# Covariance of conditional poisson random variable sequence

Suppose $$X_0,X_1,\cdots$$ are iid $$Poisson(\theta)$$ r.v.

Define $$Y_k = X_k I_{\{ X_{k-1} = 0 \}}$$ for $$k=1,2,3,\cdots$$

Find the limit of $$Var(\sqrt{n}\overline{Y_n})$$ and asymptotic distribution of $$\sqrt{n}\left(\bar{Y}_{n}-\mathrm{E} \bar{Y}_{n}\right)$$.

It's not difficult to find $$E(Y_k) = E(X_k)P(X_{k-1}=0) = \theta exp(-\theta)$$. But when it comes to variance, I know that $$Y_k$$ is independent to $$Y_{k+i}$$ for $$i>1$$. However, I dont know how to calculate the covariance of $$Y_k$$ and $$Y_{k+1}$$. Besides, since $$Y_k$$ are not i.i.d. random variables, how can I use CLT to get the asymptotic distribution?

The covariance of $$Y_k$$, $$Y_{k+1}$$ can be found directly: $$\mathbb E(Y_kY_{k+1})=\mathbb E(X_kX_{k+1}I_{\{X_{k-1}=0, X_k=0\}}) = \mathbb E(0) = 0.$$ Indeed, if $$X_k\neq 0$$ then $$I_{\{X_{k-1}=0, X_k=0\}}=0$$. If $$X_k=0$$, then $$X_kX_{k+1}=0$$. So the product $$Y_kY_{k+1}$$ is zero for every elementary event $$\omega$$. Then $$\text{Cov}(Y_k,Y_{k+1}) = 0- \mathbb E(Y_k)\mathbb E(Y_{k+1}) = -\theta^2 e^{-2\theta}.$$ So using variance of a sum we get $$\text{Var}\left(\sqrt{n}\overline{Y_n}\right) = \frac1n \text{Var}\left(\sum_{i=1}^n Y_i\right) = \frac1n\bigl(n\text{Var}(Y_1) + 2(n-1)\cdot\text{Cov}(Y_1,Y_2) \bigr) = \ldots \tag{1}\label{1}$$ Here $$\mathbb E(Y_1^2)=\mathbb E(X_1^2)\mathbb P(X_0=0)=(\theta^2+\theta)e^{-\theta}$$, and $$\text{Var}(Y_1)=(\theta^2+\theta)e^{-\theta} - \theta^2e^{-2\theta}.$$
In our case $$m=1$$ and limiting distribution is normal with zero mean and variance $$\text{Var}(Y_1) + 2\text{Cov}(Y_1,Y_2).$$ This variance is exactly the limit of \eqref{1} as $$n\to\infty$$.