Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$ Let $Y_t$ be a random variable (Not positive necesarily).
Can I make the next assumption?
$$E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$$
Thanks! I think it is correct but I do not know how to justify. 
 A: Fubini's theorem says you can interchange the order of two integrations with respect to $\sigma$-finite measures (and a sum is just an integration with respect to counting measure) if the double integral converges absolutely.
You do need that absolute convergence, in this case 
$$ E\left[\sum_{m=0}^\infty  |t|^m |Y_t|^m/m!\right] < \infty $$
EDIT: In this case the series on the left is of course that of $\exp(itY_t)$, and if $Y_t$ and $t$ are real this has absolute value $1$; the expected value on the left exists and is the characteristic function $E\left[ e^{itY_t}\right]$.  On the right we have a power series $\sum_{m=0}^\infty c_m t^m$ where $c_m = i^m E[Y_t^m]/m!$.  Of course for some random variables the moments $E[Y_t^m]$ don't all exist; even if they do exist, the series might have a radius of convergence of $0$, so the right side would undefined except at $t=0$.  If the radius of convergence $R > 0$, the series converges absolutely for $|t| < R$, and as mentioned the two sides are equal in that case.  For $|t| > R$, the series diverges and the right side is undefined.
For $t = \pm R$, matters are somewhat more delicate: the series may diverge, converge absolutely or converge conditionally.  If it does converge, even if only conditionally, Abel's theorem says the value at $t$ is the limit of the values at $rt$ as $r \to 1-$, and since $E[\exp(itY_t)]$ is a continuous function of $t \in \mathbb R$ that says your equation would be true at $t$.
For example, let $Y_t$ have a Gamma distribution with scale parameter $1$ and shape parameter $c > 0$.  Then your equation says
$$ (1 - i t)^{-c} = \sum_{m=0}^\infty \dfrac{\Gamma(m+c)}{\Gamma(c)\; m!} (it)^m$$
The series has radius of convergence $1$.  Now $\Gamma(m+c)/m! = m^{c-1} + O(m^{c-2})$ as $m \to \infty$, so at $t = \pm 1$ the series diverges if $c \ge 1$, but converges conditionally there (using Dirichlet's test) if $0 < c < 1$.  Thus if $c \ge 1$ your equation is valid for $-1 < t < 1$, while if $0 < c < 1$ it is valid for $-1 \le t \le 1$.
