A tied jump produces a nonzero cross bracket

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.