I have the following question
Let
$X_{1}$, $X_{2}$, ..., $X_{n}$ be independent and identically distributed random variables with moment generating function $M_{X}(t)$, for -h < t < h, h > 0, $S_{n}$ = $\sum_{i=1}^{n}X_i$, and $\bar{X}$ = $S_{n}/n$.
Ultimately I need to establish the following:
Use the fact that $M_{X}(0)=1$ and $dM_{X}(t)/dt$ $\vert_{t=0} = > \mathbb{E}(X)$ to show that if $\mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) \leq c^{n}$. Establish a similar bound for $P(S_{n} \leq a)$.
I have previously shown that
$P(S_{n}>a) \leq e^{-at}[M_{X}(t)]^{n}$ for $0<t<h$ , $a\geq0$ and $P(S_{n}\leq a) \leq e^{-at}[M_{X}(t)]^{n}$ for $ -h<t<0 $
As follows:
$P(S_{n} > a) = \int_a^{\infty} dF(s) \leq \int_a^{\infty} e^{-at} e^{st} dF(s) \leq \int_{-\infty}^{\infty} e^{-at} e^{st} dF(s) = e^{-at} \mathbb{E} (e^{t \sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $
And similarly:
$P(S_{n} \leq a) = \int_{-\infty}^{a} dF(s) \leq \int_{-\infty}^{a} e^{-at} e^{st} dF(s) \leq \int_{-\infty}^{\infty} e^{-at} e^{st} dF(s) = e^{-at} \mathbb{E} (e^{t \sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $
Now, I have really no clue on how to establish a bound at $c_{n}$ using the fact that $M_{X}(t) = 0 $ and $dM_{X}(t)/dt$ $\vert_{t=0} = \mathbb{E}(X)$ to show that if $\mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) \leq c^{n}$. I have no attempt to show but any ideas would be very welcome.
Thanks in advance