I am taking a course on Large deviation theory but I'm a bit stuck at the first place. The lecturer gave an example to motivate the study of large deviation.
First, if $X_1$ is standard gaussian, and $X_1,\cdots,X_n$ are i.i.d. r.v.s, then $S_n/\sqrt{n}$ is also standard gaussian, thus the event $S_n>an$ where $a>0$ is rare in the sense that its probability tends to zero when $n\rightarrow\infty$. Specifically, the speed of decay is $\sim\exp(-an^2)$
Then, if $X_1$ is not standard gaussian, we wonder whether the event $S_n>an$ still rare in the sense that it has exponential decay.
I don't get here, because as long are $X_i$ are iid then we can always have exponential tail according to CLT right? Specifially what I meant is:
Let $X_1,\cdots,X_n$ are i.i.d. r.v.s with expectation $0$ and variance $1$, then according to central limit theorem we have $$\frac{S_n}{\sqrt{n}}\sim N(0,1)$$ where $S_n=\sum_{i=1}^nX_i$.
Thus for $a>0$, we have $\mathbb P(\frac{S_n}{\sqrt{n}}>a)\leq \exp(-\frac{a^2}{2})$, i.e., $\mathbb P(S_n>a\sqrt{n})\leq \exp(-\frac{a^2}{2})$
Let $a=\sqrt{n}a$ then $\mathbb P(S_n>an)\leq \exp(-\frac{na^2}{2})$.
Thus, the event $S_n>an$ is rare in the sense that its probability tends to zero as $n\rightarrow\infty$.
Thus conclusion:
$\mathbb P(S_n\geq an)\leq \exp(-na^2)$, for iid X_i with zero expectation and 1 variance
This conclusion works for all iid random variables, not necessary being gaussian. Then why, large deviation theory is to first, answer whether this event is rare for non-gaussian case?
The lecture also gave an example where $S_n\geq an$ doesn't have exponential decay, but it seems contradicts to the upper bound above:
Assume $-X_1\overbrace{=}^{(d)} X_1$, $X_1$ has density, $\mathbb P(X_1>x)\geq cx^{-d}$ for $x\geq 1$ for some $c,d>0$ then $\mathbb P(S_n>an)\geq \mathbb P(X_n>an) \mathbb P(S_{n-1}>0)\geq c(an)^{-d}\frac{1}{2}$.
What could be wrong..