# Prove convergence in distribution of sum of non-i.i.d random variables.

$$X_1,X_2,X_3 ,\ldots$$ be independent random variables with distribution $$P(X_i=i)=P(X_i=-i)=1/2$$ for all $$i$$. Define $$S_n=X_1+X_2+X_3+\cdots+X_n$$.

And the question is to show "Does $$\{S_n/n^p\}_{n=1}^∞$$ converge in distribution? Why?"

I know can use CLT or LLN when $$X_i$$ be i.i.d random variables, but in this question, $$X_i$$ has different distribution.

I hope can get some hint. :) And I m confused on how to prove converge in distribution(I know that can use φ,E,CDF)

• What is $p$ here? Jun 2, 2020 at 5:38
• Oops! p is a number belong to real number. I should use another letter, "p" is confusing. Jun 2, 2020 at 9:38
• This question can be solved by Using lyapunov central-limit theorem Jun 2, 2020 at 9:39

can we really use Lyapunov CLT to solve this question? Here the variance of $$X_i$$ should be $$i^2$$. Lyapunov central-limit theorem need a finite variance.
It is possible to use Lindeberg's condition: since the random variables $$X_i$$ are centered and $$\operatorname{Var}\left(X_i\right)=i^2/2$$, Lindeberg's condition translates as $$\forall \varepsilon >0, \lim_{n\to\infty}\frac{1}{n^3}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf{1}\{\left\lvert X_i\right\rvert >\varepsilon n^{3/2}\}\right]=0.$$ Since $$\mathbf{1}\{\left\lvert X_i\right\rvert >\varepsilon n^3\}=0$$ for $$1\leqslant i\leqslant n$$, we derive that $$S_n/n^{3/2}$$ converges in distribution to a centered normal law.