Counting-process compensators, integrals, brackets

Fix $R>0$ and define ${\varphi }_R(x)=\max (-R,\min (x,R))$ and $G_R(x)=\{\begin{array}{ll}{x}^2\mathrm{/}2,& \mathrm{\mid }x\mathrm{\mid }\le R,\\ R\mathrm{\mid }x\mathrm{\mid }-R^2\mathrm{/}2,& \mathrm{\mid }x\mathrm{\mid }\ge R\mathrm{.}\end{array}$ The explicit piecewise formula gives $G_R^{\mathrm{\prime }}={\varphi }_R$; ${\varphi }_R$ is bounded by $R$ and is $1$-Lipschitz, while $G_R\ge 0$ and $G_R(x)=0$ exactly when $x=0$. Splitting an interval at $-R$ and $R$ if needed and using the Lipschitz bound gives, for all $x,y\in \mathbb{R}$, $\mathrm{\mid }{G}_R(y)-G_R(x)-{\varphi }_R(x)(y-x)\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }y-x{\mathrm{\mid }}^2\mathrm{.}$

Fix $t\in (0,T]$ and deterministic partitions $0=t_0^m<\cdots <t_{q_m}^m=t$ with mesh tending to zero. Put $S_m={\sum }_{j=0}^{q_m-1}{\varphi }_R(D_{t_j^m})(D_{t_{j+1}^m}-D_{t_j^m})\mathrm{.}$ Each coefficient is bounded and ${\mathcal{F}}_{t_j^m}$-measurable. The martingale identity and [F2] give $\mathbb{E}{S}_m=0$. The estimate of step 1.1 and the total-variation bound [F4] imply $\mathrm{\mid }{S}_m-G_R(D_t)+G_R(D_0)\mathrm{\mid }\le \frac{1}{2}{\sum }_j\mathrm{\mid }{D}_{t_{j+1}^m}-D_{t_j^m}{\mathrm{\mid }}^2\le \frac{1}{2}{\max }_j\mathrm{\mid }{D}_{t_{j+1}^m}-D_{t_j^m}\mathrm{\mid }{V}_T(D)\mathrm{.}$ On the probability-one continuity event in [F1], compactness [F5] and Heine--Cantor [F6] make the maximum increment tend to zero. Thus $S_m\to G_R(D_t)$ almost surely, since $D_0=0$. Also $\mathrm{\mid }{S}_m\mathrm{\mid }\le R{V}_T(D)$, an integrable bound by the hypothesis. Dominated convergence [F3] yields $\mathbb{E}{G}_R(D_t)=0$. Since $G_R$ is nonnegative and vanishes only at zero, [F7] gives $D_t=0$ almost surely.

The conclusion also holds at $t=0$ because $D_0=0$. For every rational $r\in [0,T]$, step 2.1 gives a measurable probability-one event on which $D_r=0$. Intersect these countably many events with the probability-one continuity event from [F1]. On the resulting measurable probability-one event, continuity and density of the rationals give $D_t=0$ for every $t\in [0,T]$. By [F8], $D$ is indistinguishably zero.

Let ${\pi }_n$ be nested finite partitions of $[0,T]$ containing any fixed $s,t$ with mesh tending to zero. For each ${\pi }_n$, the values of $\mathrm{\mid }M\mathrm{\mid }$ on its grid form a discrete-time nonnegative submartingale by [F1, F3]. The Doob bound gives $\mathbb{E}{\max }_{u\in {\pi }_n}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2\le 4\mathbb{E}\mathrm{\mid }{M}_T{\mathrm{\mid }}^2$. Càdlàg paths and the nested dense grids imply ${\sup }_{u\le T}\mathrm{\mid }{M}_u\mathrm{\mid }={\sup }_{u\in {\cup }_n{\pi }_n}\mathrm{\mid }{M}_u\mathrm{\mid }$; [F5] then gives $\mathbb{E}{\sup }_{u\le T}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2<\mathrm{\infty }$.

For a finite partition ${\pi }_n=\{r_0<\cdots <r_m\}$, round $\tau$ up to the first grid point ${\tau }_n\ge \tau$. This is a finite-valued stopping time because $\{{\tau }_n\le r_j\}=\{\tau \le r_j\}\in {\mathcal{F}}_{r_j}$. At adjacent grid points, $M_{r_k\wedge {\tau }_n}-M_{r_{k-1}\wedge {\tau }_n}={\mathbf{1}}_{\{{\tau }_n>r_{k-1}\}}(M_{r_k}-M_{r_{k-1}})$; the indicator is ${\mathcal{F}}_{r_{k-1}}$-measurable. Conditioning each increment on its left endpoint proves the stopped grid values have the martingale identity at grid times, in particular at $s,t$.

Refine the partitions while keeping $s,t$ as grid points. Then ${\tau }_n\downarrow \tau$, and pathwise right continuity gives $M_{s\wedge {\tau }_n}\to M_{s\wedge \tau }$ and $M_{t\wedge {\tau }_n}\to M_{t\wedge \tau }$. For $u=s$ or $u=t$, the events $\{{\tau }_n=r_j\}$ with grid value $r_j<u$ lie in ${\mathcal{F}}_{r_j}\subseteq {\mathcal{F}}_u$, and their complement $\{{\tau }_n\ge u\}$ is also in ${\mathcal{F}}_u$; expressing $M_{u\wedge {\tau }_n}$ over these finitely many events proves ${\mathcal{F}}_u$-measurability. The pointwise limits are then measurable there without completing the filtration. Both are dominated by ${\sup }_{u\le T}\mathrm{\mid }{M}_u\mathrm{\mid }$, integrable by step 1.1. For each $B\in {\mathcal{F}}_s$, apply [F4] to the event-integral martingale identity from step 1.2; it yields $\mathbb{E}[{\mathbf{1}}_B(M_{t\wedge \tau }-M_{s\wedge \tau })]=0$, hence the stopped process satisfies the all-pairs martingale identity. Also $\mathrm{\mid }{M}_{t\wedge \tau }\mathrm{\mid }\le {\sup }_{u\le T}\mathrm{\mid }{M}_u\mathrm{\mid }$ gives square integrability.

Put $f(s,\omega )=Y_s(\omega ){\lambda }_s(\omega )$. By [F1], $f$ is product-measurable, so [F2] defines $\mu$ and gives $\mu ([0,T]\times \mathrm{\Omega })=\int fd(ds\times P)=\mathbb{E}{\int }_0^{T}f(s,\omega )ds=\mathbb{E}{A}_T<\mathrm{\infty }$. Countable additivity follows from the product measure; [F5] also gives it by monotone convergence on disjoint unions. Lebesgue measure gives $\mu (\{0\}\times B)=0$ for every $B\in {\mathcal{F}}_0$.

Let $\mathcal{G}$ contain the empty set, whole space, time-zero generators, and interval rectangles from [F1]. Their finite intersections are empty or generators, so $\mathcal{G}$ is a $\pi$-system generating ${\mathcal{P}}_T$. Let $\mathcal{D}$ be the sets whose indicators lie in the $L^2(\mu )$-closure of the span of generator indicators. The whole space is in $\mathcal{D}$ because ${\mathbf{1}}_{(0,T]\times \mathrm{\Omega }}$ is an elementary step and the time-zero slice is null. Complements preserve $\mathcal{D}$ by subtraction from $\mathbf{1}$; disjoint countable unions preserve it because finite partial sums converge in $L^2(\mu )$, with squared tail norm the measure of a decreasing tail tending to zero as $\mu$ is finite. Thus $\mathcal{D}$ is a $\lambda$-system containing $\mathcal{G}$, and [F3] gives $\mathcal{D}={\mathcal{P}}_T$.

By [F4], every $L^2(\mu )$ class is an $L^2$ limit of finite simple functions on measurable sets. Each indicator in such a simple function belongs to the closure from step 2.1, so each simple function does as well; hence the span of bounded predictable step indicators is dense in $L^2(\mu )$.

The difference $D-C=(X-C)-(X-D)$ is a martingale by [F3]. It is continuous, has finite variation, starts at zero, and $\mathbb{E}\mathrm{Var}(D-C{)}_T\le \mathbb{E}\mathrm{Var}(D{)}_T+\mathbb{E}\mathrm{Var}(C{)}_T<\mathrm{\infty }$, so it satisfies [F1].

By [F1], $D-C$ is zero outside one probability-one event, which is indistinguishability by [F4]. Hence $C=D$ indistinguishably, proving uniqueness for both scalar and cross-bracket compensators in the stated continuous class.

Each summand in [F2] is ${\mathcal{F}}_t$-measurable at its current endpoint. It is square-integrable because $H_j$ is bounded and each martingale increment is in $L^2$. A finite sum is therefore adapted and square-integrable at every $t\le T$.

Fix $0\le s\le t\le T$ and $B\in {\mathcal{F}}_s$. Refine the original deterministic partition by inserting $s$ and $t$. After subtracting $I_s(H)$ from $I_t(H)$, each remaining piece has the form $H_j(M_v-M_u)$ on an interval $(u,v]$ with $u\ge s$ and $t_j\le u$. Thus ${\mathbf{1}}_B{H}_j$ is bounded and ${\mathcal{F}}_u$-measurable. Since $\mathbb{E}[M_v-M_u\mid {\mathcal{F}}_u]=0$, [F3] gives $\mathbb{E}[{\mathbf{1}}_B{H}_j(M_v-M_u)]=0$. Summing the finitely many pieces yields $\mathbb{E}[{\mathbf{1}}_B(I_t(H)-I_s(H))]=0$. The event-integral characterization of conditional expectation proves $\mathbb{E}[I_t(H)\mid {\mathcal{F}}_s]=I_s(H)$. Thus the elementary integral is a square-integrable martingale.

For finitely many distinct jump times $0<s_1<\cdots <s_q\le T$, choose disjoint intervals $(u_i,s_i]$ with $u_i\uparrow s_i$. A partition containing all their endpoints has variation at least ${\sum }_i\mathrm{\mid }{x}_{s_i}-x_{u_i}\mathrm{\mid }$, so [F2] and the left limits give ${\sum }_i\mathrm{\mid }\mathrm{\Delta }{x}_{s_i}\mathrm{\mid }\le {\mathrm{Var}}_{[0,T]}(x)$. Hence each set of jumps of size at least $1\mathrm{/}m$ is finite, in its canonical time order, and the nonzero jumps form a countable set with ${\sum }_{0<s\le T}\mathrm{\mid }\mathrm{\Delta }{x}_s\mathrm{\mid }\le {\mathrm{Var}}_{[0,T]}(x)$. Let $J_t={\sum }_{0<s\le t}\mathrm{\Delta }{x}_s$. Absolute convergence makes omitted jump tails uniformly small, so finite jump partial sums converge uniformly to a path $J$; right continuity passes by an $\epsilon \mathrm{/}3$ estimate, and their left limits are Cauchy under the uniform tail bound, so completeness of $\mathbb{R}$ gives left limits for $J$. Absolute convergence gives $\mathrm{\Delta }{J}_s=\mathrm{\Delta }{x}_s$. For any partition, each increment of $J$ is a subseries of its jumps, so ${\mathrm{Var}}_{[0,T]}(J)\le {\sum }_{0<s\le T}\mathrm{\mid }\mathrm{\Delta }{x}_s\mathrm{\mid }$; the same argument bounds the variation of any omitted jump tail by its absolute sum. Therefore $x^c:=x-J$ is continuous and of finite variation. Choose a finite retained jump sum $z$ with omitted tail $r:=J-z$ satisfying ${\mathrm{Var}}_{[0,T]}(r)<\epsilon$, and set $y=x-r=x^c+z$. By [F3] and [F4], the continuous part has squared-increment sum at most ${\omega }_{x^c}(\mathrm{\parallel }{\pi }_n\mathrm{\parallel }){\mathrm{Var}}_{[0,T]}(x^c)\to 0$, where ${\omega }_{x^c}(\delta ):={\sup }_{\mathrm{\mid }u-v\mathrm{\mid }\le \delta }\mathrm{\mid }{x}_u^c-x_v^c\mathrm{\mid }$; sufficiently fine partitions separate the finitely many retained jump times, so ${\sum }_j({\mathrm{\Delta }}_j^{n}y(t){)}^2$ tends to the sum of their squared jumps, with cross terms between each retained jump and the continuous increment in its partition cell tending to zero by the same modulus.

For every partition, expansion of squares and [F5] give $\mathrm{\mid }{\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2-{\sum }_j({\mathrm{\Delta }}_j^{n}y(t){)}^2\mathrm{\mid }\le 2{\mathrm{Var}}_{[0,T]}(y){\mathrm{Var}}_{[0,T]}(r)+{\mathrm{Var}}_{[0,T]}(r{)}^2\le 2{\mathrm{Var}}_{[0,T]}(x)\epsilon +3{\epsilon }^2$, since ${\mathrm{Var}}_{[0,T]}(y)\le {\mathrm{Var}}_{[0,T]}(x)+\epsilon$. Also $x=y+r$ gives $\mathrm{\Delta }{x}_s=\mathrm{\Delta }{y}_s+\mathrm{\Delta }{r}_s$; applying the same finite-list expansion and [F5] to the jump sums, then using their absolute convergence [F1], yields $\mathrm{\mid }[x{]}_t-[y{]}_t\mathrm{\mid }\le 2{\mathrm{Var}}_{[0,T]}(y){\mathrm{Var}}_{[0,T]}(r)+{\mathrm{Var}}_{[0,T]}(r{)}^2\le 2{\mathrm{Var}}_{[0,T]}(x)\epsilon +3{\epsilon }^2$. Since the squared-increment sums for $y$ converge to $[y{]}_t$ by step 1.1, these two error bounds show that the limsup distance between the squared-increment sums for $x$ and $[x{]}_t$ is at most $4{\mathrm{Var}}_{[0,T]}(x)\epsilon +6{\epsilon }^2$. Letting $\epsilon \to 0$ proves ${\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2\to {\sum }_{0<s\le t}(\mathrm{\Delta }{x}_s{)}^2$.

For every finite partition, telescoping differences of squares gives $x_t^2-x_0^2=2{\sum }_j{x}_{t\wedge t_j^n}{\mathrm{\Delta }}_j^{n}x(t)+{\sum }_j({\mathrm{\Delta }}_j^{n}x(t){)}^2$. Step 2.1 gives the limit of the second sum, and rearrangement gives the stated limit of the left-endpoint sums.

Since $A$ is continuous, $\mathrm{\Delta }{M}_s=\mathrm{\Delta }{N}_s$; each jump of $N$ is zero or one, so the jump-square definition gives $[M{]}_t={\sum }_{0<s\le t}(\mathrm{\Delta }{N}_s{)}^2=N_t$, using $N_0=0$.

Let $S=N+A$ and ${\tau }_n=\inf \{t:S_t\ge n\}\wedge T$ for integers $n\ge 1$. The adapted nondecreasing càdlàg process $S$ satisfies $\{{\tau }_n\le t\}=\{S_t\ge n\}$ for $t<T$, so ${\tau }_n$ is bounded stopping; at the first crossing, continuity of $A$ and unit jumps of $N$ give $S_{{\tau }_n}\le n+1$, hence $\mathrm{\mid }{M}_{t\wedge {\tau }_n}\mathrm{\mid }\le n+1$. By [F5], $M^{{\tau }_n}$ is a square-integrable martingale.

The variables $M_T$ and $A_T$ are integrable, so $N_T=M_T+A_T$ is integrable; since $M$ is a martingale starting at zero, $\mathbb{E}{N}_T=\mathbb{E}{A}_T<\mathrm{\infty }$. The pathwise total variation of $M$ is at most $N_T+A_T$, hence is integrable.

Choose nested deterministic partitions ${\pi }_m=\{0=t_0^m<\cdots <t_{q_m}^m=T\}$ with mesh tending to zero. The coefficients $H_j^{n,m}=M_{t_j^m\wedge {\tau }_n}$ are ${\mathcal{F}}_{t_j^m}$-measurable and bounded by $n+1$, so [F3] makes the step integral against $M^{{\tau }_n}$ a martingale. The partition identity [F4] gives $(M_{t\wedge {\tau }_n}{)}^2-{\sum }_j(M_{t\wedge t_{j+1}^m\wedge {\tau }_n}-M_{t\wedge t_j^m\wedge {\tau }_n}{)}^2=2I_t(H^{n,m})$. The stopped process has the same unit jumps as $N^{{\tau }_n}$, so the squared-increment sum converges to $[M^{{\tau }_n}{]}_t=N_{t\wedge {\tau }_n}$ by [F4] and step 1.1. Also $\mathrm{Var}(M^{{\tau }_n}{)}_T\le N_{T\wedge {\tau }_n}+A_{T\wedge {\tau }_n}\le n+1$, so the terms are bounded by a deterministic constant depending only on $n$; dominated convergence in the martingale event identities proves $(M_{t\wedge {\tau }_n}{)}^2-N_{t\wedge {\tau }_n}$ is a martingale.

Since $S_T=N_T+A_T<\mathrm{\infty }$ a.s., ${\tau }_n=T$ eventually on each path. On nested deterministic grids, [F6] gives $\mathbb{E}{\max }_{u\in {\pi }_m}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2\le 4\mathbb{E}\mathrm{\mid }{M}_T{\mathrm{\mid }}^2$; right continuity identifies the increasing grid limit with ${\sup }_{u\le T}\mathrm{\mid }{M}_u\mathrm{\mid }$, and [F8] gives $\mathbb{E}{\sup }_{u\le T}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus $R={\sup }_{u\le T}\mathrm{\mid }{M}_u{\mathrm{\mid }}^2+N_T$ is integrable and dominates every stopped value $\mathrm{\mid }(M_{t\wedge {\tau }_n}{)}^2-N_{t\wedge {\tau }_n}\mathrm{\mid }$. For $B\in {\mathcal{F}}_s$, [F7] and the event identities from step 2.2 let $n\to \mathrm{\infty }$, proving $M^2-N$ is a martingale.

Since $N-A=M$ is a martingale, $M^2-A=(M^2-N)+M$ is a martingale by step 3.1 and linearity, so [F2] and [F1] make $A$ a scalar bracket candidate; uniqueness gives $⟨M⟩=A$. Also $[M,M]=[M]=N$, so $[M,M]-A=N-A=M$ is a martingale, making $A$ a cross-bracket candidate too; uniqueness gives $⟨M,M⟩=A$.

For every $s$, continuity of each $A^i$ gives $\mathrm{\Delta }{M}_s^i=\mathrm{\Delta }{N}_s^i$. The no-simultaneous-jump hypothesis therefore gives $\mathrm{\Delta }{M}_s^i\mathrm{\Delta }{M}_s^j=0$ for every $s$ and every $i\mathrm{\ne }j$, outside one probability-zero event. By the optional cross-variation definition in [F2], $[M^i,M^j{]}_t={\sum }_{s\le t}\mathrm{\Delta }{M}_s^i\mathrm{\Delta }{M}_s^j=0$ for all $t\le T$.

The zero process is continuous, predictable, has finite variation, and satisfies $[M^i,M^j]-0=0$, which is a martingale. Thus it is a continuous predictable cross-bracket candidate. Uniqueness [F3] implies $⟨M^i,M^j⟩=0$. The proof uses the common filtration for the component martingale contracts and uses no independence.

On each $(t_j,t_{j+1}]$, write $\mathrm{\Delta }{M}_j=M_{t_{j+1}}-M_{t_j}$ and $\mathrm{\Delta }{A}_j=A_{t_{j+1}}-A_{t_j}$. The martingale $M^2-A$ gives $\mathbb{E}[M_{t_{j+1}}^2-M_{t_j}^2\mid {\mathcal{F}}_{t_j}]=\mathbb{E}[\mathrm{\Delta }{A}_j\mid {\mathcal{F}}_{t_j}]$; expand the left square as $(\mathrm{\Delta }{M}_j{)}^2+2M_{t_j}\mathrm{\Delta }{M}_j$. Since $M_{t_j}$ and $\mathrm{\Delta }{M}_j$ are in $L^2$, their product is integrable by [F5], so [F3] applies with the possibly unbounded factor $M_{t_j}$. Use $\mathbb{E}[\mathrm{\Delta }{M}_j\mid {\mathcal{F}}_{t_j}]=0$ to get $\mathbb{E}[(\mathrm{\Delta }{M}_j{)}^2\mid {\mathcal{F}}_{t_j}]=\mathbb{E}[\mathrm{\Delta }{A}_j\mid {\mathcal{F}}_{t_j}]$. By [F3], $\mathbb{E}[H_j^2(\mathrm{\Delta }{M}_j{)}^2]=\mathbb{E}[H_j^2\mathrm{\Delta }{A}_j]$.

If $i<j$, then $H_i\mathrm{\Delta }{M}_i{H}_j$ is ${\mathcal{F}}_{t_j}$-measurable and its product with $\mathrm{\Delta }{M}_j$ is integrable by boundedness of $H$ and [F5]; [F3] and $\mathbb{E}[\mathrm{\Delta }{M}_j\mid {\mathcal{F}}_{t_j}]=0$ give $\mathbb{E}[H_i\mathrm{\Delta }{M}_i{H}_j\mathrm{\Delta }{M}_j]=0$. Expanding the finite-sum square [F2], all off-diagonal terms vanish, and step 1.1 gives $\mathbb{E}[I_T(H{)}^2]={\sum }_j\mathbb{E}[H_j^2(\mathrm{\Delta }{M}_j{)}^2]={\sum }_j\mathbb{E}[H_j^2\mathrm{\Delta }{A}_j]=\mathbb{E}{\int }_0^T{H}_s^{2}d{A}_s$; the martingale assertion is [F4].

If two step representations agree in the compensator $L^2$ seminorm, their difference has seminorm zero; applying the isometry in step 2.1 to the refined difference representation makes the squared $L^2$ norm of their terminal-integral difference zero, so they define the same terminal class.

By [F2], choose bounded predictable step processes $H_n$ with $\mathbb{E}{\int }_0^T\mathrm{\mid }{H}_n-H{\mathrm{\mid }}^{2}dA\to 0$. For any $m,n$, [F1] gives $\mathbb{E}\mathrm{\mid }{I}_T(H_n)-I_T(H_m){\mathrm{\mid }}^2=\mathbb{E}{\int }_0^T\mathrm{\mid }{H}_n-H_m{\mathrm{\mid }}^{2}dA$, so terminal integrals are Cauchy in $L^2(P)$; the same estimate on $[0,t]$ gives the Cauchy property at every $t\le T$. [F3] gives a unique full-space limit and, since every $I_t(H_n)$ is ${\mathcal{F}}_t$-measurable, also an ${\mathcal{F}}_t$-measurable limit at each fixed $t$; uniqueness makes these the same $L^2$ class.

If $K_n$ is another step sequence converging to $H$, then $\mathbb{E}{\int }_0^T\mathrm{\mid }{H}_n-K_n{\mathrm{\mid }}^{2}dA\to 0$. Applying [F1] to $H_n-K_n$ makes the terminal limits and each fixed-time limit equal in $L^2$, so the limits in [F5] are independent of approximation.

Norm continuity and [F1] give, for every $t\le T$, $\mathbb{E}\mathrm{\mid }{I}_t(H){\mathrm{\mid }}^2={\lim }_n\mathbb{E}\mathrm{\mid }{I}_t(H_n){\mathrm{\mid }}^2={\lim }_n\mathbb{E}{\int }_0^t{H}_{n,s}^{2}d{A}_s=\mathbb{E}{\int }_0^t{H}_s^{2}d{A}_s$. This proves square integrability and the isometry.