Say I have $N$ random variables $X\_1,\cdots,X\_i,\cdots,X\_N$, with zero mean and finite variance. $X\_i$ and $X\_j$ are independent iif $|i-j|>m$, and positively correlated otherwise (say the covariance is of $\mathcal{O}(1)$). It is well-known that the sums of $N$ of these random variables are distributed as a normal distribution as $N \rightarrow \infty$, if $m$ is a finite number of $\mathcal{O}(1)$.


My question is, what if $m=\sqrt{aN}$, and we take a sum of $\sqrt{aN}$ of these random variables, randomly selected from the sequence $X\_1,\cdots,X\_i,\cdots,X\_N$. Would the sum become a normal distribution? Each time we take a sum, we resample $\sqrt{aN}$ random variables and take a sum of them.


My intuition says yes. The reason is if we randomly sample $\sqrt{aN}$ variables from the original sequence of length $N$ (and $m=\sqrt{aN}$), the expected value of the number of variables sampled from any given window of consecutive $\sqrt{aN}$ variables is $a$, based on the following calculation:


$$\text{number of samples}\times\text{probability of sampling from a given window}=\sqrt{aN} \frac{\sqrt{aN}}{N} = a$$


Now it is as if we are taking a sum of $\sqrt{aN}$ random variables that are independent if $|i-j|>a$, so the conventional '$m$-dependent CLT' still holds since our $a$ is of $\mathcal{O}(1)$. Is this intuition correct? How do I prove/disprove this more rigorously?



Weakly dependent central limit theorem
The answer is yes. Say we have a stationary process, but we observe samples at random times $\{t\_n\}$ which itself is a stochastic point process (e.g. Poisson process). The resulting sample is also a stationary process ([Ref](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.851.9334&rep=rep1&type=pdf)).


I assume that if the original sequence is m-dependent ($m=\mathcal{O(1)}$), the resulting autocovariance function (for the sampled sequence) decays fast enough that CLT holds for the sum of the sampled sequence.


