I was reading an article by Kac and he used a lemma by Markov which states:

Let $(X_n)_{n>1}$ be a sequence of centred random variables. Suppose there exists an integer $L$ such that every two collections : $X_{i_1}, \dots, X_{i_p}$ and $X_{j_1}, \dots, X_{j_q}$ are independent whenever $j_1-i_p>L,$ and $i_1<i_2<\cdots <i_p < j_1 < \cdots < j_q.$ In addition, we assume that $X_1, X_2, \dots$ are uniformly bounded and $\frac{1}{n}\left(X_1+\cdots + X_n\right)^2 \to 1,$ when $n \to \infty.$ Under these assumptions, $$\frac{1}{\sqrt{n}}\left(X_1+\cdots + X_n\right)$$ converges in distribution to $\mathcal{N}(0,1)$.

I couldn't find Markov's result as it is pretty old (1912). Could you please suggest me a reference where I can find a proof of this result?


You can use the method of moments. According to this method, since your random variables are "nice" (in this case, they are uniformly bounded), it is enough to show that the individual moments of $\frac{X_1+\cdots+X_n}{\sqrt{n}}$ converge to the corresponding moments of $\mathcal{N}(0,1)$. A standard calculation shows that when all $X_i$ are independent then this indeed happens, see for example here. In your case there will be added error terms that arise from estimating products of variables that are too close, but fortunately these terms would be asymptotically nil. Here is an example using the third and fourth moments – I'll let you figure out how to generalize it. (For the second moment we are given that it converges to 1.)

For the third moment we have $$ \mathbb{E}\left[ \frac{(X_1+\cdots+X_n)^3}{n^{3/2}} \right] = \frac{1}{n^{3/2}} \sum_i \mathbb{E}[X_i^3] + \frac{3}{n^{3/2}}\sum_{i\neq j} \mathbb{E}[X_i^2 X_j] + \frac{6}{n^{3/2}} \sum_{i<j<k} \mathbb{E}[X_i X_j X_k]. $$ The first term is $O(B^3/\sqrt{n})$, where $B$ is the uniform bound on all $X_i$. The second term we split into two parts: $|i-j| > L$ and $|i-j| \leq L$. The first part vanishes since the variables are centered. The second part is $O(LB^3/\sqrt{n})$. The third term we again split into two parts: $|i-j|,|i-k|,|j-k| \leq L$, and its complement. The second part vanishes since the variables are centered. The first part is $O(L^2B^3/\sqrt{n})$.

For the fourth moment we have $$ \mathbb{E}\left[\frac{(X_1+\cdots+X_n)^4}{n^2}\right] = \frac{1}{n^2} \sum_i \mathbb{E}[X_i^4] + \frac{4}{n^2} \sum_{i \neq j} \mathbb{E}[X_i^3 X_j] + \frac{6}{n^2} \sum_{i < j} \mathbb{E}[X_i^2 X_j^2] + \frac{6}{n^2} \sum_{i,j<k} \mathbb{E}[X_i^2 X_j X_k] + \frac{24}{n^2} \sum_{i<j<k<l} \mathbb{E}[X_i X_j X_k X_l]. $$ The first term is $O(B^4/n)$. The second term is $O(B^4L/n)$. The third term we leave for now. The fourth term is $O(B^4L^2/n)$. The fifth term is $O(B^4L^3/n)$. Finally, we break the third term into two parts: $|i-j| \leq L$ and $|i-j| > L$. The contribution of the first part is $O(B^4L/n)$. The second part is $$ \frac{3}{n^2} \sum_{|i-j|>L} \mathbb{E}[X_i^2] \mathbb{E}[X_j^2] = \frac{3}{n^2} \sum_{i,j} \mathbb{E}[X_i^2] \mathbb{X_j^2} - O\left(\frac{B^4L}{n}\right) = \\ 3 \mathbb{E}\left[\frac{X_1^2 + \cdots + X_n^2}{n}\right]^2 - O\left(\frac{B^4L}{n}\right) = 3 + o(1). $$ So the third moment tends to 0, and the fourth moments tends to 3. Similar calculations work for all other moments.

(My error estimates might contain mistakes. Please let me know and I will correct them.)

  • $\begingroup$ Thank you a lot for the solution. In the calculations of the third moment I think $\sum_{|i-j| \geq L} \mathbb{E}[X^2_i X_j] = 0.$ $\endgroup$ – Hovher Nov 12 '15 at 23:06
  • $\begingroup$ Do you think there is a way to do this using characteristic functions? $\endgroup$ – Hovher Nov 12 '15 at 23:13
  • $\begingroup$ Possibly. Try it out and let us know whether it worked. $\endgroup$ – Yuval Filmus Nov 12 '15 at 23:14
  • $\begingroup$ thank you. I will let you know. $\endgroup$ – Hovher Nov 12 '15 at 23:16

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