Let $X_1,X_2,\dots\,$ i.i.d random variables with mean zero and variance $1$. Let $S_n=\sum_{i=1}^n X_i\,,n\in \mathbb N.$ Compute the weak limes $\lim_{n\to\infty} \frac1n \sum_{i=1}^n \frac{S_i}{\sqrt n}$

Surely we will have to use the CLT. First I tried to simplify the expression, but I am not sure how to continue here. $$\lim_{n\to\infty} \frac1n \sum_{i=1}^n \frac{S_i}{\sqrt n}=\dots=\lim_{ n\to\infty}\frac{1}{\sqrt n} \frac{nX_1+(n-1)X_2+\dots+X_n}{n}$$


According to the comments, we have to verify Lindberg's condition (https://en.wikipedia.org/wiki/Lindeberg%27s_condition)

Lindberg's condition: $$\lim_{n\to\infty} \frac{1}{s_n^2} \sum_{k=1}^n E[(X_k - \mu_k)^2 \mathbb 1_{\{\mid X_k - \mu_k \mid > \epsilon s_n \}}=0,\quad \text{for all $\epsilon >0$}$$


$E(S_i) {\overset{\text{$X_i$ i.i.d}}{=}}0$ , $Var(S_i) {\overset{\text{$X_i$ i.i.d}}{=}} \sum Var( X_i) {\overset{\text{$X_i$ i.i.d}}{=}} i$ for all $i=1,2,\dots$ Furthermore $s_n^2= \sum_{i=1}^n \sigma_i^2 =Var(S_1)+Var(S_2)+\dots + Var(S_n)=1+2+\dots +n=\frac{n(n+1)}{2}$.

Plugging in: $$\lim_{n\to\infty}\frac{2}{n^2+n}\sum_{k=1}^n E(S_k)^2 1_{\{\mid S_k \mid > \epsilon {\frac{\sqrt {n^2+n}}{\sqrt 2}}\}}$$ Intuitively this does not seem correct to me. Furthermore I am not sure how to simplify this expression.

Some help is welcome and obviously needed!

  • 1
    $\begingroup$ To verify the Lindeberg's condition is satisfied. $\endgroup$
    – JGWang
    Jan 12, 2018 at 3:30
  • $\begingroup$ Here is a suggestion: type the words Lindeberg's condition in your favorite search engine and see what pops up... $\endgroup$
    – Did
    Jan 12, 2018 at 8:46
  • $\begingroup$ @Did I tried what you say, I edited my post, but how can we proceed? $\endgroup$
    – user519338
    Jan 12, 2018 at 12:03
  • $\begingroup$ Why are you checking Lindeberg's condition on the increments of another normalized sum than the one you are interested in? $\endgroup$
    – Did
    Jan 12, 2018 at 13:27
  • $\begingroup$ therefore I should do the same with $X_k=S_k $ instead, right? $\endgroup$
    – user519338
    Jan 12, 2018 at 13:33

1 Answer 1


Let $Y_{ni}=\frac{(n+1-i)X_i}{n^{3/2}}$, then $\{Y_{ni}, 1\le i\le n, n\ge1\}$ is a triangular array of independent(in row) variables, $\mathsf{E}[Y_{ni}]=0$ and \begin{gather*} \mathsf{var}[Y_{ni}]=\mathsf{E}[Y_{ni}^2]=\frac{(n+1-i)^2}{n^3},\\ \frac1n\sum_{i=1}^n\frac{S_i}{\sqrt{n}}=\sum_{i=1}^n\frac{(n+1-i)X_i}{n^{3/2}} =\sum_{i=1}^nY_{ni},\\ \begin{aligned} s_n^2&=\mathsf{var}\biggl[\frac1n \sum_{i=1}^n \frac{S_i}{\sqrt{n}} \biggr] =\sum_{i=1}^n\frac{(n+1-i)^2}{n^3}\\ &=\frac{n(n+1)(2n+1)}{6n^3}\to\frac13, \quad\text{as } n\to\infty. \end{aligned} \end{gather*}

Next we verify the Lindeberg's condition. Noting that \begin{align*} Y_{ni}^21_{\{|Y_{ni}|\ge \varepsilon s_n\}} &=\frac{(n+1-i)^2}{n^3}X_i^2I\biggl(\frac{(n+1-i)^2}{n^3}X_i^2 \ge\varepsilon^2\frac{n(n+1)(2n+1)}{n^3}\biggr)\\ &\le \frac1n X^2_iI(X_i^2\ge \varepsilon^2 n) \end{align*} we have \begin{align*} \frac1{s_n^2}\sum_{i=1}^n\mathsf{E}[Y_{ni}^21_{\{|Y_{ni}|\ge \varepsilon s_n\}}] &\le \frac{1}{ns_n^2}\sum_{i=1}^n\mathsf{E}[X_i^21_{\{X_i^2\ge \varepsilon n\}}]\\ &=\frac{1}{s_n^2}\mathsf{E}[X_1^21_{\{X_1^2\ge \varepsilon n\}}]\to 0, \quad \text{as } n\to\infty. \end{align*} At last, using Lingdeberg-Feller Theorem, we have $$ \lim_{n\to\infty}\frac1n\sum_{i=1}^n\frac{S_i}{\sqrt{n}}\stackrel{d}{=}N(0,1/3)$$

  • $\begingroup$ actually I only know Lindberg levy clt. is it possible to transform your proof in its form? $\endgroup$
    – user519338
    Jan 15, 2018 at 13:11
  • $\begingroup$ @J.D About Lindeberg-Feller Theorem, please refer to "R. Durrett, Probability: theory and examples, 2nd ed, p.116, 2.4.b, Triangular arrays" and "S. I. Resnick, A Probability Path, Birkhäuser Boston, 2005. p.321, $9.9 Ex.1". $\endgroup$
    – JGWang
    Jan 16, 2018 at 2:24

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