 Solution

Conditional on $Z=k$, the summands are independent with the common Laplace transform of a nonnegative random variable $\varphi (q)$. The empty sum for $k=0$ is zero. Therefore the compound Poisson distribution hasand applying the probability generating function from part (a) givesIf $m=\mathbb{E}{X}_1<\infty$, differentiating at zero from the right gives ${\varphi }^{\prime }(0+)=-m$ and $L_Y^{\prime }(0+)=-\lambda m$. HenceThe differentiation is justified by $(1-e^{-q{X}_1})/q\le X_1$ and the dominated convergence theorem. No finite moment was assumed in the question, so the formula must also be interpreted when $m=\infty$. For any nonnegative random variable $V$, the monotone convergence theorem givesApply this to $X_1$ and $Y$, and use $1-e^{-\lambda h}\sim \lambda h$ as $h\downarrow 0$, with $h=1-\varphi (q)$; it yields the same formula with value $+\infty$ when $m=\infty$. If $m=0$, nonnegativity makes all marks zero almost surely and the result is immediate.

For the second method, condition on the count rather than differentiating a transform. For finite $m$, the tower property of conditional expectation gives $\mathbb{E}[Y\mid Z]=Zm$, and $\mathbb{E}Z=\lambda$ gives $\mathbb{E}Y=\lambda m$. More generally, Tonelli's theorem and independence giveFor $m=\infty$, already the term $j=1$ is infinite since $\mathbb{P}(Z\ge 1)>0$. Thus the two methods agree, including in the extended-valued case.