Characteristic function of a poisson random variable Let $u_n$ be a poisson process and $X_n$ be a sequence of iid rv's independent from $u_n$.
Define the process $Y_t = \sum^{u}_{i=1} Xi$.
Then let $\phi_{Xn}(t)$ be the characteristic function of $X_n$, show:
$$
\phi_{Y_t}(t) = E[e^{itY_t}] = e^{t\lambda(\phi_{X_n}(t) -1)}
$$
What I tried to do is that:
$$
E[e^{it\sum^{u_n}Xi}] = E[E(e^{it\sum^{u_n}Xi}|u_n)] = E[E(e^{itX_i})]^{u_n}
$$
From that I don't know what to do
 A: $Y_t = \displaystyle\sum_{i=1}^{N_t}X_i$ is a compound Poisson process, i.e. a sum of iid random variables $\{X_i\}_i$ whose number of terms is itself a Poisson process $N_t$ (of intensity $\lambda$). Its characteristic function will be then given by
$$
\begin{array}{rclll}
\phi_{Y_t}(\omega) 
   &=& \displaystyle
   \mathbb{E}\left[e^{i\omega Y_t}\right] \\
   &=& \displaystyle
   \mathbb{E}\left[\left.\exp\left(i\omega \sum_{i=1}^{N_t}X_i\right)\right|N_t\right] \\
   &=& \displaystyle
   \mathbb{E}_{N_t}\left[\mathbb{E}\left[\prod_{i=1}^{N_t}e^{i\omega X_i}\right]\right] && (1) \\
   &=& \displaystyle
   \mathbb{E}_{N_t}\left[\prod_{i=1}^{N_t}\mathbb{E}\left[e^{i\omega X_i}\right]\right] && (2) \\
   &=& \displaystyle
   \mathbb{E}_{N_t}\left[\phi_X(\omega)^{N_t}\right] && (3) \\
   &=& \displaystyle
   \sum_{k=0}^\infty e^{-\lambda t}\frac{(\lambda t)^k}{k!}\phi_X(\omega)^k \\
   &=& \displaystyle
   e^{\lambda t(\phi_X(\omega)-1)}
\end{array}
$$
where we used (1) conditional expectation / law of total probability, (2) the independence and (3) the identity of the random variables $X_i$. Finally, note that the variable $t$, denoting time, shouldn't be confused with the Fourier variable $\omega$.
