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For an adapted process with right-continuous paths and a stopping time $T$, $X_t^T=X_{t\wedge T}$ is adapted. At fixed $t$, round $t\wedge T$ upward on finite partitions of $[0,t]$. Every approximating value is ${\mathcal{F}}_t$-measurable, and right continuity gives the limit. Completeness handles a null exceptional set if right continuity holds only almost surely. A martingale assumption and boundedness of $T$ are unnecessary.

An announcing sequence consists of stopping times increasing to $T$ almost surely, strictly smaller than $T$ on $\{T>0\}$. Such a lifetime can be approached through stopped intervals on which a locally defined stochastic process is an ordinary adapted process. For the usual maximal local solution of a stochastic differential equation, the exit times ${\tau }_n$ from a nested compact exhaustion of the open domain, capped as $T_n={\tau }_n\wedge n$, supply this sequence. The existence of such a sequence is an extra lifetime convention; arbitrary stopping times need not admit one.

An adapted process with integrable coordinates is a martingale exactly when the displayed equality holds for every bounded stopping time. For the forward implication, expand $M_T-M_0={\sum }_{k=1}^N{1}_{\{T\ge k\}}(M_k-M_{k-1})$ and condition each term on ${\mathcal{F}}_{k-1}$. Conversely, for $A\in {\mathcal{F}}_n$ choose $T=n+1$ on $A$ and $T=n$ elsewhere. Subtract the equality for the constant stopping time $n$ to obtain $\mathbb{E}[1_A(M_{n+1}-M_n)]=0$, which is the defining conditional expectation identity.

A filtration $({\mathcal{F}}_t)$ is a Brownian filtration for $B$ when $B$ is an adapted process and each future increment $B_t-B_s$ is independent of ${\mathcal{F}}_s$. The natural filtration of a Brownian motion has this property. Arbitrary enlargement by future information need not preserve it.

An integrable adapted process is a martingale if all its values at bounded stopping times are integrable with the same expectation as $X_0$. For the converse, compare deterministic $t$ with $U=s{1}_A+t{1}_{A^c}$, where $s<t$ and $A\in {\mathcal{F}}_s$. The resulting identity $\mathbb{E}[(X_t-X_s)1_A]=0$ is precisely the defining test for conditional expectation. In the forward direction, use the optional stopping theorem for a càdlàg martingale under the usual conditions for a filtration.

Let an integrable process $(M_n)$ be adapted to $({\mathcal{F}}_n)$. It is a martingale if and only if $\mathbb{E}[M_{n\wedge \tau }]=\mathbb{E}[M_0]$ for every $n$ and every stopping time $\tau$. For the reverse implication, fix $A\in {\mathcal{F}}_n$ and take $\tau =n$ on $A$ and $\tau =n+1$ on $A^c$. Comparing its stopped expectation at $n+1$ with the expectation obtained from the deterministic stopping time $n+1$ gives $\mathbb{E}[1_A(M_{n+1}-M_n)]=0$. Since this holds for every $A\in {\mathcal{F}}_n$, it is exactly $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$.

For a continuous adapted process $X$ and $T=\inf \{s\ge 0:X_s\ge \lambda \}$,Continuity makes the supremum on $[0,t]$ attained and equals its supremum on this countable dense set. Each displayed event belongs to ${\mathcal{F}}_t$, so $T$ is a stopping time without requiring right-continuity of the filtration.

// Target: martingale.bigb

For bounded measurable $a,b$, with $a$ symmetric positive semidefinite, an L-diffusion is a continuous adapted process $X$ such that $g(X_t)-g(X_0)-{\int }_0^{t}Lg(X_s)ds$ is a true martingale for every $g\in C_b^2$. The general local martingale problem may instead use compactly supported test functions and require only a local martingale. Specify the test class and the true or local convention when coefficients are unbounded.

For a left-continuous locally bounded adapted process $H$ and a continuous semimartingale $X$, the elementary predictable left-endpoint approximations tend pointwise to $H$. Decompose $X$ into a continuous local martingale and a continuous finite-variation process. After localization, dominated convergence with respect to the variation measure handles the finite-variation integral. The Itô isometry and Doob L2 maximal inequality handle the martingale integral. The omitted partial last interval is bounded by the local bound on $H$ times the modulus of continuity of $X$. This proves uniform convergence on compacts in probability and is the grid-sum foundation of the Itô product rule.

A continuous locally defined process is a pair $(X,T)$ specified on the stochastic interval $[0,T)$, with a lifetime stopping time $T$ approached by an announcing sequence for a stopping time $T_n$. Each stopped process is an ordinary continuous adapted process. This is the convention used for continuous local differential equations; no value at $T$ is required. Restricting a globally defined process gives examples, but finite lifetimes also allow approach to a domain boundary or explosion.

Suppose $X_n\ge 0$ is an integrable adapted process, $Y_n\ge 0$ is finite and adapted, and $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]\le (1+Y_n)X_n$. The displayed correction is a nonnegative supermartingale: its next denominator is measurable at time $n$, and its reciprocal is at most one. If ${\sum }_n{Y}_n<\infty$ almost surely, the denominators converge to a finite positive limit since $\log (1+y)\le y$. The almost sure supermartingale convergence theorem then gives finite almost sure convergence of $X_n$. No integrability of the infinite product or of ${\sum }_n{Y}_n$ is needed.

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.

The instantaneous forward rate is defined by $f(t,T)=-{\partial }_T\log P(t,T)$, with $P(t,t)=1$. Integrating in maturity gives the zero-coupon bond price and its diagonal short rate:The solution of the continuous-time bank account equation $d{B}_t=r_t{B}_{t}dt$, $B_0=1$, is $B_t=\exp ({\int }_0^t{r}_{s}ds)$; the integration variable is time. Its discounted bond value isAll these are adapted processes when the Heath-Jarrow-Morton model coefficients are adapted and the integrals exist.

For market completeness, stochastic integration over maturity givesThe Itô formula for the exponential then yieldsConsequently under a risk-neutral measure the no-arbitrage drift restriction isThe last equality needs sufficient maturity regularity and has a positive sign despite the negative definition of $\Sigma$.

Write the predictable sigma-algebra on positive times asIt is equivalently the smallest sigma-algebra making all left-continuous adapted processes measurable. Indeed, the indicator of each generating rectangle is left-continuous and adapted, and approximating any left-continuous adapted process by its values at the preceding points of deterministic partitions gives the reverse inclusion. Previsibility records information available immediately before observation. If time zero is included, add $B\times \{0\}$ with $B\in {\mathcal{F}}_0$. The displayed description in part (b) uses positive times, as indicated by its product space.

Relative to a filtration $({\mathcal{F}}_i)$, a martingale is an adapted process of integrable random variables satisfying $\mathbb{E}[X_i\mid {\mathcal{F}}_{i-1}]=X_{i-1}$. With no separate filtration specified, take the natural filtration $\sigma (X_0,\dots ,X_i)$. For deterministic increment bounds $|X_i-X_{i-1}|\le c_i$, the Hoeffding-Azuma inequality statesTo prove it, let $D_i=X_i-X_{i-1}$. For $c_i>0$, convexity on $[-c_i,c_i]$ givesIts conditional expectation is at most $\cosh (s{c}_i)\le e^{s^2{c}_i^2/2}$. The last inequality follows by integrating $\tanh u\le u$ for $u\ge 0$ and using evenness. If $c_i=0$ the bound is immediate. Conditional iteration now yields $\mathbb{E}{e}^{s(X_n-X_0)}\le \exp (s^2\sum c_i^2/2)$. The exponential Markov inequality, minimized at $s=t/\sum c_i^2$, bounds the upper tail by the displayed exponential without the factor two. Apply it to $-X_i$ and use the union bound for the two-sided assertion. When every $c_i=0$, the deviation probability is zero.

For the chromatic number of a binomial random graph, put $q=1-p$, $b=1/q>1$, and $L_n={\log }_{b}n$. We prove matching bounds; treating $0<\epsilon <1$ suffices. A colour class is an independent set, so $\chi \ge n/\alpha (G)$. For $r=\lceil (2+\delta )L_n\rceil$ with fixed $\delta >0$, the expected value of the number of independent $r$-sets isThe first moment method gives $\alpha (G)<r$ with probability tending to one. Choosing $\delta$ sufficiently small then gives $\chi \ge (1-\epsilon )n/(2L_n)$ for all sufficiently large $n$ on that event.

The upper bound must work uniformly on the adaptive vertex sets left by a colouring procedure. SetFor a specified $m$-set $U$, its complement graph has distribution $G(m,q)$. Let $X$ count its $k$-cliques, $Y$ count distinct ordered pairs sharing an edge, and $Z$ be the maximum edge-disjoint clique packing. The allowed clique-count estimates at this value of $k$ giveThe overlap quantity, if the diagonal is included, has the exact expression ${\sum }_{j=2}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)q^{-\left(\genfrac{}{}{0ex}{}{j}{2}\right)}/\left(\genfrac{}{}{0ex}{}{m}{k}\right)$ after division by $(\mathbb{E}X{)}^2$; the same bound holds. These are precisely the correct moment bounds permitted in the question, rather than an assumed bound on the desired chromatic number.

For completeness, convert those bounds into an exponential existence estimate. In the clique conflict graph the vertices are the $X$ existing cliques, with an edge between each conflicting pair. A random ordering keeps every vertex that precedes all its neighbours. No two retained vertices are adjacent. Their mean number is ${\sum }_{v}1/(d_v+1)\ge X^2/(X+Y)$ by the Cauchy-Schwarz inequality. Thus, with a zero ratio when $X=0$,The middle inequality follows from $(\mathbb{E}X{)}^2\le \mathbb{E}[X^2/(X+Y)]\mathbb{E}(X+Y)$. Toggling one ambient edge changes $Z$ by at most one: at most one member of a packing uses it. Reveal the $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edges successively and form the edge-exposure martingale for $Z$. Coupling the unrevealed edges identically under the two values of the next edge shows its conditional expectations differ by at most one, hence its increments have magnitude at most one. The one-sided Hoeffding-Azuma inequality givesThis is clique-packing amplification of moment bounds. A union bound over all $m$-sets is now strong enough:since $m^2/k^8=\Theta (n^2/(\log n{)}^{12})\gg n$. Thus every such set contains an independent set of size $k$. Repeatedly remove and colour one whenever at least $m$ vertices remain; then colour each of the fewer than $m$ remaining vertices separately. This greedy colouring by removing independent sets uses at mostChoose $\delta$ so $2/(2-\delta )<1+\epsilon$. Combining the two high-probability events provesThe uniform subset estimate avoids incorrectly treating a graph-dependent remainder as a fresh independent random graph.

For $A\in {\mathcal{F}}_n$, define the bounded stopping time $T=n+1$ on $A$ and $T=n$ on $A^c$. Before time $n$ its stopping events are empty, at time $n$ the event $\{T\le n\}=A^c$ belongs to ${\mathcal{F}}_n$, and subsequently it has stopped everywhere. Subtract the assumed mean identity for this $T$ from that for the constant stopping time $n$ to obtainSince $M_n$ is integrable and ${\mathcal{F}}_n$-measurable, this is precisely the defining property of conditional expectation giving $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus the adapted process is a martingale, completing the bounded-stopping-time characterization of a martingale.

Relative to a filtration $({\mathcal{F}}_t)$, a supermartingale is an adapted process $U_t$ with $\mathbb{E}|U_t|<\infty$ for every $t$ andBy the tower property this also gives $\mathbb{E}[U_u\mid {\mathcal{F}}_t]\le U_t$ for every $u\ge t$.

A discrete-time martingale relative to a filtration $({\mathcal{F}}_n)$ is an adapted process $(M_n)$ with $\mathbb{E}|M_n|<\infty$ andIterating conditional expectation also gives $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ for $m\ge n$. It is bounded in $L^2$ when each $M_n$ is square integrable andThis is a uniform bound on the second moments, not a pathwise bound on all sample values.

The optional stopping theorem for bounded stopping times says that, if $\sigma \le \tau \le K$ for a deterministic finite $K$, thenHere ${\mathcal{F}}_{\sigma }$ is the stopping-time sigma-algebra. The second assertion follows by taking $\sigma =0$ and then expectations. An $L^2$ bound is not needed: integrability at the finitely many deterministic times suffices.

For a proof, both stopped values are integrable since $|M_{\tau }|,|M_{\sigma }|\le {\sum }_{j=0}^K|M_j|$. ExpandIf $A\in {\mathcal{F}}_{\sigma }$, then $A\cap \{\sigma \le k\}\in {\mathcal{F}}_k$ by the definition of the stopping-time sigma-algebra, and $\{\tau >k\}\in {\mathcal{F}}_k$ because $\tau$ is a stopping time. Thus the indicator of $A\cap \{\sigma \le k<\tau \}$ is ${\mathcal{F}}_k$-measurable. Multiplying the sum by $1_A$, taking expectations, and conditioning each summand on ${\mathcal{F}}_k$ gives zero, by the martingale increment property. Therefore $\mathbb{E}[1_A{M}_{\tau }]=\mathbb{E}[1_A{M}_{\sigma }]$ for every $A\in {\mathcal{F}}_{\sigma }$. Since $M_{\sigma }$ is ${\mathcal{F}}_{\sigma }$-measurable, this is exactly the claimed conditional expectation identity.

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by $E\times \{0\}$ with $E\in {\mathcal{F}}_0$ and by $E\times (s,t]$ with $E\in {\mathcal{F}}_s$, $0\le s<t$. Equivalently it is the smallest sigma-algebra making every left-continuous adapted process measurable. A previsible process is a process measurable for $\mathcal{P}$.

A bounded simple predictable process has finitely many time intervals and bounded coefficients known at their left endpoints. One may use deterministic endpoints, or the customary more general elementary predictable process with stopping-time intervals:A bounded ${\mathcal{F}}_0$-measurable value at time zero may be added; it does not affect integration against a continuous martingale. A process with finite support is included by adding a zero coefficient on its terminal interval. Both versions of simple processes satisfy the arguments below. The elementary stochastic integral isWhen a stopping time is infinite, use the terminal value $M_{\infty }$, which exists in $L^2$ and almost surely for an L2-bounded continuous martingale. Refining intervals leaves this definition unchanged.

For a continuous local martingale $M$, its quadratic variation $[M]$ is the unique continuous adapted nondecreasing process starting at zero such that $(M_t-M_0{)}^2-[M{]}_t$ is a local martingale. It is predictable because it is continuous and adapted. To recover it from the path, use refining deterministic grids, for example $t_k^{(n)}=k{2}^{-n}$, and formOnly finitely many increments are nonzero on a compact time interval. ThenIn particular each fixed-time sum converges in probability, and a suitable subsequence converges uniformly on compact intervals almost surely. The partial last grid interval is included in this path approximation.