Covariance of conditional poisson random variable sequence

Question

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 $\sqrt{n}\left(\bar{Y}_{n}-\mathrm{E} \bar{Y}_{n}\right)$.

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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?

Answer 1

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 $$ \text{Var}(Y_1)=(\theta^2+\theta)e^{-\theta} - \theta^2e^{-2\theta}. $$

Concerning the limiting distribution, I’m not sure that one can use the well-known fact that a sequence of one-dependent random variables satisfies the CLT. Look at this paper for example.

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 $n\to\infty$.