The martingale convergence theorem

Theorem

The martingale convergence theorem

T-135Home MU-404Threads chance

Statement

Let \( (\Omega, \mathcal{F}, \mathbb{P}) \) be a probability space, let \( (\mathcal{F}_n)_{n \ge 0} \) be a filtration (an increasing sequence of sub-\(\sigma\)-algebras \( \mathcal{F}_0 \subseteq \mathcal{F}_1 \subseteq \cdots \subseteq \mathcal{F} \)), and let \( (X_n)_{n \ge 0} \) be a martingale with respect to \( (\mathcal{F}_n) \): each \( X_n \) is \( \mathcal{F}_n \)-measurable, \( \mathbb{E}[|X_n|] \lt \infty \) for all \( n \), and \( \mathbb{E}[X_{n+1} \mid \mathcal{F}_n] = X_n \) almost surely. Suppose \( (X_n) \) is \( L^1 \)-bounded, i.e. \( \sup_{n} \mathbb{E}[|X_n|] \lt \infty \) (in particular this holds if \( \sup_n \mathbb{E}[X_n^2] \lt \infty \), or if \( X_n \ge 0 \) for all \( n \)). Then there exists an integrable random variable \( X_\infty \), measurable with respect to \( \mathcal{F}_\infty := \sigma\!\left(\bigcup_n \mathcal{F}_n\right) \), such that \[ X_n \longrightarrow X_\infty \quad \text{almost surely as } n \to \infty. \]

Why it matters

The martingale convergence theorem is the single most important stability result in probability theory: it says that a "fair game" whose stake stays bounded on average cannot oscillate forever — it must settle down. This underlies almost every limiting argument in modern probability: the strong law of large numbers, Kolmogorov's 0-1 law refinements, branching process extinction theory, Doob's theorems on likelihood ratios and Radon-Nikodym derivatives, and the convergence of Markov chains and Pólya urn schemes.

Its power comes from requiring no independence and only a first-moment bound — vastly weaker hypotheses than classical limit theorems, yet it delivers pointwise (almost sure) convergence rather than merely convergence in distribution or in probability.

Hypotheses

Martingale property (conditional mean-preservation). Without \( \mathbb{E}[X_{n+1}\mid\mathcal{F}_n] = X_n \), there is no "restoring force" pulling the process back; e.g. a sequence of independent \( \mathrm{Bernoulli}(1/2) \) shifted deterministically, \( X_n = n \), is \( \mathcal{F}_n \)-measurable with finite means but drifts to \( +\infty \) with no cancellation mechanism.

\( L^1 \)-boundedness (\( \sup_n \mathbb{E}[|X_n|] \lt \infty \)). Simple symmetric random walk \( S_n = \sum_{i=1}^n \xi_i \), \( \xi_i = \pm 1 \) with probability \( 1/2 \) each, is a martingale but \( \mathbb{E}[|S_n|] \asymp \sqrt{n} \to \infty \); indeed \( S_n \) does not converge a.s. (it oscillates, by the law of the iterated logarithm it exceeds every bound infinitely often).

Adaptedness and integrability at each finite stage (\( X_n \) is \( \mathcal{F}_n \)-measurable, \( \mathbb{E}|X_n| \lt \infty \)). These are needed merely to make the martingale definition meaningful stage-by-stage; without \( \mathcal{F}_n \)-measurability the conditional expectation identity does not typecheck, and without finite means \( \mathbb{E}[X_{n+1}\mid\mathcal{F}_n] \) is not even well-defined as an ordinary conditional expectation.

Proof

\text{Fix } a \lt b \in \mathbb{R}. \text{ Define the number of upcrossings of } [a,b] \text{ by } (X_0,\dots,X_n) \text{ as } U_n[a,b].

Setup: an upcrossing is a completed excursion of the path from below \(a\) to above \(b\); we will bound the expected number of such excursions to control oscillation. A

\text{(Doob's Upcrossing Lemma) } \quad (b-a)\, \mathbb{E}[U_n[a,b]] \le \mathbb{E}\big[(X_n-a)^-\big] \le \mathbb{E}|X_n| + |a|.

Cited lemma (Doob, 1953): constructed via a predictable "buy-low, sell-high" strategy \( C_k \in \{0,1\} \) (predictable since it depends only on \( \mathcal{F}_{k-1} \)) that is 1 exactly while the path is between an excursion below \(a\) and above \(b\); the gains process \( (C \cdot X)_n = \sum_{k=1}^n C_k(X_k - X_{k-1}) \) satisfies \( (C\cdot X)_n \ge (b-a)U_n[a,b] - (X_n-a)^- \), and since \( C \) is a bounded predictable process, the Optional Stopping / martingale transform lemma gives \( \mathbb{E}[(C\cdot X)_n] = 0 \) (a predictable transform of a martingale is a martingale with mean zero), yielding the stated bound. C

\mathbb{E}[U_n[a,b]] \le \frac{\mathbb{E}|X_n| + |a|}{b-a} \le \frac{K + |a|}{b-a}, \qquad K := \sup_n \mathbb{E}|X_n| \lt \infty \text{ by hypothesis.}

Direct rearrangement of Step 2 using the \(L^1\)-boundedness hypothesis; the bound is uniform in \(n\). B

U_\infty[a,b] := \lim_{n\to\infty} U_n[a,b] \ \text{(increasing limit)} \quad \Rightarrow \quad \mathbb{E}[U_\infty[a,b]] \le \frac{K+|a|}{b-a} \lt \infty \ \text{by Monotone Convergence.}

\( U_n[a,b] \) is nondecreasing in \(n\) (more time cannot create fewer upcrossings), so the Monotone Convergence Theorem lets the expectation pass to the limit; a finite expectation forces \( U_\infty[a,b] \lt \infty \) almost surely. B

\mathbb{P}\Big(\liminf_{n} X_n \lt a \lt b \lt \limsup_n X_n\Big) \le \mathbb{P}\big(U_\infty[a,b] = \infty\big) = 0.

If \( \liminf X_n \lt a \) and \( \limsup X_n \gt b \) on an event, the path crosses from below \(a\) to above \(b\) infinitely often on that event by definition of liminf/limsup, forcing \( U_\infty[a,b]=\infty \); Step 4 shows this has probability 0. B

\mathbb{P}\Big(\liminf_n X_n \lt \limsup_n X_n\Big) = \mathbb{P}\!\left(\bigcup_{a \lt b,\, a,b \in \mathbb{Q}} \{\liminf X_n \lt a \lt b \lt \limsup X_n\}\right) = 0.

The event \( \{\liminf \lt \limsup\} \) is exactly the countable union over rational pairs \(a\lt b\) of the events in Step 5 (density of \( \mathbb{Q} \) in \( \mathbb{R} \)); a countable union of null sets is null by countable subadditivity. A

\text{Hence } \liminf_n X_n = \limsup_n X_n =: X_\infty \ \text{a.s., so } X_n \to X_\infty \ \text{a.s. in the extended reals } [-\infty,\infty].

Definition of convergence: a real sequence converges (possibly to \( \pm\infty \)) iff its liminf equals its limsup. \( X_\infty \) is \( \mathcal{F}_\infty \)-measurable as an a.s. limit of \( \mathcal{F}_\infty \)-measurable functions (each \(X_n\) is \(\mathcal{F}_n\)-measurable \(\subseteq \mathcal{F}_\infty\)-measurable). A

\mathbb{P}(X_\infty = \pm\infty) = 0, \qquad X_\infty \in L^1.

By Fatou's Lemma applied to \( |X_n| \ge 0 \): \( \mathbb{E}[|X_\infty|] = \mathbb{E}\big[\liminf_n |X_n|\big] \le \liminf_n \mathbb{E}[|X_n|] \le K \lt \infty \). A random variable with finite expectation is a.s. finite, ruling out \( \pm\infty \), so \( X_\infty \) is a genuine (a.s. finite) integrable random variable and \( X_n \to X_\infty \) a.s. in \( \mathbb{R} \). B

Result

X_n \to X_\infty \ \text{a.s.}, \qquad X_\infty \in L^1(\mathcal{F}_\infty), \qquad \text{whenever } \sup_n \mathbb{E}[|X_n|] \lt \infty

Reading. A martingale that stays bounded in average absolute size cannot wander forever — its upcrossing count of any interval is almost surely finite, which forces the liminf and limsup of the path to coincide, i.e. the path converges to a genuine (finite) random variable almost surely.

Scope. Applies to any discrete-time martingale, submartingale, or supermartingale (with the obvious sign adjustments) relative to any filtration, on any probability space — no independence, stationarity, or Markov structure is needed. It gives almost sure convergence only; it does NOT by itself give convergence in \(L^1\) or the "closing" identity \( X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n] \), which require the strictly stronger hypothesis of uniform integrability.

Corollaries & converses

Non-negative martingales always converge a.s. (since \( \mathbb{E}|X_n| = \mathbb{E}[X_n] = \mathbb{E}[X_0] \) is automatically bounded) — this is the version used for branching-process extinction probabilities and Pólya urns.
\( L^2 \)-bounded martingales (\( \sup_n \mathbb{E}[X_n^2] \lt \infty \)) converge a.s. and in \( L^2 \), by combining this theorem with orthogonality of martingale increments (Doob's \( L^2 \) maximal inequality gives the \( L^2 \) mode).
If in addition \( (X_n) \) is uniformly integrable, convergence upgrades to \( L^1 \) and \( X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n] \) a.s. for every \( n \) (Doob's martingale closure theorem) — this is a genuinely stronger conclusion, not a free corollary.
Converse fails: a.s. convergence of a martingale does NOT imply \( L^1 \)-boundedness of the limiting mechanism in the closure sense, and it certainly does not imply uniform integrability — see the gambler's ruin counterexample below, which converges a.s. to 0 while \( \mathbb{E}[X_n] \equiv 1 \ne 0 = \mathbb{E}[X_\infty] \).
The theorem does not require \( (\mathcal{F}_n) \) to be the natural filtration of \( (X_n) \); any admissible filtration works, which is essential for change-of-measure and likelihood-ratio applications.

Fails without

Drop \(L^1\)-boundedness: simple symmetric random walk \( S_n \) is a martingale with \( \mathbb{E}|S_n| \sim c\sqrt{n} \to \infty \); it is recurrent, so \( \limsup S_n = +\infty \) and \( \liminf S_n = -\infty \) almost surely — the theorem's conclusion fails completely.
Drop the martingale property (keep boundedness): let \( X_n \) alternate deterministically between \( 0 \) and \( 1 \) (an adapted, bounded, but non-martingale sequence with \( \mathbb{E}[X_{n+1}\mid\mathcal{F}_n] \ne X_n \)); it never converges even though it is bounded in every sense.
Even when it holds, \(L^1\) convergence can fail: let \( X_0 = 1 \) and evolve as a critical branching-type doubling martingale — e.g. \( X_n = 2^n \mathbf{1}_{A_n} \) where \( A_n \) are nested events with \( \mathbb{P}(A_n) = 2^{-n} \) and \( A_n \downarrow \emptyset \); then \( X_n \) is a non-negative martingale with \( \mathbb{E}[X_n]=1 \) for all \(n\), so it converges a.s. (to \(0\), since \( \mathbb{P}(A_n)\to 0 \)), yet \( \mathbb{E}[X_\infty] = 0 \ne 1 = \lim \mathbb{E}[X_n] \): convergence is a.s. but not in \(L^1\), because the family is not uniformly integrable.

Common errors

Concluding \( \mathbb{E}[X_n] \to \mathbb{E}[X_\infty] \) automatically — this needs uniform integrability, not merely \(L^1\)-boundedness; the branching-doubling example above is the standard counterexample cited to puncture this.
Confusing "\(L^1\)-bounded" (\( \sup_n \mathbb{E}|X_n| \lt \infty \)) with "\(L^1\)-convergent" (\( X_n \to X_\infty \) in \(L^1\) norm) — the theorem's hypothesis is the former and its conclusion is only a.s. convergence, a strictly weaker mode than \(L^1\) convergence in general.
Applying the theorem to a random walk without checking boundedness — students often reach for "martingale convergence" to argue \( S_n \) settles down, forgetting simple random walk is the textbook example where it manifestly does not.
Misusing the upcrossing inequality by forgetting predictability of the strategy \(C_k\) — the martingale transform argument (Step 2) is only valid because \(C_k\) is \(\mathcal{F}_{k-1}\)-measurable ("decided before you see the outcome"); using an \(\mathcal{F}_k\)-measurable (look-ahead) strategy invalidates the zero-mean identity.
Believing the limit \(X_\infty\) must equal \(X_0\) in distribution or that \(X_n\) must be "close" to \(X_\infty\) for large finite \(n\) in expectation — almost sure convergence gives no rate and no control on \( \mathbb{E}[X_n - X_\infty] \) without extra uniform integrability.

Discussion

The theorem was proved by Joseph Doob in the 1940s as part of his systematic development of martingale theory, and the upcrossing inequality is the technical engine behind almost every convergence result in the field, including the martingale central limit theorem's auxiliary lemmas and the theory of local martingales in continuous time. The proof strategy — bound expected oscillation via a combinatorial "excursion count," then let the count explode only on a null set — is a template reused throughout analysis and probability wherever one wants a.s. convergence without a direct handle on the limit itself.

Conceptually, the martingale property encodes "no free information": if you only know the past, your best guess for tomorrow is today's value. The convergence theorem says such fair processes, so long as they don't grow unboundedly in size, must eventually stop moving in any coherent sense — informally, all the "surprise" gets absorbed into smaller and smaller increments as \(n\) grows, a phenomenon made precise by the almost-sure Cauchy-like behavior forced by finite upcrossing counts.

The theorem generalizes substantially: for continuous-time martingales with right-continuous paths, an analogous upcrossing argument (applied along a countable dense set of times, using path-regularity to fill in) yields a.s. convergence as \(t \to \infty\); and submartingale/supermartingale versions follow by the same proof with sign changes, since the upcrossing lemma itself only needs the submartingale inequality \( \mathbb{E}[X_{n+1}\mid\mathcal{F}_n]\ge X_n \).

A subtler point often missed at first encounter: the theorem gives convergence relative to the ambient filtration \(\mathcal{F}_n\), and the limit \(X_\infty\) is \(\mathcal{F}_\infty\)-measurable but need not be measurable with respect to any single \(\mathcal{F}_n\) — it genuinely depends on "the entire tail" of information. This is precisely what powers applications like the Kolmogorov 0-1 law's quantitative refinements and Doob's proof of the Radon-Nikodym theorem via likelihood-ratio martingales, where \(X_\infty\) recovers exactly the Radon-Nikodym derivative between two measures once uniform integrability is separately verified.

Common misconception: that "martingale convergence theorem" is a single statement — in practice the name covers a family of related results (a.s. convergence under \(L^1\)-boundedness; \(L^p\) convergence under \(L^p\)-boundedness for \(p\gt1\); \(L^1\) convergence under uniform integrability) with genuinely different hypotheses and conclusions, and conflating them is the single most frequent error in applying the theorem correctly.

Worked examples

\text{Gambler's ruin: } X_n = \frac{S_n + n}{2} \text{ absorbed appropriately, or directly consider the classical ruin martingale } M_n.

Setup: let \( (\xi_i) \) be i.i.d. \( \pm1 \) fair coin flips, \( S_0 = k \in \{0,1,\dots,N\}\), \( S_n = k + \sum_{i=1}^n \xi_i \) stopped at \( \tau = \inf\{n : S_n \in \{0,N\}\} \), and set \( X_n = S_{n \wedge \tau} \). A

X_n \text{ is a martingale w.r.t. } \mathcal{F}_n = \sigma(\xi_1,\dots,\xi_n), \qquad 0 \le X_n \le N \ \Rightarrow \ \sup_n \mathbb{E}|X_n| \le N \lt \infty.

Stopped martingales remain martingales (optional stopping preserves the martingale property for bounded stopping); boundedness by \(N\) gives \(L^1\)-boundedness trivially, satisfying the theorem's hypothesis. A

\text{By the theorem, } X_n \to X_\infty \text{ a.s., and since } X_n \text{ takes only integer values in } [0,N] \text{ and the walk keeps moving until absorption,}

A convergent sequence of integers must be eventually constant, so a.s. convergence forces \(\tau \lt \infty\) a.s. and \(X_\infty \in \{0,N\}\) — the walk is absorbed in finite time almost surely. B

\mathbb{E}[X_\infty] = \mathbb{E}[X_0] = k \quad \text{(bounded convergence: } |X_n|\le N \text{ justifies passing the limit through the expectation)},

Bounded Convergence Theorem: since \(0 \le X_n \le N\) for all \(n\) and \(X_n \to X_\infty\) a.s., \( \mathbb{E}[X_n] \to \mathbb{E}[X_\infty] \); combined with the martingale identity \( \mathbb{E}[X_n] = \mathbb{E}[X_0] = k \) for every \(n\). B

\mathbb{P}(X_\infty = N) \cdot N + \mathbb{P}(X_\infty=0)\cdot 0 = k \ \Rightarrow \ \mathbb{P}(\text{ruin at } N) = \frac{k}{N}.

Solving the linear equation using \( X_\infty \in \{0,N\} \) and \( \mathbb{P}(X_\infty=N)+\mathbb{P}(X_\infty=0)=1\); this recovers the classical gambler's ruin probability. A

\mathbb{P}(\text{gambler starting at } k \text{ reaches } N \text{ before } 0) = \frac{k}{N}

Reading. The convergence theorem guarantees the walk actually stops (rather than oscillating forever), which is the missing rigorous step that turns the mean-preservation identity into the ruin probability formula.

\text{Pólya urn: start with 1 red, 1 blue ball. At each step draw uniformly and replace it together with one more ball of the same colour.}

Setup: let \( R_n \) be the number of red balls after \(n\) draws (so total balls \(= n+2\)), and let \( X_n = \dfrac{R_n}{n+2} \) be the red proportion. A

\mathbb{E}[X_{n+1} \mid \mathcal{F}_n] = X_n \cdot \frac{X_n(n+2)+1}{n+3} + (1-X_n)\cdot\frac{X_n(n+2)}{n+3} = X_n.

Direct computation of the conditional expectation: with probability \(X_n\) a red is drawn (increasing the red proportion to \(\frac{R_n+1}{n+3}\)), with probability \(1-X_n\) a blue is drawn (red proportion becomes \(\frac{R_n}{n+3}\)); algebra collapses this to \(X_n\), verifying the martingale property directly from the urn dynamics. B

0 \le X_n \le 1 \ \text{ for all } n \ \Rightarrow \ \sup_n \mathbb{E}|X_n| \le 1 \lt \infty.

\(X_n\) is a proportion, automatically bounded in \([0,1]\), so the \(L^1\)-boundedness hypothesis holds trivially — indeed \(X_n\) is uniformly integrable being uniformly bounded. A

\text{By the theorem, } X_n \to X_\infty \ \text{a.s. for some random variable } X_\infty \in [0,1].

Direct application of the martingale convergence theorem to the bounded martingale \(X_n\); this is the rigorous justification that the long-run proportion of red balls in a Pólya urn stabilizes. A

\text{One can further show (via exchangeability / de Finetti, beyond this theorem) that } X_\infty \sim \mathrm{Uniform}(0,1).

Cited fact (not proved here): the exact law of the limit requires the extra machinery of exchangeable sequences and de Finetti's theorem; the convergence theorem alone only guarantees existence of an a.s. limit, not its distribution. C

X_n = \frac{R_n}{n+2} \to X_\infty \ \text{a.s.}, \qquad X_\infty \sim \mathrm{Uniform}(0,1)

Reading. The Pólya urn's red-ball proportion is a bounded martingale, so the convergence theorem guarantees it settles to a genuine random limit rather than oscillating forever — the classical "urn reinforces its own random preference" phenomenon.

Problems

Show directly (without invoking uniform integrability) that if \( (X_n) \) is a martingale with \( X_n \ge 0 \) a.s. for all \(n\), then \( (X_n) \) is automatically \(L^1\)-bounded, and hence converges a.s.
Solution
Since \(X_n \ge 0\), \( \mathbb{E}|X_n| = \mathbb{E}[X_n] \). By the martingale property, \( \mathbb{E}[X_n] = \mathbb{E}[\mathbb{E}[X_n \mid \mathcal{F}_{n-1}]] = \mathbb{E}[X_{n-1}] = \cdots = \mathbb{E}[X_0] \) for every \(n\), a constant. Hence \( \sup_n \mathbb{E}|X_n| = \mathbb{E}[X_0] \lt \infty \) since \(X_0\) is integrable by the martingale definition. The \(L^1\)-boundedness hypothesis of the theorem is satisfied, so \(X_n \to X_\infty\) a.s. for some finite \(X_\infty \ge 0\).
Let \( Y_1, Y_2, \dots \) be i.i.d. with \( \mathbb{P}(Y_i = 2) = \mathbb{P}(Y_i=0) = 1/2 \), and set \( X_n = \prod_{i=1}^n Y_i \) (\(X_0=1\)). Show \( (X_n) \) is a non-negative martingale, deduce it converges a.s., and identify the a.s. limit.
Solution
\(X_n\) is \(\mathcal{F}_n=\sigma(Y_1,\dots,Y_n)\)-measurable and non-negative, hence integrable termwise (in fact \(\mathbb{E}[X_n]=1\) by independence and \(\mathbb{E}[Y_i]=1\)). Martingale check: \( \mathbb{E}[X_{n+1}\mid \mathcal{F}_n] = X_n \cdot \mathbb{E}[Y_{n+1}\mid\mathcal{F}_n] = X_n \cdot \mathbb{E}[Y_{n+1}] = X_n \cdot 1 = X_n \), using independence of \(Y_{n+1}\) from \(\mathcal{F}_n\). By Problem 1 (or directly, \(\mathbb{E}|X_n|=1\) bounded), the martingale convergence theorem applies: \(X_n \to X_\infty\) a.s. Since \(X_n = 0\) as soon as any \(Y_i=0\) occurs, and \(\mathbb{P}(\text{some } Y_i = 0 \text{ eventually}) = 1\) (infinitely many independent \(1/2\)-chance zeros occur a.s., by Borel-Cantelli II since \(\sum \mathbb{P}(Y_i=0) = \sum 1/2 = \infty\) and independence), we get \(X_n = 0\) for all sufficiently large \(n\) a.s., so \(X_\infty = 0\) a.s. (Note: \(\mathbb{E}[X_n]=1\) for all \(n\) yet \(\mathbb{E}[X_\infty]=0\) — convergence is a.s. but not in \(L^1\), illustrating lack of uniform integrability.)
Give an explicit example of a martingale \( (X_n) \) that is \(L^2\)-bounded (\(\sup_n \mathbb{E}[X_n^2] \lt \infty\)) but for which the increments do not go to zero in \(L^2\) trivially — verify the hypotheses of the convergence theorem apply and explain why \(L^2\)-boundedness is a strictly stronger and sufficient condition.
Solution
Take \(X_n = \sum_{i=1}^n 2^{-i}\xi_i\) where \(\xi_i\) are i.i.d. mean-zero, variance-1, independent increments (e.g. Rademacher \(\pm1\)), \(\mathcal{F}_n=\sigma(\xi_1,\dots,\xi_n)\). This is a martingale since increments are independent mean-zero given the past. By independence, \(\mathrm{Var}(X_n) = \sum_{i=1}^n 4^{-i} \le \sum_{i=1}^\infty 4^{-i} = 1/3 \lt \infty\), and \(\mathbb{E}[X_n]=0\), so \(\mathbb{E}[X_n^2]=\mathrm{Var}(X_n)\le 1/3\) uniformly in \(n\): \(L^2\)-bounded. By Cauchy-Schwarz, \(\mathbb{E}|X_n| \le \sqrt{\mathbb{E}[X_n^2]} \le \sqrt{1/3}\), so \(\sup_n\mathbb{E}|X_n|\lt\infty\): the \(L^1\)-boundedness hypothesis of the martingale convergence theorem holds (an \(L^2\) bound always implies an \(L^1\) bound by Cauchy-Schwarz/Jensen, but not conversely), so \(X_n \to X_\infty\) a.s.; here \(X_\infty = \sum_{i=1}^\infty 2^{-i}\xi_i\), an absolutely convergent random series.
Construct a martingale that is bounded in \(L^1\) but for which convergence in \(L^1\) fails, distinct from the doubling example in the notes, to confirm you understand the gap between the theorem's conclusion (a.s. convergence) and the stronger \(L^1\) convergence.
Solution
Let \( (B_n) \) be i.i.d. events with \( \mathbb{P}(B_n \mid B_1^c,\dots,B_{n-1}^c) = \frac{1}{n+1} \) chosen so that exactly one "success" occurs a.s. eventually (constructible via a sequential urn/exploration scheme), and define \(X_n = n \cdot \mathbf{1}\{\text{no success in first } n \text{ steps and success occurs at step } n+1\}\) scaled appropriately — concretely, the standard construction is: let \(U\) be Uniform\((0,1)\), \(\mathcal{F}_n = \sigma(\mathbf{1}\{U \le 1/n\})\) type reveal, and \(X_n = n\cdot \mathbf{1}\{U \le 1/n\}\). One checks \(\mathbb{E}[X_n] = n \cdot \frac{1}{n} = 1\) for all \(n\) (constant, so \(L^1\)-bounded) and the martingale property holds by construction of the filtration revealing whether \(U\le 1/n\). Since \(U\gt 0\) a.s., eventually \(1/n \lt U\), so \(X_n = 0\) for all large \(n\), giving \(X_n \to 0\) a.s. But \(\mathbb{E}[X_n] = 1 \not\to 0 = \mathbb{E}[X_\infty]\): a.s. convergence holds (as guaranteed by the theorem) while \(L^1\) convergence fails, because \(\{X_n\}\) is not uniformly integrable (all the mass concentrates on a shrinking-probability, growing-size event).
A branching process \( (Z_n) \) has offspring mean \(m=\mathbb{E}[\text{number of children per individual}]\) with \(Z_0=1\). Show that \(W_n := Z_n/m^n\) is a martingale w.r.t. \(\mathcal{F}_n = \sigma(Z_0,\dots,Z_n)\), and that if \(m \gt 0\) is finite, \(W_n\) converges a.s. State precisely which hypothesis of the convergence theorem must be checked and why it holds here.
Solution
Write \(Z_{n+1} = \sum_{i=1}^{Z_n} \xi_{n,i}\) where \(\xi_{n,i}\) are i.i.d. copies (independent of \(\mathcal{F}_n\)) of the offspring distribution with mean \(m\). Then \(\mathbb{E}[Z_{n+1}\mid \mathcal{F}_n] = Z_n \cdot m\) (Wald's identity / conditioning on \(Z_n\) and using independence of the offspring counts from the past), so \(\mathbb{E}[W_{n+1}\mid\mathcal{F}_n] = \mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]/m^{n+1} = Z_n m / m^{n+1} = Z_n/m^n = W_n\): the martingale property holds. Since \(W_n = Z_n/m^n \ge 0\) (population counts are non-negative), by Problem 1's argument \(W_n\) is automatically \(L^1\)-bounded (indeed \(\mathbb{E}[W_n] = \mathbb{E}[W_0] = 1\) for all \(n\)) — this is precisely the \(L^1\)-boundedness hypothesis of the martingale convergence theorem, verified here via non-negativity rather than checked directly. Hence \(W_n \to W_\infty\) a.s. for some finite \(W_\infty \ge 0\); this is the classical Kesten-Stigum-type setup, and whether \(W_\infty\) is non-degenerate (i.e. not a.s. \(0\)) is a separate, deeper question depending on second-moment conditions on the offspring law, not addressed by the convergence theorem alone.