# Sum of independent random variables having exponential tails

Suppose $$X_1, X_2,\dots, X_n$$ are iid random variable,$$P(X_i=-\infty)$$ is allowed,$$P(X_i>v)< e^{-v}\forall v>0$$, $$X$$ is distributed as $$X_i$$, if $$c$$ is a finite real number such that $$E(X), then show that there is $$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.

• you seem to really mean $P(X_i=-\infty)>0$ , then the prob you want is bounded by the probability that you haven't seen -infinity yet, which is of the right form. . But then of course E(X) = $-\inf$. – mike Sep 25 '19 at 15:15

By the Markov--Bernstein inequality (incorrectly referred to as Chernoff's) $$\begin{equation} P(X_1+\dots+X_n\ge nc)\le e^{-tnc}Ee^{t(X_1+\dots+X_n)} =e^{-tnc}(Ee^{tX})^n=e^{ng(t)} \end{equation}$$ for real $$t\ge0$$, where $$\begin{equation} g(t):=-tc+\ln Ee^{tX}. \end{equation}$$ The condition $$P(X_i>v)< e^{-v}\ \forall v>0$$ implies $$Ee^{tX}<\infty$$ and hence $$\frac d{dt}\,Ee^{tX}=EXe^{tX}\in\mathbb R$$ for real $$t<1$$. So, $$g(0)=0$$ and $$g'(0)=-c+EX<0$$, whence $$g(s)<0$$ for some real $$s>0$$. We conclude that $$\begin{equation} P(X_1+\dots+X_n\ge nc)\le r^n, \end{equation}$$ where $$r:=e^{g(s)}\in(0,1)$$, as was desired.