Martingale and predictable bracket for a finite-state chain

Fix $t>0$ and a bounded function $a:E\to \mathbb{R}$. On $[0,t]$, approximate $a(X_u)$ by the process with value $a(X_{kt\mathrm{/}{2}^n})$ on $((k-1)t\mathrm{/}{2}^n,kt\mathrm{/}{2}^n]$, $1\le k\le 2^n$, and value $a(X_0)$ at zero. Each approximation is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable by adaptedness of $X$ in [F1], and converges pointwise by right continuity. The construction uses constant paths on its exceptional explosion event, so this argument applies there too. Its time integral is the ${\mathcal{F}}_t$-measurable sum $(t\mathrm{/}{2}^n){\sum }_{k=1}^{2^n}a(X_{kt\mathrm{/}{2}^n})$. These sums converge to ${\int }_0^{t}a(X_u)du$: on a path with $J$ jumps, the error is at most $2\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}Jt\mathrm{/}{2}^n$, since only intervals containing a jump contribute. Thus the time integral is ${\mathcal{F}}_t$-measurable, without completing the filtration. Taking $a=g$ proves that $M$ is adapted, also at $t=0$. By [F1], $\mathbb{E}[g(X_{s+r})\mid {\mathcal{F}}_s]=(e^{rL}g)(X_s)$. The joint measurability just proved and boundedness allow [F5] on finite time intervals. Testing against each event in ${\mathcal{F}}_s$ gives $\mathbb{E}\text{\hspace{0.17em}⁣}[{\int }_s^{s+h}g(X_u)du|{\mathcal{F}}_s]={\int }_0^h(e^{rL}Lf)(X_s)dr=(e^{hL}f-f)(X_s),$ where the last equality follows by termwise differentiation of the absolutely convergent finite-matrix exponential series. Also $\mathbb{E}[f(X_{s+h})\mid {\mathcal{F}}_s]=(e^{hL}f)(X_s)$. Subtracting and using adaptedness proves $\mathbb{E}[M_{s+h}\mid {\mathcal{F}}_s]=M_s$ for every $s,h$. On $[0,T]$, $\mathrm{\mid }{M}_t\mathrm{\mid }\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+T\mathrm{\parallel }Lf{\mathrm{\parallel }}_{\mathrm{\infty }}$, so every $M_t$ is square-integrable and [F2] gives the all-pairs martingale assertion.

Since ${\sum }_j{L}_{ij}=0$, ${\mathrm{\Gamma }}_f(i)={\sum }_j{L}_{ij}(f(j)-f(i){)}^2={\sum }_{j\mathrm{\ne }i}{L}_{ij}(f(j)-f(i){)}^2\ge 0.$ Thus $A$ is increasing and bounded on $[0,T]$. The integral argument of step 1.1 with $a={\mathrm{\Gamma }}_f$ proves that $A$ is adapted. Also $\mathrm{\mid }{A}_t-A_s\mathrm{\mid }\le \mathrm{\parallel }{\mathrm{\Gamma }}_f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }t-s\mathrm{\mid }$, so $A$ is continuous. For $T>0$ and each $n$, the step process equal to $A_{kT\mathrm{/}{2}^n}$ on $(kT\mathrm{/}{2}^n,(k+1)T\mathrm{/}{2}^n]$ and zero at time zero is predictable by [F4]; these processes converge pointwise to $A$ by continuity. Therefore $A$ is predictable; on the horizon $T=0$ the assertion is immediate from $A_0=0$.

Fix $u$ and condition on ${\mathcal{F}}_u$, writing $X_u=i$. For $0<h\le 1$, the number of uniformization events during $(u,u+h]$ is Poisson with rate $q={\max }_i{\lambda }_i$ by the exponential-clock construction in [F1]. If there is no event, $\mathrm{\Delta }M:=M_{u+h}-M_u=-hg(i)$, whose square is $O(h^2)$. If there is exactly one event and the new state is $j\mathrm{\ne }i$, its probability is $e^{-qh}qh{R}_{ij}=e^{-qh}h{L}_{ij}=h{L}_{ij}+O(h^2)$, and $\mathrm{\Delta }M=f(j)-f(i)+O(h)$ uniformly because the drift integral has magnitude at most $h\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Its contribution to the conditional second moment is therefore $h{L}_{ij}(f(j)-f(i){)}^2+O(h^2)$. A one-event self-transition contributes $O(h^3)$. The probability of at least two events is $O(h^2)$ and $\mathrm{\Delta }M$ is uniformly bounded for $h\le 1$, so its contribution is $O(h^2)$. Since $E$ is finite, all error bounds are uniform in $i$, and $\mathbb{E}[(\mathrm{\Delta }M{)}^2\mid {\mathcal{F}}_u]=h{\mathrm{\Gamma }}_f(i)+O(h^2)\mathrm{.}$ Also $\mathbb{P}(X_{u+r}\mathrm{\ne }i\mid {\mathcal{F}}_u)\le qr$, so boundedness of ${\mathrm{\Gamma }}_f$ gives $\mathbb{E}\text{\hspace{0.17em}⁣}[{\int }_u^{u+h}{\mathrm{\Gamma }}_f(X_v)dv|{\mathcal{F}}_u]=h{\mathrm{\Gamma }}_f(i)+O(h^2)\mathrm{.}$ Consequently $\mid \mathbb{E}[(\mathrm{\Delta }M{)}^2-{\int }_u^{u+h}{\mathrm{\Gamma }}_f(X_v)dv\mid {\mathcal{F}}_u]\mid \le C{h}^2$ for a finite constant $C$ depending only on $L,f,T$.

Let $D_t=M_t^2-A_t$. On any interval $[u,u+h]$, $D_{u+h}-D_u=2M_u\mathrm{\Delta }M+(\mathrm{\Delta }M{)}^2-{\int }_u^{u+h}{\mathrm{\Gamma }}_f(X_v)dv$. On $[0,T]$, $M_u$ is bounded and ${\mathcal{F}}_u$-measurable by step 1.1, so [F7] and the mean-zero increment give $\mathbb{E}[2M_u\mathrm{\Delta }M\mid {\mathcal{F}}_u]=2M_u\mathbb{E}[\mathrm{\Delta }M\mid {\mathcal{F}}_u]=0$, while step 3.1 bounds the conditional expectation of the remaining terms by $C{h}^2$. For fixed $s<t$, divide $[s,t]$ into $n$ equal intervals of length $h=(t-s)\mathrm{/}n\le 1$. For the subdivision times $u$, ${\mathcal{F}}_s\subseteq {\mathcal{F}}_u$ and every increment of $D$ is integrable. Thus [F8], linearity and order in [F3], and the preceding bound imply $\mathrm{\mid }\mathbb{E}[D_t-D_s\mid {\mathcal{F}}_s]\mathrm{\mid }\le nC{h}^2=C(t-s{)}^2\mathrm{/}n$. Letting $n\to \mathrm{\infty }$ proves $\mathbb{E}[D_t\mid {\mathcal{F}}_s]=D_s$. Thus $M^2-A$ is a martingale.

In this item, a continuous predictable bracket means a continuous predictable increasing process $B$ with $B_0=0$, $\mathbb{E}{B}_T<\mathrm{\infty }$, and $M^2-B$ a martingale. Steps 2.1 and 4.1 show that $A$ has these properties. If $B$ is another such process, then $B-A$ is a continuous martingale starting at zero, and its total variation on $[0,T]$ is at most $B_T+A_T$, which is integrable. By [F6], $B-A$ is indistinguishably zero. Therefore $A$ is the unique bracket in this stated continuous predictable compensator class, and it has the displayed formula.