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As in the intended independent-mark construction, take the entire sequence $(X_j)$ independent of the Poisson process, not merely independent of one fixed count. Then $Y$ is a Compound Poisson process. For deterministic $s<t$, its increment $Y_t-Y_s$ is independent of ${\mathcal{F}}_s=\sigma (Y_u:0\le u\le s)$. To justify this carefully, first condition on $N_s=k$. The past of $Y$ uses only the counting history up to $s$ and marks among $X_1,\dots ,X_k$. The future count has independent rate-$\lambda$ Poisson process increments, and future marks $X_{k+1},X_{k+2},\dots$ are fresh with the common law. Their joint conditional law is the same for every $k$ and every past history, so the increment is independent even of this larger marked history, hence of ${\mathcal{F}}_s$. Marks equal to zero do not cause a problem: ${\mathcal{F}}_s$ may then reveal less than the full counting history.

By part (b), with Poisson parameter $\lambda (t-s)$,The proposed process is adapted to ${\mathcal{F}}_t$, nonnegative, andIn particular, no finite expectation of a mark is required. Also $Z_s>0$, andis independent of ${\mathcal{F}}_s$ and has mean one. Therefore the exponential Laplace martingale of a compound Poisson process satisfiesThis proves every condition for a continuous-time martingale in the requested filtration. When $q=0$, $\varphi (0)=1$ and the same conclusion reduces to the constant martingale $Z_t=1$.