Cramers theorem for large deviations

258 Views Asked by At

Let $(X_n)$ be real-valued i.i.d. random variables such that the cumulant generating function $\Lambda(t):=\log E e^{tX_1}$ is finite for all t and let $S_n:=\frac{1}{n}(X_1+...+X_n)$ denote the sample mean. Then Cramers theorem, as stated in Wikipedia (https://en.wikipedia.org/wiki/Cram%C3%A9r%27s_theorem_(large_deviations)) says: \begin{align} \lim \limits_{n \to \infty} \frac{1}{n} \log \mathbf{P}(S_n \geq x) = \Lambda^*(x) \quad \forall x\geq E(X_1), \end{align} where $\Lambda^{*}(x) := \sup_{y \in \mathbb{R}} (y x - \Lambda(y))$.

However in chapter 2.2 of Dembo's "Large Deviations Techniques and Applications" (1998) Cramers theorem says \begin{align} -\inf_{x \in A^{\circ}} \Lambda^*(x) \leq \liminf_{n \to \infty} \frac{1}{n} \log \mathbf{P}(S_n \in A) \leq \limsup_{n \to \infty} \frac{1}{n} \log \mathbf{P}(S_n \in A) \leq -\inf_{x \in \overline{A}} \Lambda^*(x) \quad \forall A\in \mathcal{B}, \end{align} in other words, the $S_n$ satisfy the LDP with rate function $\Lambda^*$.

Now I looked around in literature and never found a satisfactorily result to connect both statements. How does one get from the one from Wikipedia to the one from Dembo? Under what circumstances can one generalize Large Deviation Statements from sets $(x,\infty)$ to all measurable sets?