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Let a nonnegative martingale start at one, let $T$ be its first zero, and let $T_R$ be its first hit of $[R,\infty )$. Suppose $\mathbb{E}[(M_{n+1}-M_n{)}^2\mid {\mathcal{F}}_n]\ge {\sigma }^2{1}_{\{T>n\}}$ and $M_{T_R}\le \tau R$ on finite hits, for thresholds $R\ge 1$. Then $M_{T\wedge T_R\wedge n}^2-{\sigma }^2(T\wedge T_R\wedge n)$ is a submartingale. The bounded stopped values have mean one, so their second moment is at most $\tau R$. Consequently $\mathbb{E}(T\wedge T_R)\le (\tau R-1)/{\sigma }^2$. For $\tau \ge 1$ this implies the weaker bound ${\tau }^{2}R/{\sigma }^2$. The threshold restriction prevents an impossible overshoot requirement below the starting value.

A submartingale with ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable random variable. The positive-part bound also bounds negative parts because $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$. The result specializes to an $L^1$-bounded martingale, and hence to every nonnegative martingale. It does not by itself give convergence in L1; uniform integrability supplies that stronger conclusion.

If $(X_m{)}_{m\le n}$ is a nonnegative submartingale, then for every $a>0$,Stop at the first crossing of $a$ and use the submartingale property on that event.

For independent zero-mean square-integrable increments, their partial sums form a martingale, and their squares form a nonnegative submartingale. Apply the Doob maximal inequality to the squared process at threshold $x^2$. Independence and zero means make its terminal second moment the sum of the increment variances. Identical distributions are not required.

For a centered square-integrable martingale with $v=\mathbb{E}{M}_n^2$ and $\lambda >0$, apply the Doob maximal inequality for a nonnegative submartingale to $(M_k+c{)}^2$ for $c\ge 0$. Conditional Jensen inequality gives the submartingale property, and crossing $\lambda$ forces its square above $(\lambda +c{)}^2$. The bound $(v+c^2)/(\lambda +c{)}^2$ is minimized by $c=v/\lambda$. This extends the Cantelli inequality from one random variable to a martingale maximum, without requiring independent increments.

For a submartingale $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed upcrossings of $[a,b]$ by time $N$. The Doob upcrossing inequality givesFor a martingale $(X_n)$, a useful sharper form isHere $x^{+}=\max (x,0)$ and $x^{-}=\max (-x,0)$. To see the martingale form, use a bounded predictable process $H_k\in \{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a{)}^{-}$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a{)}^{-}$. Since $\mathbb{E}{G}_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a{)}^{+}$, itself a submartingale. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb{E}(Z_N-a{)}^{+}-\mathbb{E}(Z_0-a{)}^{+}$, and in particular by the stated right side.

The almost-sure martingale convergence theorem says that if $C={\sup }_n\mathbb{E}|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable random variable $Y$, with $\mathbb{E}|Y|\le C$. Indeed, the Doob upcrossing inequality bounds the expected total number of upcrossings of every rational interval by $(C+|a|)/(b-a)$. By monotone convergence, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a sample path differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The Fatou lemma givesso the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves expectations nor supplies this uniform first-moment bound.

For the loss of expectation, let $({\eta }_k)$ be independent variables with the fair Bernoulli distribution and setConditional on the past, survival doubles the value with probability $1/2$ and otherwise sets it to zero, so this is a nonnegative martingale. An infinite string of successes has probability zero. The process therefore eventually vanishes almost surely, giving yes: $X_0=1$ and $Y=0$ almost surely are possible. Its expectations remain one at every finite time; the process is not uniformly integrable.

A nonintegrable limit is also possible for a signed martingale. Take independent ${\xi }_k$ withand let $M_0=1$, $M_n=1+{\sum }_{k=1}^n{\xi }_k$. Every increment has mean zero and a finite absolute expectation, so each $M_n$ is integrable and $(M_n)$ is a martingale. Since ${\sum }_k\mathbb{P}({\xi }_k\ne 0)<\infty$, the Borel-Cantelli lemma makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+{\sum }_k{\xi }_k$ exists almost surely. To compute its absolute expectation, let $A_k$ be the event that only the $k$th increment is nonzero. Independence givesThe infinite product is positive because its tail logarithms are bounded in absolute value by a constant times the summable series ${\sum }_j{2}^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\ge 4^k-1$. ThereforeThus yes: a finite almost-sure martingale limit can have infinite absolute expectation, even with deterministic $M_0=1$. This is almost-sure martingale convergence to a nonintegrable limit; the missing hypothesis is the uniform first-moment bound. For a nonnegative martingale, in contrast, the Fatou lemma would force the limit to be integrable.

For the continuous adapted process $X$, the supremum on each compact time interval is attained. Let $D_t=(\mathbb{Q}\cap [0,t])\cup \{t\}$. Continuity gives ${\sup }_{s\le t}{X}_s={\sup }_{r\in D_t}{X}_r$, and henceEach event inside the union is in ${\mathcal{F}}_r\subseteq {\mathcal{F}}_t$, so this proves carefully that $T$ is a stopping time. Merely checking rational times for $X_r\ge \lambda$ would miss a path that touches the level at an isolated irrational time; the approximation from below avoids this issue. This is the closed-level hitting times of continuous adapted processes criterion. For $\lambda =0$, nonnegativity makes $T=0$ and the later inequality is immediate.

For $\lambda >0$, put $A=\{T\le t\}$ and $\tau =T\wedge t$. On $A$, continuity implies $X_{\tau }\ge \lambda$. To justify the optional sampling step without extra filtration assumptions, round $\tau$ upwards to a finite dyadic grid in $[0,t]$, obtaining stopping times ${\tau }_n\downarrow \tau$. The event $A$ belongs to ${\mathcal{F}}_{\tau }$, and hence to ${\mathcal{F}}_{{\tau }_n}$: directly, $A\cap \{\tau \le u\}=\{T\le u\}$ for $u<t$, while for $u\ge t$ it is $A$. Discrete optional sampling for the sampled submartingale givesAs $n\to \infty$, continuity gives $X_{{\tau }_n}\to X_{\tau }$. Nonnegativity and the Fatou lemma therefore yieldConsequently the Doob maximal inequality for a nonnegative submartingale takes the precise form

For the second-moment bound, suppose first that $\mathbb{E}{X}_t^2<\infty$; otherwise the inequality with an infinite right side is automatic. Set $M_R=X_t^{\ast }\wedge R$. The layer cake representation and the preceding tail estimate giveThe integrands are nonnegative, so the interchange is justified by the Tonelli theorem. Since $M_R$ is bounded, division is legitimate when its norm is nonzero, giving $\Vert M_R{\Vert }_2\le 2\Vert X_t{\Vert }_2$; the zero-norm case is immediate. Finally, monotone convergence as $R\to \infty$ givesThis truncated layer-cake proof of the L2 maximal inequality proves square integrability of the maximum rather than assuming it during the Cauchy-Schwarz inequality step.

For $t>0$, put $M_t={\sup }_{0\le s\le t}{B}_s$. For $\lambda >0$, reflect the Brownian path after its first hit of $\lambda$. The Strong Markov property and symmetry of subsequent Brownian increments make this reflection preserve the path distribution. It exchanges the events $\{M_t\ge \lambda ,B_t<\lambda \}$ and $\{B_t>\lambda \}$. Since $B_t$ has no atom at $\lambda$, the Brownian reflection principle givesThe case $\lambda =0$ follows by decreasing positive levels to zero: the probability is one. At $t=0$, $M_0=0$ and the probability is zero for every $\lambda \ge 0$.

The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the half-normal distribution of $\sqrt{t}|Z|$ with $Z\sim N(0,1)$. ThusFor $A_t={\sup }_{s\le t}|B_s|$, the pointwise inequality $A_t\ge M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda >0$. Reflection at $\lambda$ also givesSubtracting this from $\mathbb{P}(B_t<-\lambda )=\mathbb{P}(B_t>\lambda )$ givesOn this event $A_t\ge |B_t|>\lambda >M_t$, so $A_t^2-M_t^2$ is strictly positive with positive probability. Finally, $|B_t|$ is a continuous nonnegative submartingale by the conditional Jensen inequality, and part (a) gives $\Vert A_t{\Vert }_2\le 2\Vert B_t{\Vert }_2=2\sqrt{t}$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event givesThis is the strict comparison of one-sided and absolute Brownian maxima. The strict lower inequality printed with $t\ge 0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.

Let $(X_k{)}_{k=0}^n$ be a nonnegative submartingale and assume $X_n\in L^p$. Set $X^{\ast }={\max }_{0\le k\le n}{X}_k$. For $a>0$, let $A_k$ be the event that $k$ is the first index with $X_k>a$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $\{X^{\ast }>a\}$.

The submartingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_k]\ge X_k$. HenceThus part (ii) applies on the probability space with $F=X^{\ast }$ and $g=X_n$, giving the Doob Lp maximal inequalityThe same split at $a/2$ makes $X^{\ast }/2$ satisfy part (i), recovering the preliminary constant $2^{p}q$. Probability spaces automatically have finite superlevel measures. For a right-continuous nonnegative submartingale on $[0,T]$, apply these finite-time bounds to increasing finite grids containing $T$ and a dense set of times; right continuity and monotone convergence theorem extend the estimate to ${\sup }_{t\le T}{X}_t$.

A convenient precise version is the almost sure submartingale convergence theorem: if $(S_n)$ is a real submartingale and ${\sup }_n\mathbb{E}{S}_n^{+}<\infty$, then $S_n$ converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|S_n|<\infty$. It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.

Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed upcrossings by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\le a$, exit when $S_k\ge b$. Witheach completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a{)}^{-}$. Hence $G_n\ge (b-a)U_n[a,b]-(S_n-a{)}^{-}$. The submartingale property and $0\le H_k\le 1$ give $\mathbb{E}{G}_n\le \mathbb{E}(S_n-S_0)$: the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since $(S_n-a{)}^{+}\le S_n^{+}+|a|$. By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso ${\sup }_n\mathbb{E}|S_n|<\infty$. The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to $-Z_n$ gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.

For the given adapted multiplicative drift, putThe next denominator $P_{n+1}$ is ${\mathcal{F}}_n$-measurable, and $P_n\ge 1$ makes $Z_n$ integrable. ThusThe nonnegative supermartingale $Z_n$ has a finite almost sure convergence limit $Z_{\infty }$. On the probability-one event where the specified series is finite,Consequently $P_n\uparrow P_{\infty }\in [1,\infty )$ and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Use ${\mathcal{F}}_m=\sigma ({\xi }_1,\dots ,{\xi }_m)$ and $S_0=0$. Independence and zero means imply $\mathbb{E}[S_{m+1}\mid {\mathcal{F}}_m]=S_m$, so the partial sums form a square-integrable martingale. More explicitly,Thus $S_m^2$ is a nonnegative submartingale. The Doob maximal inequality at threshold $x^2$ now proves the Kolmogorov maximal inequality:Finally, expanding the terminal square and using independence and the zero means gives $\mathbb{E}{S}_n^2={\sum }_{j=1}^n\mathbb{E}{\xi }_j^2$. Identical distributions are not required.

Put $V=\mathbb{E}{X}_n^2$. For $c\ge 0$, the process $(X_m+c{)}^2$ is a nonnegative submartingale, by the conditional Jensen inequality. A crossing of $x$ by $X$ implies a crossing of $(x+c{)}^2$ by this squared process. Since $\mathbb{E}{X}_n=\mathbb{E}{X}_0=0$, the Doob maximal inequality givesFor $V>0$, differentiation of the last expression gives $2(cx-V)/(x+c{)}^3$, so its minimum over $c\ge 0$ occurs at $c=V/x$. Substitution proves the one-sided martingale maximal inequalityIf $V=0$, then $X_n=0$ almost surely and $X_m=\mathbb{E}[X_n\mid {\mathcal{F}}_m]=0$ for every $m\le n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has mean zero, variance $V$, and achieves equality.

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.

Put $C={\sup }_p\mathbb{E}|X_p|<\infty$. Since $x\mapsto x^{+}$ is an increasing convex function, the conditional Jensen inequality and the submartingale property giveFor fixed $n$ and $p\ge n$, the tower property of conditional expectation therefore impliesChoose versions for this countable family so that all these inequalities hold outside one null set. Its nonnegative increasing limit $M_n$ is ${\mathcal{F}}_n$-measurable. The monotone convergence theorem shows thatIn particular the limit is finite almost surely. Thus the increasing conditional means converge to an integrable nonnegative $M_n$, rather than merely to a possibly infinite extended value. This is the construction of the positive martingale majorant of an L1-bounded submartingale.

A filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ on a probability space is an increasing sequence of sigma-algebras contained in $\mathcal{F}$: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. A real stochastic process $(X_n)$ is a martingale for this filtration if $X_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|X_n|<\infty$, andThe first condition is adaptedness; the tower property of conditional expectation then gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$ whenever $m\le n$.

The almost-sure martingale convergence theorem states that if a real martingale satisfies ${\sup }_n\mathbb{E}|X_n|<\infty$, then some integrable random variable $X_{\infty }$ satisfies $X_n\to X_{\infty }$ almost surely. The more general almost sure submartingale convergence theorem needs only ${\sup }_n\mathbb{E}[X_n^{+}]<\infty$ for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.

The Doob maximal inequality for a nonnegative submartingale states that, for a nonnegative submartingale $(M_j{)}_{0\le j\le N}$ and $u>0$,For completeness, partition the crossing event according to its first crossing time $j$. On that event $M_j\ge u$, while $\mathbb{E}[M_N\mid {\mathcal{F}}_j]\ge M_j$. Summing the corresponding expectations over $j$ proves $u\mathbb{P}(\max M_j\ge u)\le \mathbb{E}[M_N{1}_{\{\max M_j\ge u\}}]\le \mathbb{E}{M}_N$.

If $q>p$, the exponential $r^{S_j}$, with $r=q/p>1$, is increasing as a function of $S_j$. Apply the inequality to the martingale from part (a), with $u=r^k$:Letting $N\to \infty$ and using continuity of probability measures on increasing events givesThe same event is obtained if the supremum excludes time zero, since $k\ge 1$. If $p\ge q$, the right side is at least one and the bound is automatic; the nontrivial Doob maximal inequality argument is for $q>p$.

Backward induction first proves that the Snell envelope is adapted and integrable. At each step, $|U_t|\le |Y_t|+\mathbb{E}[|U_{t+1}|\mid {\mathcal{F}}_t]$, so integrability follows from that of $Y_t$ and $U_{t+1}$. Its defining maximum immediately giveswhich proves that $U$ is a supermartingale.

If $Y$ is a submartingale, the sharper backward identity isIt holds at $T$. If it holds at $t+1$, the tower property of conditional expectation gives $\mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]=\mathbb{E}[Y_T\mid {\mathcal{F}}_t]$. Iterating the submartingale inequalities shows that this is at least $Y_t$, so the maximum chooses this continuation value. The displayed identity then makes $U$ a martingale.

For a nonnegative submartingale $(M_k{)}_{0\le k\le n}$ and $M_n^{\ast }={\max }_{0\le k\le n}{M}_k$, the weak form of the Doob maximal inequality isTo prove it, let $A_k$ be the event that $k$ is the first index with $M_k\ge u$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $A=\{M_n^{\ast }\ge u\}$. The submartingale property gives $\mathbb{E}[M_n\mid {\mathcal{F}}_k]\ge M_k$. Multiplying by $1_{A_k}$, taking expectations and summing yields $\mathbb{E}[M_n{1}_A]\ge {\sum }_k\mathbb{E}[M_k{1}_{A_k}]\ge u\mathbb{P}(A)$. Nonnegativity gives the second inequality.

The associated Doob Lp maximal inequality, for $p>1$ and $M_n\in L^p$, isIndeed $M_k\le \mathbb{E}[M_n\mid {\mathcal{F}}_k]$, so conditional Jensen inequality first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $p{u}^{p-2}du$ and using the Tonelli theorem givesThe last step is Holder inequality. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary martingale, its absolute value is a nonnegative submartingale by conditional Jensen inequality, so both corresponding absolute-maximum bounds follow.

The martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.