Central limit theorem for randon variables with exponentially decaying covariance Let $X_1,X_2,...$ be i.i.d. bounded random variables with $\mathbb{E}[X]=0$. In addition, let $C_1,C_2>0$ and  $\{d_{i,j}\}_{i,j\in \mathbb{N}}$ such that
$$d_{i,j} = C_1e^{-C_2|i-j|} $$
Now, define
$$Y_n = \sum_{j=1}^n d_{n,j} X_n\cdot X_j $$
Does
$$\frac{Y_1+...+Y_n}{\sqrt{n}} \overset{d}{\longrightarrow} N(0,\sigma^2)?$$
Generally the exponential decay of the covariance of the $Y_i$ is not enough. However, I think that for this type of random variable it should work, but not sure how to prove it.
Thanks!

Under the conditions stated above, the following conclusion is true,
\begin{equation*}   \frac{1}{\sqrt{n}}\sum_{k=1}^{n}(Y_k-C_1\mathsf{E}[X_1^2])\overset{d}{\longrightarrow} N(0,\sigma^2).  \tag{1}
\end{equation*}
This conclusion can be proved by CLT of array of MD (martingale difference).
(cf. P. Hall and C. C. Heyde, Martingale Limit Theory and Its Application, Academic Press(1980),  Th.3.2, p.58--).
The following is an outline of the proof.
Denote
\begin{equation*}
W_0=0, \quad W_{k-1}=\sum_{j=1}^{k-1}d_{k,j}X_j , \quad k\ge 2. 
\end{equation*}
Then
\begin{gather*}
 Y_k=\Big(\sum_{j=1}^{k}d_{k,j}X_j \Big)X_k
 =W_{k-1}X_k+d_{k,k}X_k^2\\
 \mathsf{E}[Y_k]=d_{k,k}\mathsf{E}[X_k^2]=C_1\mathsf{E}[X_1^2]
 \overset{\triangle}{=}m.
\end{gather*}
Let
\begin{gather*}
 Z_{n,k}=\frac{1}{\sqrt{n}}(Y_k-m), \quad 1\le k\le n, n\ge 1.\\
 \mathscr{F}_{k}=\sigma\{ X_j,1\le j\le k\}\vee\mathscr{N},\quad 1\le k\le n, n\ge 1.
\end{gather*}
Then
\begin{equation*}
 \mathsf{E}[Z_{n,k}|\mathscr{F}_{k-1}]
 =\frac{1}{\sqrt{n}}(W_{k-1}\mathsf{E}[X_k]+ 
 d_{k,k}\mathsf{E}[X_k^2]-m)=0.
\end{equation*}
and $ Z=\{ Z_{n,k}, \mathscr{F}_{k}, 1\le k\le n, n\ge 1\}$ is a MD-array(array of martingale difference). Now we verify that the $Z$ satisfy the conditions of $S_n=\sum_{k\le n}X_{n,k} \overset{d}{\to} N(0,\sigma^2)$.
At first, due to the $\{X_i,i\ge 1\} $  are bounded, the $\{W_i,i\ge 1\}, \{Y_i,i\ge 1\} $
are bounded too, and
\begin{equation*}
 \max_{1\le k\le n}|Z_{n,k}|\le \frac{C}{\sqrt{n}} \tag{2}
\end{equation*}
where and latter the $C$ is a constant, irrelated $k,n$, in different expression $C$ may be different.
Secondly, using direct calculation, the following holds,
\begin{align*}
 \lim_{k\to\infty}\frac1k\sum_{j=1}^{k}W_j&=0, \quad \text{a.s.} \tag{3}\\ 
 \lim_{k\to\infty}\frac1n\sum_{j=1}^{n}W_j^2&=b>0,\quad\text{a.s.}  \tag{4}
\end{align*}
Hence,
\begin{align*}
 &\mathsf{E}[Z_{n,k}^2|\mathscr{F}_{k-1}]\\
 &\quad =\frac1n\mathsf{E}[(W_{k-1}X_k+C_1(X_k^2-\mathsf{E}[X_k^2]))^2 | \mathscr{F}_{k-1}]\\
 &\quad =\frac1n[W_{k-1}^2\mathsf{E}[X_k^2]]+C_1^2\mathsf{E}[(X_k^2-\mathsf{E}[X_k^2])^2]\\
 &\qquad +2C_1W_{k-1}\mathsf{E}[(X_k^2-\mathsf{E}[X_k^2])X_k],\\
 &\sum_{k=1}^{n}\mathsf{E}[Z_{n,k}^2|\mathscr{F}_{k-1}]\\
 &\quad = \frac1n \sum_{k=1}^{n}W_k^2\mathsf{E}[X_1^2] + 
 C_1^2 \mathsf{E}[(X_1^2-\mathsf{E}[X_1^2])^2] \\
 &\qquad  + \frac{2C_1\mathsf{E}[(X_1^2-\mathsf{E}[X_1^2])X_1]}n \sum_{k=1}^{n}W_{k-1}\\
 &\quad \to b\mathsf{E}[X_1^2]+C_1^2 \mathsf{E}[(X_1^2-\mathsf{E}[X_1^2])^2]
  \overset{\triangle}{=}\sigma^2.  \tag{5}
\end{align*}
At last, from (2) and (5), we have
\begin{align*}
 \frac{1}{\sqrt{n}}\sum_{k=1}^{n}(Y_k-C_1\mathsf{E}[X_1^2])
 =\sum_{k=1}^{n}Z_{n,k}=S_n\overset{d}{\longrightarrow}N(0,\sigma^2).
\end{align*}
i.e. (1) is true.
