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$X_1,X_2,X_3 ,\ldots$ 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)

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

2 Answers 2

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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.

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  • $\begingroup$ You are right. I miss this requirement. :( $\endgroup$
    – Charlie F
    Jun 3, 2020 at 2:03
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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.

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