SSEP martingale for time-dependent test functions

For fixed $s$, reindexing the bond sum as in the model generator gives $L_N{F}_s^N(\eta )=⟨{\pi }^N(\eta ),{\mathrm{\Delta }}_{N}H(s,\cdot )⟩$; also ${\mathrm{\partial }}_s{F}_s^N(\eta )=⟨{\pi }^N(\eta ),{\mathrm{\partial }}_{s}H(s,\cdot )⟩$. Hence the compensating drift in the Statement is exactly $({\mathrm{\partial }}_s{F}_s^N+L_N{F}_s^N)({\eta }_s)$.

Fix $N$ and a state $i={\eta }_s$. By [F2], on an interval of length $h$, two or more potential uniformization events have probability $O_N(h^2)$, one actual transition $i\to j$ has probability $h{L}_N(i,j)+O_N(h^2)$, and no state change has probability $1-h{\lambda }_i+O_N(h^2)$, with ${\lambda }_i={\sum }_{j\mathrm{\ne }i}{L}_N(i,j)$. Taylor expansion is uniform over the finite state set and $s\in [0,T]$. If no state change occurs, then $F_{s+h}^N(i)-F_s^N(i)=h{\mathrm{\partial }}_s{F}_s^N(i)+O_N(h^2)$ and the compensating integral is $h({\mathrm{\partial }}_s{F}_s^N+L_N{F}_s^N)(i)+O_N(h^2)$, so the increment of $M^N$ is $-h{L}_N{F}_s^N(i)+O_N(h^2)$. A single transition $i\to j$ gives increment $F_s^N(j)-F_s^N(i)+O_N(h)$; summing its conditional mean over $j$ gives $h{L}_N{F}_s^N(i)+O_N(h^2)$, which cancels the no-transition term. Paths with at least two potential events contribute $O_N(h^2)$ because $M^N$ has bounded increments for fixed $N,T$. Therefore $\mathrm{\mid }\mathbb{E}[M_{s+h}^N-M_s^N\mid {\mathcal{F}}_s]\mathrm{\mid }\le C_N{h}^2$. In the conditional second moment, no transition contributes $O_N(h^2)$, one transition contributes $h{\sum }_{j\mathrm{\ne }i}{L}_N(i,j)(F_s^N(j)-F_s^N(i){)}^2+O_N(h^2)=h{\mathrm{\Gamma }}_{F_s^N}(i)+O_N(h^2)$, and two or more potential events contribute $O_N(h^2)$.

If $T=0$, the only time is $0$ and $M_0^N=0$. For $T>0$, fix $s<t$ and partition $[s,t]$ into $n$ equal intervals of length $h=(t-s)\mathrm{/}n$. Applying the conditional-mean estimate from step 2.1 on every subinterval, then the tower property, bounds $\mathrm{\mid }\mathbb{E}[M_t^N-M_s^N\mid {\mathcal{F}}_s]\mathrm{\mid }$ by $n{C}_N{h}^2\to 0$. Thus $M^N$ is an all-pairs martingale. For fixed $N$ its paths are bounded on $[0,T]$ because the state space is finite, $H$ and its time derivative are bounded, and the compensating integrand is bounded; hence it is square-integrable.

Set ${\gamma }_s(\eta )={\mathrm{\Gamma }}_{F_s^N}(\eta )$ and $A_t^N={\int }_0^t{\gamma }_s({\eta }_s)ds$. If $T=0$, both processes are zero. For $T>0$, the second-moment expansion in step 2.1 and continuity of ${\gamma }_s(i)$ in $s$ imply $\mathbb{E}[A_{s+h}^N-A_s^N\mid {\mathcal{F}}_s]=h{\gamma }_s({\eta }_s)+O_N(h^2)$: without a state change the integrand differs from ${\gamma }_s({\eta }_s)$ by $O_N(h)$, and a state change has probability $O_N(h)$ and changes the bounded integrand over an interval of length $h$. Hence $\mathbb{E}[(M_{s+h}^N-M_s^N{)}^2-(A_{s+h}^N-A_s^N)\mid {\mathcal{F}}_s]=O_N(h^2)$. For fixed $s<t$, partition $[s,t]$ into $n$ equal intervals and write ${\mathrm{\Delta }}_{j}M=M_{t_{j+1}}^N-M_{t_j}^N$, ${\mathrm{\Delta }}_{j}A=A_{t_{j+1}}^N-A_{t_j}^N$. Since $M^N$ is a martingale, $\mathbb{E}[M_{t_j}^N{\mathrm{\Delta }}_{j}M\mid {\mathcal{F}}_s]=0$ by the tower property. Expanding the square, conditioning each remainder ${\mathrm{\Delta }}_j{M}^2-{\mathrm{\Delta }}_{j}A$ first on ${\mathcal{F}}_{t_j}$, and using the bound just proved gives an absolute conditional expectation at most $n{C}_N{h}^2\to 0$. Therefore $(M_t^N{)}^2-A_t^N$ is an all-pairs martingale.

The integrand ${\gamma }_s({\eta }_s)$ is nonnegative, so $A^N$ is increasing; it is continuous and adapted. If $T=0$, the zero process is predictable. Otherwise, for each nested deterministic partition, the process equal to $A^N$ at the left endpoint on each interval $(a,b]$ and zero at time zero is predictable by [F4]; these processes converge pointwise to $A^N$ by continuity. Thus $A^N$ is predictable. At a bond $\{x,x+e_i\}$, the test-function jump is $N^{-d}({\eta }_{x+e_i}-{\eta }_x)(H(s,x\mathrm{/}N)-H(s,(x+e_i)\mathrm{/}N))$. Multiplying its square by the rate $N^2$ and summing over all bonds yields the displayed formula for ${\gamma }_s({\eta }_s)$. The fundamental theorem of calculus gives $\mathrm{\mid }H(s,(x+e_i)\mathrm{/}N)-H(s,x\mathrm{/}N)\mathrm{\mid }\le N^{-1}\mathrm{\parallel }{\mathrm{\partial }}_{i}H(s,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}$; using $({\eta }_{x+e_i}-{\eta }_x{)}^2\le 1$ and $N^d$ bonds per coordinate gives the stated bound for $A_T^N$.