 Solution

The PDF defines multiplication by an indicator: $X_t=U{1}_{\{V\ge e^{-t}\}}$. PutThen $T>0$ almost surely, $\{T\le t\}=\{X_t>0\}\in {\mathcal{F}}_t$, and $T$ is a stopping time. It has exponential distribution of rate one and is independent of $U$. The path stays at zero until $T$, jumps to $U$ at $T$, and remains there.

Let $p=\mathbb{P}(U\in B)$ and $N_t=1_{\{T\le t,U\in B\}}$. We verify the martingale property directly. On $\{T\le s\}$ both increments $N_t-N_s$ and $(T\wedge t)-(T\wedge s)$ vanish. On $\{T>s\}$, every observation up to $s$ is zero, so the past reveals only that survival event. Conditional on it, $R=T-s$ is rate-one exponential by memorylessness and remains independent of the mark $U$. With $h=t-s$,whileSubtracting $p$ times the second identity from the first gives zero. The process is adapted and integrable, since its absolute value is bounded by $1+pt$ on a finite horizon. ThereforeOn $\{T\le t\}$ we have $X_t=U$, so this is exactly the requested process. This first-principles argument uses the natural filtration, which does not disclose the future mark or lifetime. It describes a single marked exponential jump process.