Counting-process compensators, integrals, brackets — Examples

The process $N$ is adapted, càdlàg, has one unit jump, and starts at zero. On dyadic time grids on $[0,\mathrm{\infty })$, left-step processes with value $N_{t_j}$ on $(t_j,t_{j+1}]$ are predictable by [F1] and converge pointwise to $N_{t-}$ for each $t>0$, so the left-limit process on $(0,\mathrm{\infty })$ is predictable. Since $t\mapsto {\mathbf{1}}_{\{t>L\}}$ is deterministic Borel, the displayed formula makes $Y$ predictable on $(0,\mathrm{\infty })$; its separately assigned value $Y_0=0$ is predictable on the time-zero slice. This also gives $Y_L=0$ and $Y_U=1$.

Direct integration gives $A_t=\lambda ((t-L{)}^{+}\wedge W)$, so $A$ is continuous and increasing, $A_0=0$, and $A_t\le \lambda T$. Left Riemann sums make $A_t$ ${\mathcal{F}}_t$-measurable; its continuous adapted left-step approximations are predictable by [F1]. The exponential density gives $\mathbb{E}{A}_T=\lambda ({\int }_0^{T-L}u\lambda e^{-\lambda u}du+(T-L)e^{-\lambda (T-L)})=1-e^{-\lambda (T-L)}<\mathrm{\infty }$, and for every $t>L$ the same calculation with $T$ replaced by $t$ gives $\mathbb{E}{A}_t=1-e^{-\lambda (t-L)}=\mathbb{P}(U\le t)=\mathbb{E}{N}_t$. Thus the compensator integrability requirements in [F2] hold.

If $s<L$ and $t\le L$, then $N$ and $A$ are zero on $[s,t]$; if $s<L<t$, then ${\mathcal{F}}_s$ is trivial and step 1.2 gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=0=M_s$. If $s\ge L$ and $U\le s$, the martingale increment is zero. On $\{U>s\}$ the event-free history gives $W>s-L$, and the residual survival ratio is $\mathbb{P}(W-(s-L)>u\mid W>s-L)=e^{-\lambda u}$. With $h=t-s$, the conditional expected event increment is $1-e^{-\lambda h}$, and the same residual density gives $\lambda \mathbb{E}[\min (h,W-(s-L))\mid W>s-L]=\lambda ({\int }_0^{h}u\lambda e^{-\lambda u}du+h{e}^{-\lambda h})=1-e^{-\lambda h}$. Thus every conditional increment has mean zero. Finally $\mathrm{\mid }{M}_t\mathrm{\mid }\le 1+\lambda T$, so $M$ is square-integrable and satisfies [F2].

Each $N^i$ has one unit jump and $Y_0^i=1$. For $t>0$, $Y_t^i=1-(N^i{)}_{t-}$ is predictable by dyadic left-step approximation using [F4]; its time-zero value is predictable on the time-zero generator. Direct integration gives $A_t^i=\lambda (t\wedge W)$; left Riemann sums make it adapted, and continuous left-step approximations make it predictable by [F4]. The exponential density gives $\mathbb{E}{A}_T^i=\lambda ({\int }_0^{T}u\lambda e^{-\lambda u}du+T{e}^{-\lambda T})=1-e^{-\lambda T}<\mathrm{\infty }$. If $W\le s$, the martingale increment is zero; if $W>s$, the residual waiting time $R=W-s$ is exponential and, for $h=t-s$, $\mathbb{E}[N_t-N_s\mid {\mathcal{F}}_s]=1-e^{-\lambda h}=\mathbb{E}[A_t-A_s\mid {\mathcal{F}}_s]$ on that event, since $\lambda ({\int }_0^{h}u\lambda e^{-\lambda u}du+h{e}^{-\lambda h})=1-e^{-\lambda h}$. Hence each $M^i$ is an all-pairs martingale; $\mathrm{\mid }{M}_t^i\mathrm{\mid }\le 1+\lambda T$ makes it square-integrable and verifies [F1].

The two component martingales are identical, so [F3] gives $[M^1,M^2]=[M,M]$. By [F2], $[M,M]=N$ and $⟨M,M⟩=A$, hence the pair has the same cross-bracket values. For every $t>0$, $\mathbb{P}(N_t=1)=1-e^{-\lambda t}>0$ and $A_t=\lambda (t\wedge W)>0$ almost surely because $W>0$ almost surely; neither cross-bracket process is identically zero when simultaneous jumps are allowed.

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 $A$.