Can someone please explain how to find the Laplace transform of $Z_t$ in the following question? I am a bit confused about how to start.
Electrical pulses with iid amplitudes $X_1, X_2,...$ arrive at a detector at random times $S_1, S_2, S_3,...$ according to a Poisson process with rate $\lambda$. The detector output for the kth pulse at time t is: $$\theta_k(t) = \begin{cases} 0 &t < S_k \\ X_k\exp(-\alpha(t-S_k)) &t \geq S_k \end{cases}$$ This means that the amplitude measured by the detector start with $X_k$ and then decreases exponentially at rate $\alpha$. The total amplitude measured by the detector is then: $$Z_t = \sum_{j=1}^N\theta_k(t)$$
Let $F_X$ be the distribution function of $X_1$ and $N(t) = \sup\{k>0: S_k\leqslant t\}$ for $t\geqslant 0$. Let $S_0=0$. Then the distribution of the interarrival times $S_j-S_{j-1}$ is exponential with parameter $\lambda$ so the distribution of the arrival times $S_k = \sum_{j=1}^k (S_j-S_{j-1})$ is Erlang with parameters $\lambda$, $k$ - that is, $F_k(t):=\mathbb P(S_k\leqslant t) = \sum_{n=k}^\infty \frac{(\lambda t)^n}{n!}e^{-\lambda t}$. Now, for each $t\geqslant 0$ and nonnegative integer $k$, we have the equivalence $\{N(t) = k\}=\{S_k\leqslant t<S_{k+1}\}$. It follows that \begin{align} \mathbb P(N(t) = n) &= F_n(t) - F_{n+1}(t)\\ &= \sum_{n=k}^\infty \frac{(\lambda t)^n}{n!}e^{-\lambda t} - \sum_{n=k+1}^\infty \frac{(\lambda t)^n}{n!}e^{-\lambda t}\\ &= \frac{(\lambda t)^n}{n!}e^{-\lambda t}, \end{align} that is, $N(t)$ has Poisson distribution with mean $\lambda t$.
We now compute the Laplace transform of $Z_t$: \begin{align} \mathcal L\{Z_t\} &= \mathbb E[e^{-sZ_t}]\\ &= \sum_{k=0}^\infty \mathbb E\left[e^{-sZ_t}\mid N(t)=k\right]\mathbb P(N(t)=k)\\ &= \sum_{k=0}^\infty \mathbb E\left[e^{-s \sum_{j=1}^k X_je^{-\alpha(t-S_j)}}\mid N(t)=k\right] \frac{(\lambda t)^k}{k!}e^{-\lambda t}\\ &=\sum_{k=0}^\infty \frac{(\lambda t)^k}{k!}e^{-\lambda t}\prod_{j=1}^k \mathbb E\left[ e^{-sX_j e^{-\alpha(t-S_j)} } \right] \\ \end{align} From here I am unsure how to proceed, since the distribution of $X_1$ is not given.