= Solution
Write $Z_t=\sum_{j=1}^{N_t}Y_j$, with an empty sum equal to zero. Conditioning on the independent <Poisson random variable> $N_t$, the <moment-generating function> is
$$
\mathbb E e^{\theta Z_t}=\sum_{n=0}^\infty e^{-\lambda t}\frac{(\lambda t)^n}{n!}M_Y(\theta)^n
=\boxed{\exp\{\lambda t(M_Y(\theta)-1)\}}.
$$
This is the <compound Poisson distribution>. Differentiating its log moment-generating <function> at zero, when the first two moments are finite, gives
$$
\boxed{\mathbb E Z_t=\lambda t\mathbb EY,\qquad\operatorname{Var}(Z_t)=\lambda t\mathbb E(Y^2)}.
$$
The <variance> includes variability in the number of coaches, and so uses the second moment, not merely $\operatorname{Var}(Y)$.
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