= Solution
Conditional on $Z=k$, the summands are <independent> with the common <Laplace transform of a nonnegative random variable> $\phi(q)$. The empty sum for $k=0$ is zero. Therefore the <compound Poisson distribution> has
$$
\mathbb E[e^{-qY}\mid Z=k]=\phi(q)^k,
$$
and applying the <probability generating function> from part (a) gives
$$
\boxed{L_Y(q)=\mathbb Ee^{-qY}=\exp\{\lambda[\phi(q)-1]\}\qquad(q\geq0).}
$$
If $m=\mathbb EX_1<\infty$, differentiating at zero from the right gives $\phi'(0+)=-m$ and $L_Y'(0+)=-\lambda m$. Hence
$$
\boxed{\mathbb EY=\lambda\mathbb EX_1.}
$$
The differentiation is justified by $(1-e^{-qX_1})/q\leq X_1$ 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> gives
$$
\mathbb EV=\lim_{q\downarrow0}\frac{1-\mathbb Ee^{-qV}}q\in[0,\infty].
$$
Apply this to $X_1$ and $Y$, and use $1-e^{-\lambda h}\sim\lambda h$ as $h\downarrow0$, with $h=1-\phi(q)$; 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 $\mathbb EZ=\lambda$ gives $\mathbb EY=\lambda m$. More generally, <Tonelli's theorem> and <independence> give
$$
\mathbb EY=\sum_{j=1}^\infty\mathbb E[X_j\mathbf1_{\{Z\geq j\}}]
=m\sum_{j=1}^\infty\mathbb P(Z\geq j)=m\mathbb EZ.
$$
For $m=\infty$, already the term $j=1$ is infinite since $\mathbb P(Z\geq1)>0$. Thus the two methods agree, including in the extended-valued case.
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