# Proving that the Lindeberg condition Fails

I am having trouble proving the following result thanks for any help in advance:

Let {$X_{n}, n\geq 1$} be a sequence of independent random variables with $P(X_{n} = -n) = 1-1/n^{2}$ and $P(X_{n} = n^{3} - n) = 1/n^{2}, n \geq 1$ Prove that {$X_{n}, n\geq1$} does not obey the Lindeberg condition.

There are a few things that I have noticed:

1.$\sum_{j = 1}^{n} X_{j} / n \rightarrow -\infty$ almost certainly

2.$EX_{n}^{2} /\sum_{j = 1}^{n}EX_{j}^{2}\rightarrow 0$ and $\sum_{j = 1}^{n}EX_{j}^{2} \rightarrow \infty$, so if I assume the Lindeberg condition is satisfied then by the Lindeberg-Feller CLT there will be some contradiction stemming from $\sum_{j = 1}^{n} X_{j}/\sqrt(\sum_{j = 1}^{n}EX_{j}^{2})$ converging in distribution to a $\mathcal{N} (0,1)$ random variable.

1. I imagine that I am trying to get some contradiction with 1.
• You may verify directly that $\{X_n,n\ge1\}$ does not obey the Lindeberg condition. – JGWang Apr 21 '17 at 3:20

## 1 Answer

Observe that $$X_n$$ is centered and that for all $$n$$, \begin{align} \mathbb E\left[X_n^2\right]&=n^2\left(1-n^{-2}\right)+\left(n^3-n\right)^2n^{-2}\\ &= n^2-1+n^2\left(n^2-1\right)^2n^{-2}\\ &= \left(n^2-1\right)\left(1+n^2-1\right) \\ &= n^2\left(n^2-1\right) \end{align} hence there exists constants $$c$$ and $$C$$ such that $$cn^5\leqslant\sum_{i=1}^n\operatorname{Var}\left(X_i\right)\leqslant Cn^5$$ and the Lindeberg condition is equivalent to the following: $$\forall \varepsilon\gt 0, \frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert>\varepsilon n^{5/2}\right\}\right]=0.$$ We will show that this fails for $$\varepsilon=1$$. Let us denote by $$I_n$$ the set of integers $$i$$ such that $$2\leqslant i\leqslant n$$ and $$i^3-i\gt n^{5/2}$$. Then \begin{align} \frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right]&\geqslant \frac 1{n^5}\sum_{i\in I_n}\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right]\\ &\geqslant \frac 1{n^5}\sum_{i\in I_n}\mathbb E\left[X_i^2\mathbf 1\left\{ X_i=i^3-i\right\}\right]\\ &= \frac 1{n^5}\sum_{i\in I_n}\left(i^3-i\right)^2i^{-2}\\ &\geqslant \frac 1{n^5}\sum_{i\in I_n}\left(i^2-1\right)^2. \end{align} Observe that for $$n$$ large enough event, $$I_n$$ contains the set of integers between $$n/2$$ and $$n$$ hence $$\frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right] \geqslant \frac 1{n^5}\sum_{i=n/2}^n\left(i^2-1\right)^2\geqslant \frac 1{2n^4}\left(\left(\frac n2\right)^2-1\right)^2.$$