Central Limit Theorem/Weak Convergence for a sequence of Dependent Identically Distributed Random Variables

Question

I am familiar with the usual CLT's, but I was wondering whether there would be a corresponding CLT or for the following sort of sum of random variables:

Let $\{u_t\}_{t\in\mathbb{N}}\sim $ iid $(0,\sigma^2)$. Then let us observe that $\forall j=1,...,n-1$ the random variable $u_n^2$ is independent of $u_j^2$ and thus: \begin{equation*} \mathbb{E}\left[ u_nu_j\right] =\mathbb{E}\left[ u_n\right] \mathbb{E}\left[ u_j\right] =0 \qquad \text{ and } \qquad \text{Var}(u_nu_j) =\mathbb{E}\left[ u_n^2u_j^2\right] = \mathbb{E}\left[ u_n^2 \right]\mathbb{E}\left[u_j^2\right] = \sigma^4 \end{equation*} So that the sequence of random variables $\{u_{n+1}u_j\}_{j=1}^n$ for each $n\in\mathbb{N}$ is identically distributed with mean 0 and variance $\sigma^4$ but the sequence of random variables across $j$ is dependent. Then does the sum of random variables: \begin{equation*} \sum_{j=1}^{n} u_{n+1}u_j \quad \overset{\mathcal{D}}\longrightarrow \quad??? \end{equation*}

i.e. After some appropriate normalisation/rescaling converge to something meaningful? (if it does at all...)as $n\rightarrow\infty$)

Thanks for any help in advanced!

Answer 1

If you rescale with $\sqrt{n},$ you get $u_{n+1}\frac{1}{\sqrt{n}}\sum_{j=1}^nu_j.$

$u_{n+1}$ is from it's distribution and $\frac{1}{\sqrt{n}}\sum u_j$ converges in distribution to an independent normal so their distribution is whatever the product of independent samples from a normal and the parent distribution is.

Answer 2

A normalization is required. Let $X$ be a random variable independent of the sequence $\left(u_i\right)_{i\in\mathbb N}$ and having the same distribution as $u_0$ (enlarging the probability space if needed). Then for any positive integer $n$, the random variable $Y_n$ and $Z_n$ defined by $$Y_n=u_ {n+1 } \frac 1 { \sqrt n} \sum_{i=1}^nu_i \mbox{ and }Z_n=X \frac 1 { \sqrt n} \sum_{i=1}^nu_i$$ have the same distribution. The sequence $ \left(Z_n\right)_{n\geqslant 1}$ converges in distribution to $X\cdot N$, where $N$ is a zero-mean Gaussian random variable with variance $\sigma^2$.