# Showing that the Lindeberg CLT Condition Holds

Suppose we have a sequence of random variables, $\{X_{n}\}_{n\geq 1}$ satisfying:

$\mathbb{P}(X_{j} = 2^{j}) = \mathbb{P}(X_{j} = -2^{j}) = \frac{1}{2}$

Then is it true that the CLT holds? Or sufficiently, does the following Lindeberg condition hold?

$\lim\limits_{n\rightarrow\infty}\dfrac{1}{s_{n}^{2}}\displaystyle\sum\limits_{j=1}^{n}E[(X_{j}-E[X_{j}])^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}] = 0 \;\; \forall \epsilon>0$

where $s_{n}^{2} = \displaystyle\sum\limits_{j=1}^{n}Var[X_{j}]$. Specifically, we have $E[X_{j}] = 0$ and $s_{n}^{2} = \displaystyle\sum\limits_{j=1}^{n}4^{j}$. So we need to show:

$\lim\limits_{n\rightarrow\infty}\frac{1}{\sum\limits_{j=1}^{n}4^{j}}\displaystyle\sum\limits_{j=1}^{n}E[X_{j}^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}] = 0 \;\; \forall \epsilon>0$.

I'm not sure how to deal with the set $\{|X_{j}| > \epsilon s_{n}\}$.

What happens if Lindeberg condition does not hold? I don't think the Lindeberg condition is necessary since $\{X_{n}\}_{n\geq 1}$ does not satisfy Feller's criterion:

$\lim\limits_{n\rightarrow\infty}\max\limits_{1\leq j\leq n}\dfrac{Var[X_{j}]}{s_{n}^{2}} = \lim\limits_{n\rightarrow\infty}\dfrac{4^{n}}{\displaystyle\sum\limits_{j=1}^{n}4^{j}} = \dfrac{3}{4} \neq 0$

Back for a second try. Running a couple of simulations suggests that the normalized sum $S_n/s_n$ indeed converges, but strangely the histogram of $S_n/s_n$ seems more rectangular than bell-shaped. To understand what the limiting distribution is, calculate the characteristic function of the partial sum $S_n=X_1+\cdots+X_n$: \begin{align} E\exp(itS_n) & = \prod_{j=1}^nE \exp(itX_j)\\ &= \prod_{j=1}^n\left( {\exp(it2^j)+\exp(-it2^j)\over 2}\right)\\ &= \prod_{j=1}^n\cos(2^jt)\\ &= \prod_{j=1}^n\cos\left({2^{n+1}t\over 2^j}\right)\\ &= {\sin(2^{n+1}t)\over 2^n\sin(2t)} \end{align} where in the last step we use the identity $$\prod_{j=1}^n\cos\left({x\over2^j}\right)={\sin x\over 2^n\sin(x/2^n)}\;.$$ Now normalize by $s_n=c_n2^n$, where $c_n$ tends to a constant $c=\sqrt{4/3}$ as $n\to\infty$. We have $$E\exp(it(S_n/s_n)) = {\sin(2^{n+1}t/c_n2^n)\over 2^n\sin(2t/c_n2^n)} = {\sin(2t/c_n)\over 2^n\sin(2t/c_n2^n)}.$$ As $n\to\infty$, this last tends to $\displaystyle{\sin(2t/c)\over 2t/c}$ , which is the CF of a uniform$[-2/c,2/c]$ random variable, i.e. $S_n/s_n$ converges to a uniform$[-\sqrt3,\sqrt3]$ distribution. (Please check my algebra.)
• Wow, this is great, thank you! Can I ask how you arrived at the equality $cos(2^{j}t) = cos(\frac{2^{n+1}t}{2^{j}})$? – Quicksilver Apr 24 '15 at 22:17
• Apply the change of index $j\mapsto n+1-j$ (in preparation for applying the identity). We are just iterating in reverse order. – grand_chat Apr 24 '15 at 22:19
If you want to show the CLT holds, you'll need a criterion other than the Lindeberg condition. Note that $s_n^2< 4^{n+1}$. Use this inequality to bound the Lindeberg expression away from zero: $$\sum\limits_{j=1}^{n}E[X_{j}^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}]= \sum_{j=1}^n 4^jP(|X_j|>\epsilon s_n)\ge 4^nP(|X_n|>\epsilon s_n) \ge 4^nP(|X_n|>\epsilon 2^{n+1})\;.$$ For every small $\epsilon$ the last probability equals $1$, so $$\frac{1}{s_n^2}\displaystyle\sum\limits_{j=1}^{n}E[X_{j}^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}]\ge{4^n\over4^{n+1}}=\frac14\;,$$ i.e., the Lindeberg condition doesn't hold for your sequence.
• Thank you. How did you arrive at the conclusion that: $s_{n}^{2} = \sum\limits_{j=1}^{n}4^{j} < 4^{n+1}$? Perhaps it should be $s_{n}^{2} < n4^{n}$? – Quicksilver Apr 24 '15 at 21:27
• Sorry, it seems like a simple induction argument to prove $s_{n}^{2} < 4^{n+1}$. Thank you again. – Quicksilver Apr 24 '15 at 21:35
• I used $s_n^2=\frac43(4^n-1)$ and tried to find a nice upper bound . Your inequality is also valid, but it makes it harder to argue that $P(|X_n|>\epsilon s_n)=1$ – grand_chat Apr 24 '15 at 21:37