# CLT for Bernoulli RV with negative correlation

Suppose $X_1,X_2,...$ are Bernoulli random variables with $P(X_i=1)=p_i$ and $X_i$ have negative correlation. Is there a CLT in this case, i.e. does $\frac{Z_n-(\Sigma^n_{i=1}p_i)}{\sqrt{n}}$ converge to a Gaussian in distribution?

And if we add the assumption that $\underset{n \longrightarrow \infty}{lim} \frac{\Sigma^n_{i=1}p_i}{n}=p\in(0,1)$, is it true then?

• Just a remark "negatively correlated brenoulli RV" is quite an underspecified statement (unless n=2) as it is not possible for all the the bernoulli's variables to be negatively correlated because at least two of them must exhibit positive correlation otherwise the joint law is not properly defined. Regards – The Bridge Jan 26 '18 at 14:22

No, the CLT need not hold under these assumptions. Consider the following example: take $p=1/2$ for definiteness, and divide the (discrete) time into intervals $I_1=[1,2]$, $I_n=(2^{n-1}, 2^n]$, $n\geq 2$. On each interval, let $(X_k, k\in I_n)$ be the i.i.d. Bernoulli($1/2$) conditioned on $\sum_{k\in I_n}X_k=|I_n|/2$ (think about placing $|I_n|/2$ balls into $|I_n|$ urns at random); clearly, $(X_i,X_j)$ are negatively correlated when $i,j\in I_n$ for some $n$. Then the CLT doesn't hold since the sum becomes deterministic from time to time.
The correlations are not strictly negative though (because the $X$'s are independent if belong to different intervals), but you can probably make them so by some "small" perturbation.