Problem Setting.
Suppose $\{X_n\}_{n\in\mathbb{N}}$ are independent, identically distributed random variables with mean $m$ and variance $\sigma^2$. Consider the sequence of random variables $\{Y_n\}_{n\in\mathbb{N}}$ given by
$$
Y_n = \sum_{j=1}^n \sum_{i=1}^{j-1} X_i X_j, \qquad n\in\mathbb{N}
$$
where we take $Y_1 =0$.
Problem. Determine centering constants $\{\alpha_n\}_{n\in\mathbb{N}}$ and norming constants $\{\beta_n\}_{n\in\mathbb{N}}$ such that the sequence of random variables $\{(Y_n – \alpha_n)/\beta_n\}_{n\in\mathbb{N}}$ converges in distribution to some non-degenerate random variable $\xi$
$$
\frac{Y_n – \alpha_n}{\beta_n} \Longrightarrow \xi
$$
My Attempt.
I thought of rewriting the random variables $Y_n$ as
$$
Y_n = \sum_{j=1}^n \underbrace{X_j \sum_{i=1}^{j-1}X_i}_{=Z_j} = \sum_{j=1}^n Z_j, \qquad Z_j = X_j\sum_{i=1}^{j-1}X_i
$$
Written in this way, the desired convergence resembles the central limit theorem (even though we are not restricted to $\xi \sim N(0,1)$).
The difficulty I'm encountering is the fact that the sequence of random variables $\{Z_j\}_{j\in\mathbb{N}}$ are not independent, which means that any of the versions of the CLT that I'm familiar with (e.g., with Lindeberg condition, Lyapunov condition, or triangular arrays) do not immediately apply. Additionally, it makes computing the characteristic function of $(Y_n – \alpha_n)/\beta_n$ very difficult since it doesn't factor into a product of characteristic functions:
$$
\varphi_{(Y_n – \alpha_n)/\beta_n}(\lambda) = \mathbb{E}e^{i\lambda(Y_n – \alpha_n)/\beta_n} = e^{-i\lambda\alpha_n/\beta_n}\mathbb{E}\prod_{j=1}^ne^{i\lambda Z_j/\beta_n}
$$
I'm thinking of working instead with a version of $\{Z_j\}_{j\in\mathbb{N}}$, say $\{W_j\}_{j\in\mathbb{N}}$, on a different probability space that are independent and such that $Z_j$ and $W_j$ are identically distributed. I just have no clue how to proceed!
Thank you for your help 🙂
Best Answer
Please use the following equality: $$ Y_n=\frac12\Bigl[\Bigl(\sum_{i=1}^nX_i\Bigr)^2-\sum_{i=1}^nX_i^2\Bigr] $$