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The required condition is uniform integrability, written entirely in terms of the given probability laws:Together with the given almost sure convergence, this implies L1 convergence to an integrable $M_{\infty }$. Passing to the limit in $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$, using the contraction property of conditional expectation in L1 space, gives $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For every bounded stopping time $S$, the same identity holds with ${\mathcal{F}}_S$: split an event of its stopping-time sigma-algebra according to $\{S=k\}$ and use the deterministic-time identity. In particular $M_{T\wedge n}=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_{T\wedge n}]$. These conditional expectations are uniformly integrable: for an integrable $Y$ and $Z=\mathbb{E}[Y\mid \mathcal{G}]$, the event $A=\{|Z|>K\}$ satisfies $\mathbb{P}(A)\le \mathbb{E}|Y|/K$, andFirst choose $R$, then $K$, uniformly in $\mathcal{G}$. Define $M_T=M_{\infty }$ on $\{T=\infty \}$. Now $M_{T\wedge n}\to M_T$ almost surely and in L1 space, so the bounded-time mean identities imply $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ even when $T$ can be infinite.

The exact condition is again ${\sum }_k{a}_k^2<\infty$, with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case ${\sup }_n\mathbb{E}{M}_n^2<\infty$, andproves uniform integrability. The preceding unbounded optional stopping argument yields $\boxed{\mathbb{E}{M}_T=0}$.

For necessity, suppose $v_n\to \infty$. For every fixed $R>0$, the normal distribution of $M_n$ gives $\mathbb{P}(M_n\ge R)\to 1/2$. Hence $\mathbb{P}({\sup }_n{M}_n\ge R)\ge 1/2$, and intersecting over positive integer $R$ shows $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. Unboundedness above is unchanged by altering finitely many of the independent $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a tail event; the Kolmogorov zero-one law makes its probability one. Consequently $T<\infty$ almost surely and $M_T\ge 1$. Its expectation is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in threshold stopping of an independent Gaussian series.

The almost sure submartingale convergence theorem states that a discrete-time submartingale $Y_n$ satisfying ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable limit. In particular, a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ converges almost surely to an integrable $M_{\infty }$, and $\mathbb{E}|M_{\infty }|\le {\sup }_n\mathbb{E}|M_n|$ by Fatou's lemma. This is the martingale convergence theorem. Uniform integrability gives the stronger conclusion of L1 convergence; it is not implied by bounded first moments alone.

A family $\mathcal{R}$ is uniformly integrable ifThis implies integrability and a uniform bound on $E|X|$, since $E|X|\le K+E[|X|1_{\{|X|>K\}}]$. Convergence in probability means $P(|X_n-X|>\epsilon )\to 0$ for every $\epsilon >0$; convergence in $L^1$ means $E|X_n-X|\to 0$. The latter implies the former by Markov's inequality. Conversely, convergence in probability together with uniform integrability of $\{X_n\}$ implies that $X$ is integrable and $X_n\to X$ in $L^1$. Moreover an $L^1$-convergent sequence is uniformly integrable. This is the criterion that excludes rare very large values from carrying a nonvanishing first-moment error.

For the sum family, use the pointwise boundIndeed, if the left indicator is one then the larger of $|X|,|Y|$ exceeds $K/2$, and twice that larger number bounds $|X+Y|$. Taking expectations and the supremum over ${\mathcal{R}}_1,{\mathcal{R}}_2$ bounds the sum-family tail by the sum of their two uniformly vanishing tails. Thus the whole family of sums is uniformly integrable, with no independence assumption.

Use conditional monotone convergence and the tower property of conditional expectation in the preceding construction:This proves the martingale property. The construction gives $M_n\ge 0$ and ${\sup }_n\mathbb{E}{M}_n\le C$, so $M$ is bounded in $L^1$. Taking $p=n$ in the increasing family also gives $M_n\ge X_n^{+}\ge X_n$. Define $Y_n=M_n-X_n$. Then $Y$ is nonnegative and adapted, andThus $Y$ is a supermartingale, with $\mathbb{E}|Y_n|\le \mathbb{E}{M}_n+\mathbb{E}|X_n|\le 2C$. The requested conclusion isThis positive martingale majorant of an L1-bounded submartingale is not asserted to have uniform integrability; boundedness in $L^1$ alone does not imply that stronger property.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.

Fix $x>0$ and let $A=\{M_0<x\}\in {\mathcal{F}}_0$. On $A$, continuity of the paths implies $0\le M_{t\wedge T_x}\le x$, and $M_{T_x}=x$ on $\{T_x<\infty \}$. Bounded-time optional stopping gives, for every $t$,On $A\cap \{T_x=\infty \}$, the stopped process is $M_t\to 0$; on $A\cap \{T_x<\infty \}$ it eventually equals $x$. The bound by $x$ permits conditional dominated convergence, yieldingOn $A^c$, $T_x=0$, so the conditional hitting probability is one. Finally, a continuous nonnegative path tending to zero attains any positive supremum: after a sufficiently late time it is below half that supremum, and its maximum on the earlier compact interval is attained. Thus $\{M^{\ast }\ge x\}=\{T_x<\infty \}$, including equality at the level. We have proved the conditional form of the maximal identity for a continuous nonnegative local martingale tending to zero:The restriction to $A$ before taking the limit is important: $M_0$ need not be bounded, and the unstopped martingale need not have uniform integrability.

A Markov jump process, considered up to its explosion time, is an adapted càdlàg piecewise-constant process whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. Describe its rates by an off-diagonal kernel $q(x,dy)$ with finite total rate $q(x,E)$. In state $x$ the holding time has the exponential distribution with this rate, followed by a destination distributed as $q(x,dy)/q(x,E)$; a zero-rate state is absorbing. Its Markov jump-process generator acts byThe Markov property must hold relative to the specified filtration, not merely the natural filtration: future-revealing information could change the jump compensator.

Let $\mu$ count jumps, marked by their destination states. For this jump measure of a Markov jump process, the predictable compensator isThe compensated random-measure theorem says that a predictable $H(s,y)$ with $\mathbb{E}{\int }_0^t\int |H|d\nu <\infty$ gives a true martingaleThe same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $1_A{1}_{(s,t]}H$, for $A\in {\mathcal{F}}_s$, shows that each increment has zero conditional expectation. Taking $H(s,y)=f(y)-f(X_{s-})$ gives the generator local martingale $f(X_t)-f(X_0)-{\int }_0^{t}Qf(X_s)ds$. This states the integrability conditions in the compensated jump-measure local martingale result explicitly.

For the birth-process claim first take the customary fixed initial state $X_0=i$. Write $N_t=X_t-i$, ${\Lambda }_t={\int }_0^t\lambda (X_s)ds$ and $a=e^{\theta }-1$. The compensator of $N$ is $\Lambda$, whose integrand may equivalently use $X_{s-}$ because jump times have zero Lebesgue measure. A birth multiplies $e^{\theta X}$ by $e^{\theta }$, so the jump product rule givesFor an explicit localization, stop at ${\tau }_n=\inf \{t:X_t\ge n\}\wedge n$ with $n>i$. Up to that time the statewise rates are bounded by the maximum of the finitely many rates below $n$, the jump count is bounded, and $M$ and its compensated-integral integrand are bounded on the finite stopped horizon. The compensated-measure theorem therefore makes each stopped process a true martingale. The times increase to the explosion lifetime (or infinity if there is no explosion), proving $M$ is a local martingale on $[0,\zeta )$ for every real $\theta$.

Now suppose $\lambda (j)\le C$ uniformly. For $C>0$, propose candidate births at the times of a rate-$C$ Poisson process, accepting each with probability $\lambda (X_{s-})/C$ using fresh independent uniform marks. While the state is fixed, Poisson thinning gives a holding time with the required exponential distribution of rate $\lambda$; accepted births have exactly the prescribed law. Thus there can be no explosion, and the true birth count $N_T$ is stochastically dominated by a Poisson random variable of mean $CT$. In particular,The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,This is one integrable dominating variable for the entire stopped family. Conditional dominated convergence removes the localization and provesso the exponential martingale of a pure birth process is a genuine martingale for uniformly bounded rates. Only uniform integrability on each finite horizon is asserted; it need not be uniformly integrable over infinite time.

For a random initial state, the same proof requires $\mathbb{E}{e}^{\theta X_0}<\infty$. The source does not explicitly specify the initial distribution, so that qualification cannot be dropped: take constant rate one and an independent initial law $\mathbb{P}(X_0=n)=6/({\pi }^2{n}^2)$. For $\theta >0$, already $\mathbb{E}{M}_0=\infty$, precluding even the integrability in the definition of a martingale. The fixed-initial-state interpretation above supplies the intended complete proof; bounded rates alone do not remedy an arbitrary heavy-tailed initial law.

For a fixed $c>0$, the functions $|f_n|1_{\{|f_n|\le c\}}$ tend pointwise to zero and are dominated by the Lebesgue integrable constant $c$, since $\mu (\Omega )<\infty$. The dominated convergence theorem therefore gives convergence of their integrals to zero. SplitTake the upper limit in $n$ and then send $c\to \infty$. The specified uniform integrability condition makes the right side zero. It also ensures integrability of each $f_n$, by the same split for a sufficiently large $c$. Hence $\int |f_n|d\mu \to 0$, and in particular $\int f_{n}d\mu \to 0$.

For a counterexample without the tail condition, use Lebesgue measure on $(0,1)$ and $f_n(x)=n{1}_{(0,1/n)}(x)$. Every fixed positive $x$ eventually lies outside the shrinking interval, so $f_n(x)\to 0$, but ${\int }_0^1{f}_{n}dx=1$ for every $n$.

A martingale is an adapted Lebesgue integrable process with $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. A stopping time $T$ has $\{T\le n\}\in {\mathcal{F}}_n$ for every $n$. A precise bounded-time optional sampling theorem says: if $S\le T\le N$ are stopping times, then $\mathbb{E}[M_T\mid {\mathcal{F}}_S]=M_S$. To prove it, writeFor $A\in {\mathcal{F}}_S$, the event $A\cap \{S\le k<T\}$ lies in ${\mathcal{F}}_k$. Every summand multiplied by its indicator therefore has expectation zero. Summing proves the conditional expectation identity. All terms are Lebesgue integrable since there are finitely many. Unbounded stopping requires an additional hypothesis such as uniform integrability of the stopped martingale; it is not justified solely by almost-sure finiteness.

The martingale convergence theorem ensures a finite almost-sure limit if ${\sup }_n\mathbb{E}|M_n|<\infty$. Uniform integrability also gives convergence in $L^1$. In particular bounded or nonnegative martingales satisfy the almost-sure convergence criterion. A symmetric random walk is a martingale but cannot have a finite limit, since each increment is $\pm 1$ and does not tend to zero. Some hypothesis is therefore necessary.

If there are $j$ pessimists, the count increases with probability $(K-j)j/[K(K-1)]$ and decreases with the identical probability. The remaining outcomes leave it unchanged, so $P_n=j_n/K$ is a bounded martingale. It converges almost surely. From any nonunanimous state, there is a fixed positive probability, bounded uniformly over the finitely many states, of reaching unanimity within $K-1$ steps by letting each other individual copy a chosen seed. Repeating blocks shows eventual absorption has probability one. Thus the finite voter-model consensus probability isThe last equality follows from bounded convergence and the constant martingale expectation. The population eventually becomes wholly pessimistic or wholly optimistic, with probabilities equal to their respective initial proportions.

Take a complete, right-continuous filtration $({\mathcal{F}}_t)$, an increasing family of sigma-algebras. Adapted means that $X_t$ is ${\mathcal{F}}_t$-measurable at each $t$; càdlàg means right-continuous with left limits. A local martingale is an adapted càdlàg process $X$, with integrable initial value, for which there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $X_t^{{\tau }_n}=X_{t\wedge {\tau }_n}$ is a true martingale. Such a sequence is a localizing sequence. A true martingale is integrable at each time and satisfies $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_s$ for $s\le t$.

The necessary and sufficient condition is that $X$ is a Class DL process:A family $\mathcal{Z}$ is uniformly integrable if its members are integrable and ${\sup }_{Z\in \mathcal{Z}}\mathbb{E}(|Z|1_{\{|Z|>K\}})\to 0$ as $K\to \infty$. A stopping time $\tau$ satisfies $\{\tau \le t\}\in {\mathcal{F}}_t$ for all $t$.

For necessity, on $[0,T]$ the optional stopping theorem gives $X_{\tau }=\mathbb{E}(X_T\mid {\mathcal{F}}_{\tau })$; conditional expectations of one integrable variable form a uniformly integrable family. For sufficiency, $X_{t\wedge {\tau }_n}\to X_t$ almost surely, and Class DL upgrades this convergence to $L^1$ on each finite horizon. Pass to the limit in $\mathbb{E}(X_{t\wedge {\tau }_n}\mid {\mathcal{F}}_s)=X_{s\wedge {\tau }_n}$ to obtain the martingale identity. Requiring uniform integrability of all times on the entire half-line would be stronger than necessary; Brownian motion is already a counterexample to that stronger requirement.

Brownian scaling gives $|B_t|\stackrel{d}{=}\sqrt{t}R$, where $R=|B_1|$ has Rayleigh distribution with density $r{e}^{-r^2/2}$, $r>0$. For every $p>0$,Near zero this is bounded by ${\int }_0^1|\log r{|}^{p}rdr={\int }_0^{\infty }{u}^p{e}^{-2u}du$, which is finite; near infinity the Gaussian tail dominates any logarithmic power. The inequality $|a+b{|}^p\le C_p(|a{|}^p+|b{|}^p)$, valid also for $0<p<1$, now yieldsLet $c=\mathbb{E}\log R$, a finite constant. The same scaling identity givesAn integrable martingale has constant expectation, so the logarithm of planar Brownian radius is a strict local martingale on $[1,\infty )$. Fixed-time integrability, even of every positive power, does not imply the stopping-time uniform integrability required in part a.

For the Cox-Ross-Rubinstein diffusion approximation, fix calendar maturity $T$, take $h=T/n$, and choose $u=e^{\sigma \sqrt{h}}$, $d=e^{-\sigma \sqrt{h}}$, and the one-step bank factor $b=e^{\rho h}$. For sufficiently fine mesh, $d<b<u$. The risk-neutral probability expands asEach stock log-increment is $\pm \sigma \sqrt{h}$. Its mean under $Q$ is $(\rho -{\sigma }^2/2)h+O(h^2)$ and its variance is ${\sigma }^{2}h+O(h^2)$. Independent increments, their vanishing maximum size, and the Donsker invariance principle for this triangular array give the path limitExponentiation gives the geometric Brownian motion of the Black-Scholes model, and the bank account tends to $e^{\rho t}$.

The physical model is matched by taking $p_h=(e^{\mu h}-d)/(u-d)$. Its corresponding limit has log drift $\mu -{\sigma }^2/2$, so $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t^P$. The likelihood ratios from part (ii) have the matching continuous-time limit. With $\lambda =(\mu -\rho )/\sigma$, a Taylor expansion of their two possible log-multipliers giveswhere ${\epsilon }_k$ is $+1$ or $-1$. Thus the limiting Radon-Nikodym derivative is the Girsanov theorem density $\exp (-\lambda W_T^P-{\lambda }^{2}T/2)$, turning the physical stock drift into $\rho$.

The binomial risk-neutral pricing formula converges to $e^{-\rho T}{\mathbb{E}}_Q[f(S_T)]$ for bounded continuous terminal payoffs by convergence in distribution. For ordinary call and put payoffs, uniform stock second-moment bounds give the required uniform integrability and hence convergence despite unboundedness. Bounded payoffs with discontinuities only at events of zero limiting probability are handled by the same weak-convergence criterion. Corresponding continuous path functionals can be treated at the process level. Arbitrary discontinuous path claims do not follow from weak convergence without further conditions. In the smooth limiting model, the discrete hedging difference quotient becomes the delta hedge $V_x$, and its value satisfies the Black-Scholes equation.

Relative to a filtration $({\mathcal{F}}_t)$, a martingale is an adapted integrable process satisfying $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for $s\le t$. The martingale convergence theorem gives an almost-sure finite limit for a nonnegative martingale, with $\mathbb{E}[M_{\infty }]\le \mathbb{E}[M_0]$ by Fatou's lemma. In continuous time take the usual right-continuous version and filtration hypotheses. Convergence in $L^1$, and preservation of expectation at infinity, additionally require uniform integrability.

Uniform integrability means ${\sup }_n\mathbb{E}[|X_n|1_{\{|X_n|>K\}}]\to 0$ as $K\to \infty$. In particular the first absolute moments are uniformly bounded. By the Portmanteau theorem, $\mathbb{E}|X|\le {\mathrm{liminf}}_n\mathbb{E}|X_n|<\infty$. Let $T_K(x)=\max (-K,\min (x,K))$, a bounded continuous truncation. Convergence in distribution gives $\mathbb{E}{T}_K(X_n)\to \mathbb{E}{T}_K(X)$. ThereforeThe first term tends to zero by uniform integrability, the second by integrability of $X$. Hence $\boxed{\mathbb{E}{X}_n\to \mathbb{E}X}$.

For $a<b$, let $U_N[a,b]$ be the number of completed upcrossings by the observations $X_1,\dots ,X_N$: buy when the value is at most $a$, sell at the first subsequent value at least $b$, and repeat. With $z^{-}=\max \{-z,0\}$, one form of the Doob upcrossing inequality for a martingale isFor completeness, hold either zero or one unit according to this rule. The holding at each step is measurable before the next martingale increment, so the total gain $G_N$ has expected value zero. Completed trades earn at least $b-a$ each; an unfinished trade loses at most $(X_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(X_N-a{)}^{-}$, proving the inequality.

Suppose $C={\sup }_n\mathbb{E}|X_n|<\infty$. For each rational pair $a<b$ the inequality and monotone convergence giveEvery rational interval therefore has only finitely many upcrossings, simultaneously outside a countable union of null sets. If $\mathrm{liminf}{X}_n<\mathrm{limsup}{X}_n$, there is a rational interval strictly between them and infinitely many upcrossings, a contradiction. Thus the limit $X_{\infty }$ exists in the extended real line almost surely. Fatou's lemma now givesThe limit is finite almost surely and integrable. This proves the martingale convergence theorem rather than using it as the deduction.

The exact extra condition for convergence in L1 is uniform integrability:To see sufficiency, first discard the tails of $X_n$ uniformly and the tail of the integrable $X_{\infty }$. On the set where both absolute values are at most $K$, their difference is bounded by $2K$ and tends to zero almost surely; dominated convergence finishes the proof. Conversely, if $X_n\to X_{\infty }$ in $L^1$, thenMake the first term uniformly small for all sufficiently large $n$, then control the finitely many remaining integrable variables by increasing $K$. Hence uniform integrability is necessary too. In this case the conditional expectation identity $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ follows by taking the $L^1$ limit in $X_n=\mathbb{E}[X_m\mid {\mathcal{F}}_n]$ for $m\ge n$.

The coin-doubling martingale shows why an $L^1$ bound is not enough. Let ${\xi }_1,{\xi }_2,\dots$ be independent fair coin indicators and setConditional on the first $n$ coins, the next value is either $2X_n$ or zero with equal probabilities, so this is a nonnegative martingale. Its expected value is always $1$, while a first failure occurs almost surely, giving $X_n\to 0$ almost surely. Yet $\mathbb{E}|X_n-0|=1$ for every $n$. Indeed, whenever $2^n>K$, its tail expected value above $K$ is still $1$. Thus convergence in L1 fails because uniform integrability fails.