Characteristic Function of a Conditioned Random Varaible Let $(X_j)$ be iid random variables and $N \sim \mathrm{Poisson}(\lambda)$, independent of $X_j\, \forall j$. Define $S_n:=\sum_j^n\,X_j$ and consider $S_N$.

Find the characteristic function of $S_N$. Then find $\mathbb{E}S_N$ and $\operatorname{Var}(S_N)$.

I can find $\mathbb{E}S_N$ and $\operatorname{Var}(S_N)$ from the first and second moments, once I have the characteristic function. The last bit is what is tripping me up. Here is what I have so far:

\begin{align*}
\varphi_{S_N}(t) &amp;= \varphi_{\sum_j^N\,X_j}(t)\\
  &amp;= \mathbb{E}e^{it\sum_j^N\,X_j}\\
  &amp;= \mathbb{E}e^{it\sum_j^N\,X_j}\sum_j^N\mathbb{1}_{\{N=n\}}\\
  &amp;= \sum_{n=0}^\infty\,\mathbb{E}e^{it\sum_{j-1}^N\,X_j}\mathbb{P}(N=n)\\
  &amp;= \sum_{n=0}^\infty\,(\varphi_{X_1}(t))^n\lambda^n\frac{e^{-\lambda}}{n!}\\
  &amp;= e^{-\lambda}\sum_{n=0}^\infty\,(\varphi_{X_1}(t))^n\frac{\lambda^n}{n!}\\
  &amp;= e^{-\lambda} + \sum_{n=1}^\infty\,(\varphi_{X_1}(t))^n\frac{\lambda^n}{n!}
\end{align*}

Have I made an error, or gone off in the wrong direction? Thanks in advance.
I think earlier in your calculation some things got confusing. You should just find:

$$\phi_S(t)=E[e^{it \sum_{j=1}^N X_j}]=\sum_{n=0}^\infty E \left [e^{it \sum_{j=1}^N X_j} \mid N=n \right ] P[N=n] \\
= \sum_{n=0}^\infty E \left [ e^{it \sum_{j=1}^n X_j} \right ] P[N=n] \\
= \sum_{n=0}^\infty \phi_X(t)^n P[N=n]$$

using the total expectation formula and the fact that $X_j$ are iid.

Now the point is to make the result look more familiar:

$$\phi_S(t)=\sum_{n=0}^\infty \frac{e^{-\lambda} \phi_X(t)^n \lambda^n}{n!} = e^{-\lambda} \sum_{n=0}^\infty \frac{(\phi_X(t) \lambda)^n}{n!}.$$

You should recognize sums like that one.
You derailed on the third line. &nbsp; Backtrack and try this:

$$\begin{align}
\varphi_{S_N}(t) = &amp;~ \mathsf E(\mathscr e^{\imath t S_N})
\\[1ex]=&amp;~ \mathsf E(\mathscr e^{\imath t \sum_{k=1}^N X_k})
\\[1ex]=&amp;~ \mathsf E(\prod_{k=1}^N\mathscr e^{\imath t X_k}) &amp; \because~e^{a+b+c+\ldots}=e^a\cdot e^b\cdot e^c\cdots
\\[1ex]=&amp;~ \mathsf E(\mathsf E(\prod_{k=1}^N \mathscr e^{\imath t X_k}\mid N)) &amp; \textsf{Law of Iterated Expectation}
\end{align}$$

Then just apply independence and identicallity of your random variables and everything should fall into place.