Characteristic exponential for a continuous local martingale with deterministic clock

First handle exceptional paths without imposing completeness on the original filtration. There is a measurable null set $D\in \mathcal{F}$ outside which $M$ is continuous and $M_0=0$. Put ${\mathcal{G}}_t=\{A\mathrm{△}N:A\in {\mathcal{F}}_t,N\in \mathcal{F},N\subseteq D\}$. This is a sub-sigma-algebra of $\mathcal{F}$: complements preserve the symmetric difference, and a countable union differs from the union of the $A$'s by a measurable subset of $D$. AC permits choosing representations for such a countable family. The ${\mathcal{G}}_t$ are increasing and contain $D$. Set ${\tilde{M}}_t=M_t$ off $D$ and $0$ on $D$. It is ${\mathcal{G}}_t$-adapted, starts at zero everywhere, and has everywhere continuous paths. Each original localizer ${\tau }_k$ is a ${\mathcal{G}}_t$-stopping time. The process ${\tilde{M}}^{{\tau }_k}$ agrees off $D$ with $M^{{\tau }_k}-M_0$, and its stopped values are measurable: a continuous adapted process evaluated at $t\wedge {\tau }_k$ is the pointwise limit of the finite sums obtained by rounding that time upward on deterministic grids of $[0,t]$. Every ${\mathcal{G}}_s$ event differs from an ${\mathcal{F}}_s$ event on $D$, so their integrals agree. Thus ${\tilde{M}}^{{\tau }_k}$ is a ${\mathcal{G}}_t$ martingale. The clock assumption is unchanged by agreement off $D$. We work with this normalized process and filtration until the final descent.

Here is the stopping argument needed for these continuous martingales. Let $X$ be an everywhere continuous martingale for ${\mathcal{G}}_t$, and $\rho$ a stopping time. Fix $0\le s<t$; take finite deterministic grids of $[0,t]$ containing $s$ whose mesh tends to zero, and round $\rho \wedge t$ upward to a grid point ${\rho }_n$. Its grid stopping test is $\{{\rho }_n\le u\}=\{\rho \le u\}$ for grid points $u<t$, so finite-grid optional sampling applies. Write $V_n=X_{{\rho }_n}$ and $W_n=X_{s\wedge {\rho }_n}$. Each is a conditional expectation of the single terminal variable $X_t$ with respect to its grid stopped sigma-algebra; hence each sequence is uniformly integrable. Continuity gives $V_n\to X_{t\wedge \rho }$ and $W_n\to X_{s\wedge \rho }$ pointwise, therefore in probability and then in $L^1$. The finite-grid stopped martingale identity is $E[1_A{V}_n]=E[1_A{W}_n]$ for $A\in {\mathcal{G}}_s$: telescope the increments after $s$, multiplied by $1_{\{{\rho }_n>u\}}$, whose factor is measurable at the left grid endpoint $u$. Passing to $L^1$ limits gives the same identity for $X^{\rho }$. Its adaptedness follows by the upward-grid approximation on each $[0,t]$. Thus $X^{\rho }$ is a martingale without right continuity of the filtration.

Suppress the tilde for now. Define ${\sigma }_m=\inf \{u\ge 0:\mathrm{\mid }{M}_u\mathrm{\mid }\ge m\}\wedge m$, $m\ge 1$. For $t<m$, its stopping event is $\{{\max }_{0\le u\le t}\mathrm{\mid }{M}_u\mathrm{\mid }\ge m\}$; for $t\ge m$ it is $\mathrm{\Omega }$. The maximum is attained and equals the supremum over a countable dense set together with the endpoints, so the event belongs to ${\mathcal{G}}_t$; at $t=0$ it is empty. Continuity and $M_0=0$ give $\mathrm{\mid }{M}_{t\wedge {\sigma }_m}\mathrm{\mid }\le m$. Compact boundedness gives ${\sigma }_m\uparrow \mathrm{\infty }$ on every path. Apply step 1.2 to each martingale $M^{{\tau }_k}$ and $\rho ={\sigma }_m$: $M^{{\tau }_k\wedge {\sigma }_m}$ is a martingale and is bounded by $m$. Dominated convergence as $k\to \mathrm{\infty }$ proves $N:=M^{{\sigma }_m}$ is a bounded continuous martingale. The original ${\tau }_k$, not ${\sigma }_m\wedge {\tau }_k$, is the sequence that localizes this stopped process.

For a deterministic partition of $[0,T]$, the partial-increment square sum of $N$ at $u$ is exactly the partial-increment square sum of $M$ at $u\wedge {\sigma }_m$. Hence its uniform error against $u\wedge {\sigma }_m$ is bounded by the original uniform clock error. Its step version differs by at most the squared maximal oscillation of $N$ on partition intervals, which tends to zero pathwise. Thus the stopped clock is $u\wedge {\sigma }_m$ uniformly in probability. A partition sequence on $[s,t]$ can be extended by vanishing-mesh deterministic partitions on $[0,s]$ and $[t,T]$; subtracting the sums at $s$ gives the same assertion there, with clock $q(u)=(u\wedge {\sigma }_m)-(s\wedge {\sigma }_m)$.

Fix $0\le s<t\le T$. On a partition $s=u_0<\cdots <u_J=t$ put $z_j=N_{u_{j+1}}-N_{u_j}$, $h_j=(u_{j+1}\wedge {\sigma }_m)-(u_j\wedge {\sigma }_m)$, and $A(u)=\exp (i\theta (N_u-N_s)+{\theta }^{2}q(u)\mathrm{/}2)$. The process $A$ is continuous on $[s,t]$, adapted there and bounded in modulus by $K=e^{{\theta }^{2}T\mathrm{/}2}$. In particular $A(s)=1$ and $A(u_{j+1})=A(u_j)e^{i\theta z_j+{\theta }^2{h}_j\mathrm{/}2}$.

To justify weighted clock convergence, first take a fixed deterministic grid $s=r_0<\cdots <r_l=t$ and bounded random coefficients ${\lambda }_a$. Assign coefficient ${\lambda }_a$ when the left endpoint $u_j$ lies in $[r_a,r_{a+1})$. For sufficiently fine partitions each cell crosses at most one of the finitely many distinct block boundaries. Inserting the boundaries changes each affected squared increment by at most twice the squared oscillation of $N$ on that cell, by $(a+b{)}^2-a^2-b^2=2ab$. Reassigning split increments to their blocks costs at most another constant times that squared oscillation. The total weighted error is at most $C_l{\max }_a\mathrm{\parallel }{\lambda }_a{\mathrm{\parallel }}_{\mathrm{\infty }}{\omega }_N(\mathrm{mesh}({\pi }_n){)}^2\to 0$ pathwise, where ${\omega }_N$ is the modulus of continuity on $[s,t]$. On the refined partition each block sum converges in probability to $q(r_{a+1})-q(r_a)$ by step 3.1. Finite addition and bounded multiplication therefore prove convergence of the weighted sums to ${\sum }_a{\lambda }_a(q(r_{a+1})-q(r_a))$. This argument does not require the coefficients to be independent of the increments.

Write $Q_n={\sum }_j{z}_j^2$, $X_j=N_{u_j}-N_s$, and $L_n={\sum }_j{X}_j{z}_j$. The identity $Q_n=(N_t-N_s{)}^2-2L_n$ is a finite telescope. For $j<k$, the factors $X_j{z}_j{X}_k$ are bounded and ${\mathcal{G}}_{u_k}$-measurable; testing the centered increment $z_k$ proves orthogonality. Similarly $E{z}_j{z}_k=0$. Therefore $E{Q}_n=E(N_t-N_s{)}^2\le 4m^2$ and $E{L}_n^2={\sum }_{j}E(X_j^2{z}_j^2)\le 4m^{2}E{Q}_n\le 16m^4$. Squaring the telescope with $(a+b{)}^2\le 2a^2+2b^2$ yields $E{Q}_n^2\le 160m^4$. The maximal increment $d_n={\max }_j\mathrm{\mid }{z}_j\mathrm{\mid }$ tends to zero pathwise and is at most $2m$, so $E{d}_n^2\to 0$. Cauchy--Schwarz gives $E(d_n{Q}_n)\to 0$. Since $0\le h_j\le \mathrm{mesh}({\pi }_n)$ and ${\sum }_j{h}_j\le T$, the four remainder sums of [F5] are bounded respectively by $d_n{Q}_n$, $T{d}_n$, $T\mathrm{mesh}({\pi }_n)$ and $\mathrm{mesh}({\pi }_n)Q_n$. Consequently $E\mathrm{\mid }{\sum }_{j}A(u_j)R(z_j,h_j)\mathrm{\mid }\to 0$.

Now take the left-endpoint staircase $A_l$ of the continuous process $A$ on deterministic grids with mesh tending to zero. Put $e_l={\sup }_{[s,t]}\mathrm{\mid }{A}_l-A\mathrm{\mid }$, with the endpoint $t$ assigned $A(t)$. Then $e_l\to 0$ pathwise and $e_l\le 2K$. The error in the weighted square sums is at most $e_l{Q}_n$ and the error in their proposed limits is at most $T{e}_l$. Explicitly, for $R>0$, $P(e_l{Q}_n>\epsilon )\le P(Q_n>R)+P(e_l>\epsilon \mathrm{/}R)\le 4m^2\mathrm{/}R+P(e_l>\epsilon \mathrm{/}R)$. Choose $R$ first, then $l$, then $n$ for the fixed staircase convergence of step 4.2. This proves ${\sum }_{j}A(u_j)z_j^2\to {\int }_s^{t}A(u)1_{\{u<{\sigma }_m\}}du$ in probability. The integral is the ordinary pathwise integral up to $t\wedge {\sigma }_m$, zero when ${\sigma }_m\le s$; the sums ${\sum }_{j}A(u_j)h_j$ converge to it pathwise, their error being at most $T{\omega }_A(\mathrm{mesh}({\pi }_n))$. Thus $S_n={\sum }_{j}A(u_j)(h_j-z_j^2)\to 0$ in probability. Since $\mathrm{\mid }{S}_n\mathrm{\mid }\le K(T+Q_n)$, its second moments are uniformly bounded. For each $\epsilon >0$, Cauchy--Schwarz gives $E\mathrm{\mid }{S}_n\mathrm{\mid }\le \epsilon +(E\mathrm{\mid }{S}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}P(\mathrm{\mid }{S}_n\mathrm{\mid }>\epsilon {)}^{1\mathrm{/}2}$. Taking $n\to \mathrm{\infty }$, then $\epsilon \downarrow 0$, gives $E\mathrm{\mid }{S}_n\mathrm{\mid }\to 0$.

Let $Y$ be any bounded ${\mathcal{G}}_s$-measurable real or complex variable. For each $j$, the factor $YA(u_j)$ is bounded and ${\mathcal{G}}_{u_j}$-measurable, so $E[YA(u_j)z_j]=0$. Telescope the identity in step 4.1 and apply [F5]. The sum of the linear terms has zero expectation; the expectation of the compensator term tends to zero by step 6.1 and that of the remainders by step 5.1. The left side does not depend on the partition, so $E[Y\exp (i\theta (N_t-N_s)+{\theta }^2((t\wedge {\sigma }_m)-(s\wedge {\sigma }_m))\mathrm{/}2)]=E[Y]\mathrm{.}$

Let $m\to \mathrm{\infty }$ in step 7.1. The stopped increments and clocks converge almost surely to ${\tilde{M}}_t-{\tilde{M}}_s$ and $t-s$, while the exponential modulus is at most $e^{{\theta }^2(t-s)\mathrm{/}2}$. Dominated convergence proves the conditional increment identity for $\tilde{M}$ and ${\mathcal{G}}_s$. In particular it holds for indicators of original ${\mathcal{F}}_s$ events. Since $M=\tilde{M}$ off $D$, and the original fixed-time values are ${\mathcal{F}}_t$-measurable, these event tests establish exactly clause 1 for the original process and filtration.

The original $Z_t$ is ${\mathcal{F}}_t$-measurable with deterministic modulus $e^{{\theta }^{2}t\mathrm{/}2}$. Using the bounded known factor $Z_s$ in clause 1 gives $E[Z_t\mid {\mathcal{F}}_s]=Z_s$. Real and imaginary parts give clause 2. For $s=t$, including $s=t=0$, the increment exponential is $1$; for $\theta =0$, $Z\equiv 1$. The unit clock is the stated normalization; the identically zero process is excluded by the positive-time clock hypothesis. No converse is asserted and no general stochastic integral is introduced. AC has the uses in [F7] and step 1.1.