Suppose we have $X_1,\ldots,X_n\stackrel{{\rm i.i.d.}}\sim N(0,1)$, then we have $$ \dfrac{(X_1+\cdots+X_n)^2}{n} = \big(\sqrt{n} \bar{X}\big)^2 \sim \chi^2_1. $$ Applying law of large numbers to the mean of square terms and using Slutsky, the degenerate U-statistic $Z_{n,n}$ satisfies $$ Z_{n,n}=\dfrac{2\sum_{1\leq i<j\leq n} X_iX_j}{n}\sim {\rm(approximately)}\ \chi^2_1-1, $$ where define $$ Z_{n,k_n}:=\dfrac{2\sum_{1\leq i<j\leq k_n} X_iX_j}{n}. $$ The main point is that the limiting distribution of $Z_{n,n}$ is not normal.
But now I'm confused when checking why this doesn't (as it shouldn't) satisfy the conditions of the martingale central limit theorem (Theorem 6.2, bottom of page 24 in Gao & Lafferty (2017) https://arxiv.org/pdf/1704.06742.pdf).
- Check: for each $n$, $Z_{n,k_n}$ with $k_n$ varying is indeed a martingale, with martingale difference $Y_{n,m} = Z_{n,m}-Z_{n,m-1} = 2X_m(X_1+\cdots+X_{m-1})/n$.
- Check: the variance condition: $\sum_{m=1}^n\mathbb{E}[Y_{n,m}^2|X_1,\ldots,X_{m-1}] = n^{-2} \color{red}{\sum_{m=2}^{n-1}} ( 4\mathbb{E}[X_m^2] (X_1+\cdots+X_{m-1})^2 ) \to 2$ (Thanks to @davide-giraudo for the correction in red)
- Check: the Lindeberg condition: $\sum_{m=1}^n\mathbb{E}[Y_{n,m}^2 \cdot 1_{[|Y_{n,m}|>\epsilon]}|X_1,\ldots,X_{m-1}]$. Using the same argument as in Gao & Lafferty (2017), we have \begin{align} & \Big\{\mathbb{E}[Y_{n,m}^2 \cdot 1_{[|Y_{n,m}|>\epsilon]}|X_1,\ldots,X_{m-1}]\Big\}^2\\ \leq & \Big\{\mathbb{E}[Y_{n,m}^4|X_1,\ldots,X_{m-1}]\Big\} \cdot \Big\{\mathbb{P}[|Y_{n,m}|>\epsilon|X_1,\ldots,X_{m-1}]\Big\}\\ \leq & \Big\{\mathbb{E}[Y_{n,m}^4|X_1,\ldots,X_{m-1}]\Big\} \cdot \Big\{\mathbb{P}[|X_m|>\dfrac{n\epsilon}{|X_1+\cdots+X_{m-1}|}|X_1,\ldots,X_{m-1}]\Big\}\\ \ll & \Big\{\mathbb{E}[Y_{n,m}^4|X_1,\ldots,X_{m-1}]\Big\} \cdot \exp\Bigg\{ - \dfrac{n^2\epsilon^2/\log n}{2|X_1+\cdots+X_{m-1}|^2}\Bigg\}. \end{align} With overwhelming probability, we have $$ \dfrac{n^2\epsilon^2/\log n}{2|X_1+\cdots+X_{m-1}|^2}\gg n^{1/2} $$ So the Lindeberg condition is checked. Then the martingale CLT implies the limiting distribution of $Z_{n,n}$ should be some normal distribution?
There must be something wrong with the above derivation, but I was unable to locate it...
Please advise me. Thank you so much for your help!
The sum of conditional variances is $$ \frac 4{n^2}\sum_{m=2}^n\left(\sum_{i=1}^{m-1}X_i\right)^2 $$ (the sum over $m$ was missing in the opening post ). It is not clear whether it converges in probability to a constant, so this is the direction you should look at.