 Solution

For $0\le s\le 1$, condition on $N=n$. Its independence from the $X_i$ preserves their independent common distribution, so part (ii) givesFor $n=0$, this is $1$, agreeing with $Y=0$. By the law of total expectation,This proves the probability generating function of an independent random sum. For nonnegative $s$ outside $[0,1]$, the same argument is valid whenever the right side is finite, by Tonelli's theorem for nonnegative sums. For signed or complex $s$, a sufficient absolute-convergence condition is $G_N(G_X(|s|))<\infty$; it holds automatically on $|s|\le 1$. Outside this domain, mere evaluation of an analytically extended expression is not a substitute for existence of the probability generating function.