# Law of Large Numbers Deduction

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

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

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

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

And deduce the most general law of large numbers you can from the previous inequality.

• Hint: As a first step to deriving the general law of large numbers, you could for example calculate, $L$, $L^{*}$, for $\pm 1$ or Cauchy random variables and see what happens in the inequality. – Chris Apr 2 '13 at 23:56
• math.stackexchange.com/questions/349587/… is related. – Chris Apr 3 '13 at 1:05

$$\mathbb P\left(\left|\frac{S_n}{n}-\mu\right|\geq\alpha\right)=\mathbb P\left(\frac{S_n}{n}-\mu\geq \alpha\right)+\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)$$
$$\mathbb P\left(\frac{S_n}{n}-\mu\geq \alpha\right)+\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}+e^{-nL^{*}(\mu-\alpha)}$$
$$\mathbb P\left(\frac{S_n}{n}-\mu\geq\alpha\right)\leq e^{-nL^{*}(\mu+\alpha)}$$
$$\mathbb P\left(-\frac{S_n}{n}+\mu\geq \alpha\right)\leq e^{-nL^{*}(\mu-\alpha)}$$