# Does CLT hold for joint distribution of two dependent binomial variables?

Let $S_n$ and $T_m$ be two binomial variables satisfying $S_n\sim B(n,\frac12)$ and $T_m\sim B(m,\frac12)$. Define $\tilde{S}_n=\frac{2S_n-n}{\sqrt{n}}$ and define $\tilde{T}_m$ similarly. For any fixed $s$ and $t$, it is well-known that (Central Limit Theorem) $$\mathbb \lim_{n,m\rightarrow\infty}(\mathbb P(\tilde S_n\le s),\mathbb P(\tilde T_m\le t))=(\Phi(s),\Phi(t)),$$ where $\Phi(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{x}\exp(-\frac{z^2}{2})dz$ is the distribution function of a standard normal variable.

Let $c=c_{n,m}=\mathbb E(\tilde{S}_n\tilde{T}_m)$ for short. Here $c$ may depend on $n$ and $m$.

Suppose that $\alpha\le c_{n,m}\le\beta$ holds uniformly for two absolute constants $\alpha,\beta\in(0,1)$. For fixed $s$ and $t$, it "seems reasonable" to conjecture that the joint distribution of $\tilde S_n$ and $\tilde T_m$ is close to corresponding 2-dimensional normal distribution: \begin{equation}\label{close} \mathbb P(\tilde S_n\le s,\tilde T_m\le t)=(1+o(1))\hat\Phi_c(s,t),~~~~~~~~~~~~~~~~~~~~~~~~~~(*) \end{equation} where $o(1)$ tends to 0 as $n,m$ tend to infinity and $\hat\Phi_c(s,t)$ is defined by $$\hat\Phi_c(s,t)=\frac1{2\pi\sqrt{1-c^2}}\iint_{x\le s,~y\le t}\exp\left[-\frac{x^2-2cxy+y^2}{2(1-c^2)}\right]dxdy.$$

Here are my questions:

$1.$ Is it sufficient to guarantee $(*)$ hold that $\alpha\le c\le\beta$. How about $c=\alpha$?

$2.$ If $1.$ is ture, can $o(1)$ term in $(*)$ be improved ($O(\frac1{\sqrt{n}}+\frac1{\sqrt{m}})$ for example)?

$3.$ If $1.$ is not true, can we get a good estimation of $\mathbb P(\tilde S_n\le s,\tilde T_m\le t)$ for fixed $s$ and $t$?

$\bf{Remark:}$ This problem is one of my attempts to solve Link. Any hint or reference will be appreciated. :)

## 1 Answer

I think this is much too much ask for.

Here is a construction that is not a complete counter-example, but indicates why it shouldn't be true. let $(Z^1_n)$ and $(Z^2_n)$ be two i.i.d. sequences of Bernoulli(1/2) trials. Now let $U_n$ and $V_n$ be i.i.d sequences of random variables taking the value 1 with probability 1/2 and 2 with the probability 1/2. Finally, set \begin{align*} S_n&=\sum_{i=1}^n Z^{U_n}_i\text{ and }\\ T_n&=\sum_{i=1}^n Z^{V_n}_i. \end{align*} These have the correct marginal distributions. Now $c_{n,n}=\frac 12$ for each $n$. The reason this is not a counterexample is that $c_{m,n}$ does not stay bounded away from 0 and 1 if $m$ and $n$ are not comparable.

Fairly clearly, $(\tilde S_n,\tilde T_n)$ is approximately a mixture of a 2-dimensional normal random variable (where the components are independent) and a 1-dimensional normal random variable, supported on the diagonal.

In particular, if $a$ is very negative, you have $\mathbb{P}(\tilde S_n<a,\tilde T_n<a)\approx \frac 12 \mathbb{P}(\tilde S_n<a)$, which is far bigger than the estimate you are looking for.