# Central limit theorem for asymptotically i.i.d. random variables

I observe a sequence of r.v. $$X_1, X_2, \dots$$ where each $$X_i$$ is a function of the sample size $$n$$.

When $$n \rightarrow \infty$$ I have the following result: $$X_1 \rightarrow^d E_1, X_2 \rightarrow^d E_2, \dots$$ where $$E_i$$ are i.i.d random variable with mean $$\mu$$, variance $$\sigma^2 < \infty$$ and same density $$f_E$$. Moreover,

• $$\lim_{n \rightarrow \infty}E(X_i) = \mu$$
• $$\lim_{n \rightarrow \infty}V(X_i) = \sigma^2$$
• $$\textrm{Cov}(X_i, X_j) = O(n^{-1})$$

Denote $$\bar X_n = n^{-1}\sum_{i = 1}^n X_i$$, can I claim that (Linderberg-Levy CLT type)

$$\sqrt{n}\left(\bar{X}_{n}-\mu\right) \stackrel{d}{\rightarrow} \mathcal{N}\left(0, \sigma^{2}\right) ?$$

In other words: if the dependence of the elements of the summation fades only asymptotically, does the CLT still holds?

• "X_1 converges in distribution to E_1" doesn't make sense as X_1 isn't a sequence, it's a single random variable. Same thing for X_2, X_3, etc. The question doesn't make sense as it stands.
– Paul
Aug 11, 2022 at 19:20
• @Paul Strictly speaking you are correct, but let's help the new contributor,: they obviously imply that each $X_i$ is a function of $n$, hence the convergence result. Standard example: the $X$'s are in reality residuals from a regression with i.i.d. errors. Then each residual converges to the corresponding true error, and the question acquires meaning and substance: if dependence of the elements of the sum vanish only asymptotically, does the CLT still holds? Under perhaps additional conditions? Aug 13, 2022 at 13:58
• @AlecosPapadopoulos thank you! It is indeed my question. Aug 13, 2022 at 14:15
• I don't find the missing information obvious. The question cannot accept answers until it is edited to make sense. Now it seems to be OK so I voted to reopen.
– Paul
Aug 13, 2022 at 14:20
• Now that the question makes sense, the problem here is that convergence in distribution is too weak and provides no insurance against correlations between the X_i. The condition that the E_i are independent does not help. In fact, the X_i could all be the exact same variable and still converge in distribution to "iid E_i". You will need a stronger form of convergence to get the X_i to become "more independent" as n increases.
– Paul
Aug 13, 2022 at 14:27

The problem here is that convergence in distribution is too weak and provides no insurance against correlations between the $$X_i$$. The condition that the $$E_i$$ are independent does not help. In fact, the $$X_i$$ could all be the exact same variable and still converge in distribution to the iid $$E_i$$. You will need a stronger form of convergence to get the $$X_i$$ to become less dependent as $$n$$ increases - convergence in probability at minimum, or perhaps something stronger.
The condition of $$O(n^{-1})$$ decaying correlation unfortunately doesn't help either, as the central limit theorem does not hold on merely uncorrelated random variables.