It is known that the characteristic function of a random variable can be used to find the moments. We have that $\frac{d\phi^{(k)}(t)}{dt}|_{t=0}=i^kE(x^k)$, so given a sequence of moments, how can that information be used to recover the characteristic function?
For example, an exercise I am trying to solve asks for the characteristic function given the following sequence of moments $E(x^k)=\frac{2k!}{3\lambda^k}$ any ideas would be appreciated.
On
The fact that $\phi^{(k)}_X(0)=i^k E(X^k)$ comes from the Taylor series of $\phi_X(t)$, which as you might have read or been told before is $$\phi_X(t)=1+iE(X)t+\frac{i^2}{2!}E(X^2)t^2+\frac{i^3}{3!}E(X^3)t^3+\cdots.$$
From there, you can use what you know about the moments and maybe get to an expression for $\phi_X$ "nicer" than the power series.
I am posting the answer for the sake of completeness.
According to Alejandro Nasif post, you can write
$\phi_x(t)=1+\sum_{k=1}^{\infty}E(x^k)\frac{i^kt^k}{k!}$
Replacing $E(x^k)=\frac{2k!}{3\lambda^k}$ you get
$\phi_x(t)=1+\sum_{k=1}^{\infty}\frac{2}{3}(\frac{it}{\lambda})^k=\frac{1}{3}+\sum_{k=0}^{\infty}\frac{2}{3}(\frac{it}{\lambda})^k$
Hence, you obtain that $\phi_x(t)=\frac{2}{3}\frac{1}{1-\frac{it}{\lambda}}+\frac{1}{3}$.
Now, using the fact that the convex combination of characteristic functions is a characteristic function you can write
$ \phi_x(t)=\frac{2}{3}\phi_y(t)+\frac{1}{3}\phi_z(t)$
Where $\phi_y(t)=\frac{1}{1-\frac{it}{\lambda}}$ is the characteristic function of an exponential random variable and $\phi_z(t)=1$ is the characteristic function of the point mass function at $0$. Hence $x_t$ has a mixture distribution.