Let's suppose that $\mathbb X$ ~ Poiss$(\lambda)$ (simple sample). I need to find $\mathbb E[C^\bar X]$, where $\bar X= \frac{1}{n}\sum_1^n X_i$.
I know that $\sum_1^n X_i$ ~ Poiss$(n\lambda)$, but in this line below:
$\mathbb E[C^{\frac{1}{n}\sum X_i}] = \sum (C^{\frac{1}{n}k } *\mathbb P(\sum X_i = k))$ I don't really know how was this transition between $\mathbb E[C^{\frac{1}{n}\sum X_i}]$ and $\sum C^{\frac{1}{n}k }$ made.
My first guess was that $\mathbb E[C^{\frac{1}{n}\sum X_i}] = \mathbb E[C^{\frac{1}{n}X_1}...C^{\frac{1}{n}X_n}]= \mathbb E[C^{\frac{1}{n}X_1}]...[C^{\frac{1}{n}X_n}]$ and based on the simple sample I can say that it equals to $(\mathbb E[C^{\frac{1}{n}X_1}])^n$ but does it lead me anywhere? Or maybe there is a better way to deal with it?
What you have done is correct. To complete the computation proceed as follows:
$EC^{\frac 1 n X_1}=\int_0^{\infty} C^{\frac 1 n t} \lambda e^{-\lambda t}dt$. You can write this as $\int_0^{\infty} e^{\frac 1n t D} \lambda e^{-\lambda t}dt$ where $D =\ln C$. Finally $\int_0^{\infty} e^{\frac 1n t D} \lambda e^{-\lambda t}dt=\lambda \int_0^{\infty} e^{-t(\lambda-\frac 1 n D)}dt=\frac {\lambda } {\lambda-\frac 1 n D}$. hence the answer is $(\frac {\lambda } {\lambda-\frac 1 n D})^{n}$.