One exponential event: compensator and isometry

The process $N$ has one unit jump and $Y_0=1$. For $t>0$, $Y_t=1-N_{t-}$ is predictable by dyadic left-step approximation, while its time-zero value is predictable on the time-zero generator. Therefore $A_t=\lambda (t\wedge W)$ is continuous, increasing, starts at zero, and is adapted by left Riemann sums; continuous adapted left-step approximations make it predictable by [F1]. The exponential density gives $\mathbb{E}{A}_T=\lambda ({\int }_0^{T}u\lambda e^{-\lambda u}du+T{e}^{-\lambda T})=1-e^{-\lambda T}<\mathrm{\infty }$, verifying the compensator hypotheses in [F2].

Fix $0\le s<t\le T$. On $\{W\le s\}$ both $N$ and $A$ have stopped changing, so $M_t-M_s=0$. On $\{W>s\}$, $\mathbb{P}(W-s>u\mid W>s)=\mathbb{P}(W>s+u)\mathrm{/}\mathbb{P}(W>s)=e^{-\lambda u}$; with $h=t-s$, the conditional event increment is ${\mathbf{1}}_{\{W>s\}}(1-e^{-\lambda h})$, and the residual exponential density gives the same compensator increment ${\mathbf{1}}_{\{W>s\}}\lambda ({\int }_0^{h}u\lambda e^{-\lambda u}du+h{e}^{-\lambda h})={\mathbf{1}}_{\{W>s\}}(1-e^{-\lambda h})$. Thus $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for every pair, and $\mathrm{\mid }{M}_t\mathrm{\mid }\le 1+\lambda T$ makes $M$ square-integrable and verifies [F2].

The process $H=1$ on $(0,T]$ and $H_0=0$ is a bounded predictable step integrand with elementary integral $M_T-M_0=M_T$. By [F3], $\mathbb{E}{M}_T^2=\mathbb{E}{A}_T=1-e^{-\lambda T}$; by [F4], its optional and predictable brackets are $N$ and $A$.