I just came across Cramer's theorem for large deviations.
If $X_1,..,X_n$ are i.i.d and $S_n$ is the mean of the first $n$, then:
$$\lim_{n \rightarrow \infty} \frac{\log P(S_n >x)}{n}=-I(x)\qquad\text{ for }x>E[X]$$
, where $I(x)$ the rate function can be computed easily from the common distribution of the $X$ variables.
At first, I thought such a result could be derived from the central limit theorem, which says:
$$\sqrt{n}(S_n-\mu) \xrightarrow[]{\text{d}} Z(0,\sigma^2)$$
intuitively we know therefore what is happening to the distribution of $S_n$ as $n$ grows.
My questions:
1) Is it true that Cramer's theorem can be derived by the CLT? If not, what is the different piece of information that they bring ?
2) If yes, how to prove it? Starting with $\mu=0$ for example I would start arguing:
$$S_n\xrightarrow[]{\text{d}} Z(0,\frac{\sigma^2}{n})\tag{1}$$
, but maybe this is already a mistake, since the CLT implies that:
$$\lim_{n \rightarrow \infty} P(\sqrt{n}S_n >a)= \lim_{n \rightarrow \infty} P(Z(0,\sigma^2) >a)$$
and the $n$ factor in this expression is locked on the left member, whereas in $(1)$ it has been moved to the right. Maybe $(1)$ is not valid ?
Cramér's theorem and the CLT are statements about different regimes of deviations, so I don't think one can be used to prove the other. Specifically (for $\mu=0$ and $\sigma^2=1$), "the large deviation $P(S_n > x)$ decays to zero exponentially like $e^{-nI(x)}$" is a different statement than "the ordinary deviation $P(S_n > x/\sqrt{n})$ converges to $\Phi(-x)$."
For the specific direction of whether "CLT implies Cramér's theorem" is possible, note that the only information about the common distribution of the $X_i$ used in the CLT is its mean and variance, whereas the rate function in Cramér's theorem uses "all" the information about the common distribution (via its MGF). So the CLT could not possibly imply Cramér's theorem.
For the reverse direction, again Cramér's theorem only focuses on characterizing the rate of exponential decay of the large deviation $P(S_n > x)$, and doesn't say anything about the ordinary deviation $P(S_n > x/\sqrt{n})$.