Digits: Convergence of number of $00$ Let $n\ge2$ and $X_1,X_2,\ldots,X_{n+1}$ be i.i.d  Bernoulli sequence $\mathcal{B}(\frac1{2}).$
Let $N=\sum_{i=1}^n Y_i$ be the number of $00$ in the sequence i.e. $Y$ is a discrete random variable such that $$Y_i=1 \;\mbox{if}\; X_{i}=0\;\mbox{and}\; X_{i+1}=0; \quad Y_i=0\; \mbox{otherwise}.$$

Does $$\frac{N-E(N)}{\sqrt{V(N)}}\overset{d,n\to\infty}{\to} \mathcal{N}(0,1)?$$

I cannot apply the CLT because $Y_i$ are dependent.
 A: Yes.  I'm not sure the best way to prove it, here is one approach (perhaps someone else can give a better/shorter proof?).  The idea is that $(Y_1+Y_2+Y_3)$  and $(Y_5+Y_6+Y_7)$ are i.i.d. since $Y_4$ is "missing." Well if we pull out every 4th $Y_i$ that is a significant change, but what if we pull out every $k$th $Y_i$ and let $k$ be large? 

Fix $k$ as an integer larger than 2 and assume $n=Mk$.  So
$$ N := \sum_{i=1}^n Y_i = \sum_{m=0}^{M-1}\sum_{j=1}^k Y_{mk+j} = \left(\sum_{m=0}^{M-1}\sum_{j=1}^{k-1} Y_{mk+j}\right) + \sum_{m=1}^{M} Y_{mk} $$
So 
$$ R := N-E[N]=\sum_{i=1}^n (Y_i - E[Y_i]) = \left(\sum_{m=0}^{M-1}\underbrace{\sum_{j=1}^{k-1} (Y_{mk+j}-E[Y_{mk+j}])}_{i.i.d.}\right) + \sum_{m=1}^{M} \underbrace{(Y_{mk}-E[Y_{mk}])}_{i.i.d.}$$
So $Var(N) = E[R^2]$. Define 
\begin{align*}
Z_m &= \sum_{j=1}^{k-1} (Y_{mk+j}-E[Y_{mk+j}])\\
V_m &= Y_{mk}-E[Y_{mk}]
\end{align*}
So $\{Z_m\}$ are i.i.d. zero mean, $\{V_m\}$ are i.i.d. zero mean, $Z_m$ and $V_j$ are independent for $|m  -j|\geq 2$,  and
$$ R = \sum_{m=0}^{M-1} Z_m + \sum_{m=1}^M V_m $$
By the central limit theorem we have
$$ \frac{R}{\sqrt{ME[Z_1^2]}} = G_M + H_M $$
where $G_M$ is zero mean and unit variance, $H_M$ is zero mean and variance $\frac{E[V_1^2]}{E[Z_1^2]}$, and the distributions of $G_M$ and $H_M$ converge to Gaussian as $M\rightarrow\infty$ (unfortunately the $G_M$ and $H_M$ are not necessarily independent). So
$$ \frac{N-E[N]}{\sqrt{Var(N)}} =\frac{R}{\sqrt{ME[Z_1^2]}}\frac{\sqrt{ME[Z_1^2]}}{\sqrt{Var(N)}}= (G_M+H_M)\frac{\sqrt{M E[Z_1^2]}}{\sqrt{Var(N)}} $$
Further, it can be shown that for all $M$ we have
\begin{align*}
&\lim_{k\rightarrow\infty} \frac{E[V_1^2]}{E[Z_1^2]}=0\\
&\lim_{k\rightarrow\infty} \frac{M E[Z_1^2]}{Var(N)} = 1
\end{align*}
So if $M$ and $k$ are both large we have
$$ \frac{N-E[N]}{\sqrt{Var(N)}} = (G_M+H_M)\underbrace{\frac{\sqrt{M E[Z_1^2]}}{\sqrt{Var(N)}}}_{\approx 1} $$
where $G_M$ is approximately Gaussian in distribution with mean 0 and variance 1, while $H_M$ is negligible since it is approximately Gaussian in distribution with mean 0 and variance that goes to 0 with large $k$.  

Hopefully that was convincing, here are some pesky details I am skipping: 


*

*This only considers $n$ of the form $n=Mk$.  A detail is to show that for general integers $n=Mk+r$, the affects of the "residual" $r$ is asymptotically negligible. 

*Details on the "it can be shown" parts. 

*Details on being precise about taking "both $M$ and $k$ large." 
