Large deviation problem

Question

Let $s_1,s_2,\cdots,s_n$ be n i.i.d r.v. drawn from a probability distribution $p$ with bounded support. Show that, to leading exponential order, $$P\{s_1+\cdots+s_n\leq0\}=\{\inf_{z\geq0}E[e^{-zs_1}]\}^n$$

Answer 1

What have you tried so far? Consider using the Markov-Inequality for some function like $\varphi (x)=e^{-zx}$. Also keep in mind that $\mathbb{E}[e^{s_1+...}]$ is easily calculated using independence.

$\mathbb{P}[S_n\geq0]=\mathbb{P}[e^{-z(s_1+...)}\geq1]\leq\frac{\mathbb{E}[e^{-z(s_1+...)}]}{1}=\mathbb{E}[e^{-zs_1}]^n$