Large Deviations Question

Question

Let $\left(X_n\right)_{n\geq 1}$ be i.i.d random variables on $\left(\Omega,\mathcal A, \mathbb P\right)$, $X_1$ with mean $\mu$, and

$$ L(\lambda) = \begin{cases} \log\mathbb E\left(e^{\lambda X_1}\right)<\infty, & \text{if }\mathbb E\left(e^{\lambda X_1}\right)<\infty \\ +\infty, & \text{otherwise, } \end{cases} $$

If $$\displaystyle L^{*}(x)=\sup\left(x\lambda-L(\lambda)|\lambda\in\mathbb R\right)$$

Prove that for any $\alpha >0$ and $n \geq 1$,

$$\mathbb P\left(\frac{X_1+...+X_n}{n}-\mu\geq \alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}$$

Answer 1

Without loss of generality take $\mu = 0$ and fix any $\lambda \in \mathbb{R}$. Call $S_n =\sum_{i=1}^n X_i$. Then $$ P\left( \frac{1}{n}S_n \geq \alpha\right) = P\left(e^{\lambda S_n} \geq e^{n \lambda \alpha}\right)$$ $$ \leq e^{-n \lambda \alpha} E[e^{\lambda X_1}]^n$$ $$ = e^{-n(\lambda \alpha - L(\lambda))} $$

Now take the supremum over $\lambda$ to get the result.