Let $\{X_n\}$ be a sequence of independent random variables such that $\mathbb P(X_n=\pm 1)=\frac 14$, $\mathbb P(X_n=\pm n)=\frac 1{4n^2}$ and $\mathbb P(X_n=0)=\frac 12 - \frac 1{2n^2}$ for all $n\ge 1$. Define the triangular array $\{X_{nj}:1\le j\le n\}_{n\ge 1}$ by setting $X_{nj}=\frac{X_j}{\sqrt n}$. Check whether the above triangular array satisfies the Lindeberg condition.

I have calculated $$s_n:=\sum_{j=1}^n \frac{X_j}{\sqrt n}$$ and $$\sigma_{nj}^2:=\mathbb E[X_{nj}^2]=\frac 1n$$ and hence $$S_n^2:=\sum_{j=1}^n \sigma_{nj}^2 = 1$$ So, I need to prove $$\sum_{i=1}^n E[X_i^2\mathbf{1}_{|X_i|>\epsilon}]\to 0\;\; \forall \epsilon>0$$ to check the Lindeberg condition. which is clearly false as the expression is a sum.

I must have made some mistake somewhere which I can't figure out. Please help me.

  • $\begingroup$ Have you considered applying the Lindeberg Condition directly to this array? $\endgroup$
    – whuber
    Commented May 3, 2023 at 21:40
  • $\begingroup$ @whuber yes, but couldn't make anything of it $\endgroup$ Commented May 3, 2023 at 21:53
  • $\begingroup$ @whuber I tried using Lindeberg directly, and this is what I got- check the edit $\endgroup$ Commented May 3, 2023 at 22:48

1 Answer 1


The general Lindeberg condition is \begin{align} \lim_{n \to \infty}\sum_{j = 1}^n \frac{1}{s_n^2}E[X_{nj}^2I_{[|X_{nj}| \geq \epsilon s_n]}] = 0. \tag{1} \end{align}

In your case, as you correctly demonstrated, $s_n^2 = 1$. But $X_{nj} = \frac{X_j}{\color{red}{\sqrt{n}}}$ instead of $X_j$. Therefore, $(1)$ should become \begin{align} \lim_{n \to \infty}\sum_{j = 1}^n \frac{1}{n}E[X_{j}^2I_{[|X_{j}| \geq \epsilon\sqrt{n}]}] = 0. \tag{2} \end{align} For fixed $\epsilon > 0$ and sufficiently large $n$, we have \begin{align} n^{-1}\sum_{j = 1}^n E[X_{j}^2I_{[|X_{j}| \geq \epsilon\sqrt{n}]}] = n^{-1}\sum_{j = \lceil\epsilon\sqrt{n}\rceil}^n j^2 \times \frac{1}{2j^2} = \frac{n - \lceil\epsilon\sqrt{n}\rceil + 1}{2n} \to \frac{1}{2} \end{align} as $n \to \infty$. Hence the Lindeberg condition does not hold for this triangular array.

  • $\begingroup$ Great answer, thanks (+1)! $\endgroup$ Commented May 4, 2023 at 18:19

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