How do I transform this Probability in weak law of large number form? Let $X_{1},\dots$ be a sequence of independent random variables. Suppose, for $k=1,2,\dots$ $$P\left(X_{2k-1}=1\right)=P\left(X_{2k-1}=-1\right)=\frac{1}{2}$$ and the probability density function of $X_{2k}$ is $$f(x)=\frac{1}{\sqrt{2}}e^{-\sqrt{2}|x|},\,\,\ -\infty<x<\infty$$ Then $\lim_{n\to \infty} P\left(\frac{X_{1}+\dots+X_{2n}}{\sqrt{2n}} \ge 1 \right)$ is 


*

*$\Phi(1)$ 

*$\frac{1}{2}$ 

*$\Phi(-1)$ 

*$1$


where $\Phi(a)$ denotes the cdf of standard normal at $a$.
The question looks like it related to WLLN but I am not able to figure how to use that theorem we can separate the odd and even random variables but they suggested probabilities for odd ones only for $1$ and $-1$ (do they can not take any other value, I think they can) and the distribution given from it the $E(X_{2k}) =0$ since it is symmetric what to do next how to proceed with this problem. I am having trouble in understanding the context.
 A: This is an application of the Lyapunov (or Lindeberg) CLT for independent but not necessarily identically distributed random variables. To bring the expression in the form of the CLT note that the distribution of $X_{2k}$ it the Laplace distribution with parameters $b=\frac{1}{\sqrt{2}}$ and $μ=0$ (following the notation here). Hence $E[X_{2k}]=μ=0$ and $Var(X_{2k})=2b^2=1$. An easy calculation shows that $E[X_{2k-1}]=0$ and $Var(X_{2k-1})=1$. So, although the $X_k$ are not identically distributed they all have mean $μ=0$ and standard deviation $σ=1$. Hence, turning back to the given expression write $$s^2_{2n}=\sum_{i=1}^{2n}σ_i^2=2n \implies s_{2n}=\sqrt{2n}$$ and use the Lyapunov CLT which states that $$\frac{1}{s_{2n}}\sum_{i=1}^{2n}(X_i-0)\overset{d}\to \mathcal N(0,1)\implies \frac{X_1+\dots+X_{2n}}{\sqrt{2n}}\overset{d}\to \mathcal N(0,1)$$ giving in your case that $$\lim_{n\to+\infty}P\left(\frac{X_1+\dots+X_{2n}}{\sqrt{2n}}\ge 1\right)=1-\Phi(1)=\Phi(-1)$$
Of course it remains to check the Lyapunov (or the weaker Lindeberg) condition.

To check the Lyapunov condition, note that $|X_{2k}|\sim\exp(b^{-1})\implies |X_{2k}|\sim\exp(\sqrt{2})$. So $E[|X_{2k}|^m]=\frac{m!}{\sqrt{2}^m}$ and therefore choosing $δ=2$ yields $$\frac{1}{s_{2n}^{2+δ}}\sum_{i=1}^{2n}E\left[|X_i-μ_i|^{2+δ}\right]=\frac{1}{2n\cdot 2n}\left[\sum_{i=2k-1}^{2n}1+\sum_{i=2k}^{2n}\frac{4!}{\sqrt{2}^4}\right]=\frac{n+6n}{2n\cdot 2n}=\frac{7}{4n}\to 0$$ as $n\to+\infty$.
