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?


1 Answer 1


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: $E|X| < \infty$.


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