I'm trying to prove an equality about the generating function of a compoung Poisson process and I don't know how to continue it.
I'm using this definition:
A process $\{X_t, t\geq 0 \}$ is a compound Poisson process if we can write it as $$X_t=\sum_{i=1}^{N_t}Y_i$$
where $\{N_t,t\geq 0 \}$ is a Poisson process and $\{Y_n\}_{n\geq 1}$ is a family of random independent v. with the same distribution wich are independent from $\{N_t,t\geq 0 \}$ as well.
I'm proving that
$m_{X_t}(u)=\exp(\lambda t (m_{Y_1}(u)-1))$
And here it is my aproche:
\begin{align}
m_{X_t}(u) & = E(e^{uX_t}) = E(\exp(u\sum_{i=1}^{N_t}Y_i)) \\
& & \\
& = \sum_{m=1}^\infty E\left(\exp \left(u\sum Y_i \right) \right) \cdot P(N_t=m)\\
& \\
& = \sum_{m=1}^\infty E\left(\exp \left(u\sum Y_i \right) \right) \cdot \dfrac{(\lambda t)^m e^{-\lambda t}}{m!} \\
& \\
& = \sum_{m=1}^\infty E\left( e^{uY_1} \right)^m \cdot \dfrac{(\lambda t)^m e^{-\lambda t}}{m!}\\
& \\
& = \sum_{m=1}^\infty m_{Y_1}(u)^m \cdot \dfrac{(\lambda t)^m e^{-\lambda t}}{m!}\\
& \\
& = \sum_{m=1}^\infty e^{m\ln(m_{Y_1}(u))} \cdot \dfrac{(\lambda t)^m e^{-\lambda t}}{m!}
\end{align}
I know I have to be very close of the solution but my brain don't work. Maybe a silly thing is missing. Can you help me please?
Best Answer
$$\sum_{m=0}^\infty m_{Y_1}(u)^m \cdot \frac{(\lambda t)^m \mathrm e^{-\lambda t}}{m!}=\sum_{m=0}^\infty \frac{(m_{Y_1}(u)\lambda t)^m}{m!}\mathrm e^{-\lambda t}=\mathrm e^{m_{Y_1}(u)\lambda t}\mathrm e^{-\lambda t}$$ (Watch out for the summations, which should start at $m=0$, not $m=1$.)