Sufficient Condition for Exponential Decay in Chernoff Bound (Large Deviations) Let $X_i$ ($i=1,...,n$) be a sequence of independent and identically distributed random variables.  Denote $\mu=\mathbb{E}[X_i]$ and  $S_n=\frac{1}{n}\sum_{i=1}^nX_i$.  This question concerns the tail probability $\Pr[S_n \ge \mu+\epsilon]$, where $\epsilon>0$ is a constant.
In the case that $|X_i|$ is bounded with probability one, Hoeffding's Inequality states that the tail probability decays to exponentially fast in $n$ for any $\epsilon > 0$.  [EDIT: The paper http://arxiv.org/pdf/1209.6396v1.pdf (Theorem 1.1) used to state a theorem which made it seem that finite variance was sufficient, but this is not true, and the paper has now been edited to state the sufficient condition more precisely].
A more general sufficient condition is that each variable is sub-Gaussian - see Exercise 6 of http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/
With no assumptions at all on $X_i$, the probability may fail to decay exponentially.  For example, under the Cauchy distribution (which has no first or second moment), the probability is bounded away from zero.  However, is there a sufficient condition which is less restrictive than that of boundedness?
This question is related to Large Deviations Theory and the Chernoff bound.  The following upper bound is straightforward:
$$\Pr[S_{n}\ge\mu+\epsilon]=\Pr[e^{\tau\sum_{i=1}^{n}X_{i}}\ge e^{\tau n(\mu+\epsilon)}]\le\frac{\mathbb{E}[e^{\tau\sum_{i=1}^{n}X_{i}}]}{e^{\tau n(\mu+\epsilon)}}=\frac{\mathbb{E}[e^{\tau X_{1}}]^{n}}{e^{\tau n(\mu+\epsilon)}}$$
where $\tau$ is an arbitrary positive constant, and the inequality is Markov's inequality.  For the optimal $\tau$, and under some technical assumptions, one can obtain a lower bound with the same exponential behavior (see "Large Deviations Techniques and Applications" by Dembo/Zeitouni).
Is there a simple sufficient condition for the above Chernoff bound to decay exponentially, which is stronger than boundedness or sub-Gaussian?
[EDITED SINCE ORIGINAL POST - The original answer of "no" was in response to the questions "Does a bounded absolute first moment suffice?".  I certainly agree with this answer of no, but I am still interested in finding a condition as general possible]
 A: Note that Cramer's theorem applies in dimension 1 w/out any moment assumptions, but of course the rate function may vanish (for this version see theorem 2.2.3 in Dembo-Zeitouni). So the question is equivalent to the question ``when does the rate function vanish'' (the rate function is $\sup_{\lambda\in R}
(\lambda x-\Lambda(\lambda))$, where $\Lambda(\lambda)= \log E(e^{\lambda X_1}))$. So if 
$\Lambda(\lambda)=\infty$ for all $\lambda\neq 0$, clearly there is no exponential convergence, not just a failure of the upper bound.
A: Theorem 1.1. in the paper is clearly false - for something like that to be true (two sided bound), usually one requires the variables to be bounded from both sides with probability 1 and then information about the variance gives you more information. Also, it would be enough that the generating function of each X_i is bounded by a generating function of some normal random variable (i.e. all X_i's are sub-Gaussian).
Counterexample for Theorem 1.1:
Assume n=1 and EX=0 we, by Chebyshev's inequality P(|X|>x) < VarX/x^2 and it is optimal since the symmetric random variable that takes the values {-x,0,x} with P(X=x)=VarX/(2x^2) provides the equality. Theorem 1.1. would imply that VarX/x^2 is bounded above by an exponential function in x for all x, which is not the case. 
In general it is known that if X_i have only second moments then you will not get a bound of better order than Var(X_1+...+X_n)/x^2 for large x for the probability in question. Please let me know if you want me to expand on this. 
