# Tail bound for sum of i.i.d. random variables with common moment generating function

Suppose $$\{X_n\}_{n\in \mathbb{N}}$$ is a sequence of independent and identically distributed random variables and $$S_n:=X_1+...+X_n$$. Assume that each $$X_i$$ has mean $$0$$ and that all $$X_i$$ have a common moment generating function $$M(\theta)$$ which is bounded for all $$\theta$$ in a small neighbourhood $$(-\delta, \delta)$$ of $$0$$. For any $$a > 0$$, show that

$$\mathbb{P}(S_n>an) \leq \left( \frac{M(\theta)}{e^{a\theta}}\right)^n,\quad \theta > 0.$$

My idea to solve this exercise was to use Markov's inequality and the fact that the moment generating function of a sum of independent random variables is the product of the individual moment generating functions.

$$\mathbb{P}(S_n > an) \leq \mathbb{P}(S_n\geq an) = \mathbb{P}\left(e^{\theta S_n}\geq e^{\theta an}\right) \leq \frac{\mathbb{E}e^{\theta S_n}}{e^{\theta an}} = \frac{M(\theta)^n}{e^{\theta an}} = \left( \frac{M(\theta)}{e^{\theta a}} \right)^n$$

However this solution neglects both the fact that the $$X_i$$ have zero mean and that $$M(\theta)$$ is bounded in a neighbourhood around 0. Are these facts not needed or can you point out the flaw in my attempted solution?

Your solution is correct. I believe these are just regularity conditions which are needed because moment generating functions do not always exist. For example, having $$0$$ mean ensures that the first moments exist: $$E|X| < \infty$$.