2$ convergence theorem (Exercise 8) gives a.s. and $L^2$ convergence, $\mathbb{E}[W_{\infty }]=1$, and $\mathbb{E}[W_{\infty }^2]=1+{\sigma }^2/(m^2-m)$, whence $\mathrm{Var}(W_{\infty })={\sigma }^2/(m^2-m)$. $\square$

Proposition (a.s. convergence does not imply $L^1$ convergence). There is a nonnegative martingale $(X_n)$ with $X_n\to X_{\infty }$ a.s., $X_{\infty }\in L^1$, yet $\mathbb{E}[X_{\infty }]<\mathbb{E}[X_0]$ and $X_n\ne \mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.

Proof. Take the product martingale $X_n={\prod }_{k=1}^n{\xi }_k$ with $({\xi }_k)$ i.i.d., $\mathbb{P}({\xi }_k=2)=\mathbb{P}({\xi }_k=0)=1/2$, and $X_0=1$ (Exercise 7). It is a nonnegative martingale, $\mathbb{E}[X_n]=1$. The event $\{{\xi }_k=0\text{ for some }k\}$ has probability $1$, and on it $X_n=0$ eventually, so $X_n\to X_{\infty }=0$ a.s. Then $\mathbb{E}[X_{\infty }]=0<1=\mathbb{E}[X_0]$, and $\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]=0\ne X_n$ on $\{{\xi }_1=\cdots ={\xi }_n=2\}$, an event of probability $2^{-n}>0$ where $X_n=2^n$. The family is not uniformly integrable: $\mathbb{E}[X_n{1}_{\{X_n>c\}}]=1$ for every $c<2^n$, so ${\sup }_n\mathbb{E}[X_n{1}_{\{X_n>c\}}]=1\nrightarrow 0$. $\square$

Connections Master

Discrete-time martingales, stopping times, and optional stopping 37.04.01 are the immediate prerequisite and the source of the proof machinery: the upcrossing inequality is proved by applying the martingale-transform / supermartingale-stability lemma of that unit to the predictable "buy low, sell high" strategy, and the convergence dichotomy here is the limit companion of the optional-sampling theorem there — both are governed by the single dividing line of uniform integrability.

Absolute continuity and the Radon-Nikodym theorem 02.07.08 supply the meaning of closure: a uniformly integrable martingale is exactly the Radon-Nikodym density process $X_n=d(X_{\infty }d\mathbb{P})/d\mathbb{P}{|}_{{\mathcal{F}}_n}$, so the convergence theorem in its closed form is the time-asymptotic Radon-Nikodym identification of a consistent system of finite-horizon densities with a single limit density on ${\mathcal{F}}_{\infty }$; the failure of closure is the failure of this density to integrate to the right total mass.

Fatou's lemma and dominated convergence 02.07.05 are the analytic engine: integrability of the a.s. limit $X_{\infty }$ is Fatou applied to $|X_n|$, the $L^1$-convergence half of the dichotomy is Vitali (a uniform-integrability strengthening of dominated convergence), and the $L^p$ convergence theorem closes by dominated convergence against the Doob-maximal majorant $X^{\ast }\in L^p$.

The strong law of large numbers 37.02.02 is both a downstream application and an alternative vantage point: the reversed-martingale proof of the SLLN runs the convergence theorem on the decreasing filtration ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$, where $S_n/n=\mathbb{E}[{\xi }_1\mid {\mathcal{G}}_n]$ converges a.s. and in $L^1$ to the constant $\mathbb{E}[{\xi }_1]$ by the Hewitt-Savage $0$-$1$ law.

Historical & philosophical context Master

The almost-sure convergence theorem and the upcrossing inequality are due to Joseph Doob, who introduced the systematic measure-theoretic theory of martingales and proved both results in his 1940 paper Regularity properties of certain families of chance variables (Trans. Amer. Math. Soc. 47, 455) ^{[Doob 1940]} and codified them in Stochastic Processes (Wiley, 1953). The upcrossing argument — counting completed band-crossings via a predictable betting strategy — is Doob's signature device, replacing the earlier compactness and oscillation arguments of Lévy with a single combinatorial-probabilistic estimate. The term "martingale" for the underlying fair-game object was attached by Jean Ville in his 1939 Étude critique de la notion de collectif, building on Paul Lévy's 1930s study of conditioned sums.

The branching-process application has its own lineage: Galton and Watson posed the family-extinction problem in 1874, and the normalised-population martingale $W_n=Z_n/m^n$ with its almost-sure limit was studied by Harris in The Theory of Branching Processes (1963); the sharp uniform-integrability criterion $\mathbb{E}[Z_1{\log }^{+}{Z}_1]<\infty$ for $\mathbb{E}[W_{\infty }]=1$ is the Kesten-Stigum theorem (Ann. Math. Statist. 37, 1966, 1211). Polya's urn dates to Eggenberger and Polya (1923), and the identification of its limit with the directing measure of an exchangeable sequence is de Finetti's representation theorem (de Finetti, 1931). The gap between almost-sure and $L^1$ convergence — a limit that exists pathwise while the mean leaks to infinity — is the precise probabilistic content of the distinction between pointwise and dominated convergence, and uniform integrability is the named hypothesis under which the two coincide ^{[Williams 1991]}.

Bibliography Master

@book{williams1991,
  author    = {Williams, David},
  title     = {Probability with Martingales},
  publisher = {Cambridge University Press},
  series    = {Cambridge Mathematical Textbooks},
  year      = {1991}
}

@article{doob1940,
  author  = {Doob, Joseph L.},
  title   = {Regularity properties of certain families of chance variables},
  journal = {Transactions of the American Mathematical Society},
  volume  = {47},
  number  = {3},
  pages   = {455--486},
  year    = {1940}
}

@book{doob1953,
  author    = {Doob, Joseph L.},
  title     = {Stochastic Processes},
  publisher = {John Wiley \& Sons, New York},
  year      = {1953}
}

@book{durrett2019,
  author    = {Durrett, Rick},
  title     = {Probability: Theory and Examples},
  edition   = {5th},
  series    = {Cambridge Series in Statistical and Probabilistic Mathematics},
  publisher = {Cambridge University Press},
  year      = {2019}
}

@book{harris1963,
  author    = {Harris, Theodore E.},
  title     = {The Theory of Branching Processes},
  publisher = {Springer-Verlag, Berlin},
  year      = {1963}
}

@article{kestenstigum1966,
  author  = {Kesten, Harry and Stigum, Bernt P.},
  title   = {A limit theorem for multidimensional Galton-Watson processes},
  journal = {Annals of Mathematical Statistics},
  volume  = {37},
  number  = {5},
  pages   = {1211--1223},
  year    = {1966}
}

@book{neveu1975,
  author    = {Neveu, Jacques},
  title     = {Discrete-Parameter Martingales},
  publisher = {North-Holland, Amsterdam},
  year      = {1975}
}