which probability inequality to apply in this situation? Suppose $X_1, X_2,\dots, X_n$ are iid random variable,$P(X_i=-\infty)>0$,$P(X_i>v)< e^{-v}\forall v>0$, $X$ is distributed as $X_i$, if $c$ is  a finite real number such that $E(X)<c$, then show that there is $A>0, r<1$ such that $P(X_1+\dots+X_n>nc)< Ar^n\forall n$. Can apply Chernoff or Hoefding bound here, to apply do I need to know what distribution $X_i$ is following? I am a bit confused. what is the role of $X$ here? Thanks for helping.
 A: One way to do it would be to say that $X$ is a sub-exponential random variable (check for example Vershynin's book), and to use Bernstein inequality.
Another way is to use ideas from sub-exponential random variables, for example as below.
Let denote $\mu=\mathbb E[X]$. We have, for $\mathbb{P}(X-\mu > v)=\mathbb{P}(X>v+\mu)\leq e^{-\mu}e^{-v}$ and $\mathbb{P}(X-\mu < v)=\mathbb{P}(X<v+\mu)\leq e^{\mu}e^{v}$.
Consequently, for $v\geq0$, $\mathbb{P}(|X-\mu|>v)\leq (e^{\mu}+e^{-\mu})e^{-v}$.
We note $Y\doteq X-\mu$ and $a\doteq (e^{\mu}+e^{-\mu})$. We have $\mathbb{P}(|Y|>v)\leq a \exp(-v)$.
We are interested in computing, noting $b=c-\mu>0$ and taking $\lambda \in (0,1)$,
\begin{align*}
\mathbb{P}(X_1+\cdots+X_n>nc)&=\mathbb P (\sum_{i=1}^n X_i-\mu>nb)\\
&\leq\dfrac{\mathbb E\exp(\lambda\left(\sum_{i=1}^n X_i-\mu\right))}{\exp(\lambda n b)}\\
&\leq \dfrac{\prod_{i=1}^n\mathbb E \exp(\lambda(X_i-\mu))}{\exp(\lambda n b)}\\
&\leq \dfrac{(\mathbb E \exp(\lambda(X_i-\mu)))^n}{\exp(\lambda n b)}.
\end{align*}
Now compute $\mathbb  E \exp(\lambda Y)$.
First, notice that
\begin{align*}
\mathbb E |Y|^k&=\int_{0}^{\infty} \mathbb{P}(|Y|^k\geq u)\mbox{d}u\\
&=\int_{0}^{\infty} \mathbb{P}(|Y|\geq t)kt^{k-1}\mbox{d}t\\
&\leq \int_{0}^{\infty} a\exp(-t)kt^{k-1}\mbox{d}u \\
&\leq ak \Gamma(k)=a\Gamma(k+1)=a k!.
\end{align*}
Consequently,
\begin{align*}
\mathbb E \exp(\lambda Y)&=\mathbb E \sum_{k\geq 0} \dfrac{\lambda^k X^k}{k!} \\
&\leq \sum_{k\geq 0} \dfrac{\lambda^k \mathbb E |X|^k}{k!}
&\leq a \sum_{k\geq0}\lambda^k = \dfrac{a}{1-\lambda}.
\end{align*}
Finally, taking for instance $\lambda=1/2$,
$$
\mathbb{P}(X_1+\cdots+X_n>nc) \leq (2a e^{-b/2})^n.
$$
A better value for $r$ can be found by optimizing on $\lambda$.
