Let $\{X_n\}_{n \in \mathbb{N}}$ be a sequence of of IID random variables taken for simplicity with mean zero and variance one.
The Central Limit Theorem give us that
$$
\frac{X_1 + \dots + X_n}{\sqrt{n}} \xrightarrow {d} N\left(0,1\right)
$$
If one constructs a new sequence $\{Y_n\}_{n \in \mathbb{N}}$ from the first one given by the square of the sum of two consecutive terms, i.e.
$$Y_1 = (X_1 + X_2)^2, Y_2 = (X_2 + X_3)^2, \dots , Y_n = (X_n + X_{n+1})^2 $$
do there exist two sequences $\{\mu_n\}_{n \in \mathbb{N}}$ and $\{\sigma_n\}_{n \in \mathbb{N}}$ s.t.
$$\frac{1}{\sigma_i^2} \sum_{i=1}^n ( Y_i – \mu_i) \rightarrow N(0,1) $$
I was thinking of using some Lyapunov type central limit theorem to prove this but there is an obvious (weak) dependence in the sequence. Is it possible to show this or is it not true?
Best Answer
Let $\left(X_i\right)_{i\geqslant 1}$ be an i.i.d. sequence and let $f\colon \mathbb R^2\to \mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.
Let $n$ be a fixed integer and $q\in\left\{1,\dots,n\right\}$. We write \begin{align} \sum_{i=1}^nY_i&= \sum_{i=1}^{q\left\lfloor \frac nq\right\rfloor}Y_i+ \sum_{q\left\lfloor \frac nq\right\rfloor}^nY_i\\ &= \sum_{k=1}^{\left\lfloor \frac nq\right\rfloor} \sum_{i=(k-1)q+1}^{kq}Y_i+\sum_{q\left\lfloor \frac nq\right\rfloor}^nY_i\\ &= \sum_{k=1}^{\left\lfloor \frac nq\right\rfloor} \sum_{i=(k-1)q+2}^{kq}Y_i+\sum_{k=1}^{\left\lfloor \frac nq\right\rfloor}Y_{(k-1)q+1}+\sum_{q\left\lfloor \frac nq\right\rfloor}^nY_i. \end{align} Denoting $Z^q_k:= \sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $\left(Z^q_k\right)_{k\geqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}\sum_{k=1}^{\left\lfloor \frac nq\right\rfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.