1
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ You may verify directly that $\{X_n,n\ge1\}$ does not obey the Lindeberg condition. $\endgroup$ – JGWang Apr 21 '17 at 3:20
0
$\begingroup$

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 $$ \frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right] \geqslant \frac 1{n^5}\sum_{i=n/2}^n\left(i^2-1\right)^2\geqslant \frac 1{2n^4}\left(\left(\frac n2\right)^2-1\right)^2. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.