Prove that $X_n = e^{(\theta S_n-n \ln \phi(\theta))}$ is martingale. Let $\xi_1 \neq const.$ be a random variable with moment-generating function $\phi(\theta) = Ee^{(\theta \xi_1)}$. Let $S_n = \xi_1 + \dots \xi_n$. Prove that $X_n = e^{(\theta S_n-n \ln \phi(\theta))}$ is martingale. 
I think that if $\xi_i$ are idd this is obviously because in this case $X_n=1$.
So in my opinion I should consider case if $\xi_i$ are not iid. But in this case I dont have any idea.
 A: I think that you should be specific about probability spaces and filtrations and that you forgot integrability and adaptability.
Consider a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^{\xi} = \sigma(\xi_1, ..., \xi_n)$.
We want to show that $X_n = \exp{(\theta S_n-n \ln \phi(\theta))}$ is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)-$martingale.



*

*$\exp{(\theta S_n-n \ln \phi(\theta))}$ is adapted to the filtration $\sigma(\xi_1, ..., \xi_n)$:


$\exp{(\theta S_n-n \ln \phi(\theta))}$ is a Borel-measurable function of $S_n$, which is $\sigma(\xi_1, ..., \xi_n)$-measurable.


*$\exp{(\theta S_n-n \ln \phi(\theta))}$ is integrable:


$E[|\exp{(\theta S_n-n \ln \phi(\theta))}|] = E[\exp{(\theta S_n-n \ln \phi(\theta))}] = 1 < \infty$


*$\exp{(\theta S_n-n \ln \phi(\theta))}$ satisfies the martingale property:


Let m < n. Then
$$E[\exp{(\theta S_n-n \ln \phi(\theta))}|\mathscr F_m]$$
$$=E[\exp{(\theta (\xi_1 + ... + \xi_n)-n \ln \phi(\theta))}|\mathscr F_m]$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} E[\exp{(\theta (\xi_{m+1} + ... + \xi_n)-n \ln \phi(\theta))}|\mathscr F_m]$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} E[\exp{(\theta (\xi_{m+1} + ... + \xi_n)-n \ln \phi(\theta))}|\mathscr F_m]$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} \exp{(-n \ln \phi(\theta))} E[\exp{(\theta (\xi_{m+1} + ... + \xi_n)})|\mathscr F_m]$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} \exp{(-n \ln \phi(\theta))} E[\exp{(\theta (\xi_{m+1} + ... + \xi_n)})]$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} \exp{(-n \ln \phi(\theta))} (E[\exp(\theta\xi_{m+1})] + ... + E[\exp(\theta \xi_{n}))])$$
$$=\exp{(\theta (\xi_1 + ... + \xi_m))} \exp{(-m \ln \phi(\theta))} \exp{(-(n-m) \ln \phi(\theta))} (E[\exp(\theta\xi_{m+1})] + ... + E[\exp(\theta \xi_{n}))])$$
$$=X_m \exp{(-(n-m) \ln \phi(\theta))} E[\exp(\theta\xi_{m+1})] + ... + E[\exp(\theta \xi_{n}))]$$
$$=X_m \exp{(-(n-m) \ln \phi(\theta))} E[\exp(\theta\xi_{m+1})] + ... + E[\exp(\theta \xi_{n}))]$$
$$=X_m \phi(\theta)^{-(n-m)} (E[\exp(\theta\xi_{m+1})] + ... + E[\exp(\theta \xi_{n}))])$$
$$=X_m \phi(\theta)^{-(n-m)} (\phi(\theta)^{(n-m)})$$
$$=X_m$$
QED
