Let $(N_t)_{t\ge 0}$ be a poisson process with intensity $\alpha > 0$. Let $(X_n)_{n \in \mathbb N}$ be iid real valued random variables that are independent of $N_t$. Let $Y_t = \sum^{N_t}_{k=1}X_k$, a compound poisson process. I am trying to derive the PGF of $Y_t$ in terms of the PGF of $X_t$ when $X_1$ has density: $\rho(n) = \dfrac{-p^n}{n\ln(1-p)}, n \in \mathbb N$
Attempt: Let the PGF of $Y_t$ be denoted by $\phi_{Y_t}$, then: \begin{align*} \phi_Y(s) &= E[s^{Y}]\\ &=\sum^{\infty}_{i=1}s^iP(Y_t = i)\\ &=\sum^{\infty}_{i=1} \sum^{\infty}_{j=1}s^iP(Y_t = i \mid N_t=j)P(N_t=j)\\ &=\sum^{\infty}_{j=1} P(N_t=j) \sum^{\infty}_{i=1} s^iP(Y_t = i \mid N_t=j)\\ &=\sum^{\infty}_{j=1} P(N_t=j) \sum^{\infty}_{i=1} s^iP(X_1+\cdots+X_j=i)\\ &=\sum^{\infty}_{j=1} P(N_t=j) (\phi_x(s))^j\\ &=\phi_N(\phi_X(s)) \end{align*} Thus, $$ \phi_Y(s)=e^{\alpha t(\phi_{X_1}(s)-1)} $$
Now, I've worked out that $\phi_{X_1} = \frac{\ln(1-ps)}{\ln(1-p)}$, therefore:
$$ \phi_Y(s)=e^{\alpha t\left(\frac{\ln(1-ps)}{\ln(1-p)}-1\right)} $$
Could anyone verify this result?