Delayed entry with a predictable risk process

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].