Probability Theory – Problem 55: Prove that there exist constants and such that converges in distribution to the standard normal distribution

Let $\xi_{k},k\in Z$ be independent and identically distributed with $P(\xi_{1}=1)=P(\xi_{1}=0)=1/2$. Define $$\eta_{n}=\left\{\begin{array}{l l}{{1,}}&{{\mathrm{if}\ \xi_{n}=1,\xi_{n+1}=0;}}\\ {{0,}}&{{\mathrm{otherwise.}}}\end{array}\right.$$ Let $\displaystyle S_{n}=\eta_{1}+\cdots+\eta_{n}$. Prove that there exist constants $\mu$ and $\sigma$ such that $(S_{n}-n\mu)/(\sigma{\sqrt{n}})$ converges in distribution to the standard normal distribution.

Step 1. Proof that $\{\eta_k\}$ is a stationary sequence. By hypothesis, $\xi_k\ (k\in\mathbb{Z})$ are i.i.d., and $\eta_k = I(\xi_k=1, \xi_{k+1}=0)$ (where $I(\cdot)$ denotes the indicator function, taking value 1 when the condition holds and 0 otherwise). For any integer $m\geq1$ and $k_1<k_2<\cdots<k_m$, the joint distribution of $(\eta_{k_1}, \eta_{k_2}, \dots, \eta_{k_m})$ depends only on the joint distribution of $\xi$: $\eta_{k_i} = I(\xi_{k_i}=1, \xi_{k_i+1}=0)$, and its joint probability is determined by the i.i.d. property of $\xi_{k_1},\xi_{k_1+1},\dots,\xi_{k_m},\xi_{k_m+1}$. After a time shift, $(\eta_{k_1+t}, \eta_{k_2+t}, \dots, \eta_{k_m+t})$ corresponds to $\xi_{k_1+t},\dots,\xi_{k_m+t+1}$; since $\xi$ is stationary, the two joint distributions are identical. Therefore, $\{\eta_k\}$ is a stationary sequence. Step 2. Compute the constant $\mu = \mathbb{E}[\eta_k]$. Since $\eta_k$ is a 0-1 random variable, $\mathbb{E}[\eta_k] = P(\eta_k=1) = P(\xi_k=1, \xi_{k+1}=0)$. By the i.i.d. property of $\xi_k$ with $P(\xi_1=1)=P(\xi_1=0)=1/2$: $$P(\xi_k=1, \xi_{k+1}=0) = P(\xi_k=1) \cdot P(\xi_{k+1}=0) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$ Hence $\mu = \frac{1}{4}$. Step 3. Compute the autocovariance function $\gamma(j) = \text{Cov}(\eta_1, \eta_{1+j})$ and verify absolute summability. The covariance is $\gamma(j) = \mathbb{E}[\eta_1\eta_{1+j}] - \mathbb{E}[\eta_1]\mathbb{E}[\eta_{1+j}]$. Cases for $j\geq0$: - When $j=0$: $\gamma(0) = \text{Var}(\eta_1)$. Since $\eta_1^2 = \eta_1$, $\mathbb{E}[\eta_1^2] = 1/4$, so $$\gamma(0) = \frac{1}{4} - \left(\frac{1}{4}\right)^2 = \frac{3}{16}$$ - When $j=1$: $\eta_1 = I(\xi_1=1, \xi_2=0)$, $\eta_2 = I(\xi_2=1, \xi_3=0)$. If $\eta_1=1$ then $\xi_2=0$, forcing $\eta_2=0$, so $\eta_1\eta_2=0$. Thus $$\mathbb{E}[\eta_1\eta_2] = 0$$ and $\gamma(1) = -\frac{1}{16}$. - When $j\geq2$: The index sets $\{\xi_1,\xi_2\}$ and $\{\xi_{1+j},\xi_{2+j}\}$ do not overlap; by independence, $\eta_1$ and $\eta_{1+j}$ are independent, so $\gamma(j) = 0$. Absolute sum: $$\sum_{j=-\infty}^\infty |\gamma(j)| = \frac{3}{16} + 2 \cdot \frac{1}{16} = \frac{5}{16} < \infty$$ Step 4. Proof that $\{\eta_k\}$ is ergodic. Since $\xi_k$ is i.i.d., it is ergodic. Since $\eta_k = I(\xi_k=1, \xi_{k+1}=0)$ is a measurable function of $(\xi_k, \xi_{k+1})$, and a measurable transformation of a stationary ergodic sequence remains ergodic, $\{\eta_k\}$ is ergodic. Step 5. Application of the CLT for stationary ergodic sequences. For the stationary ergodic sequence $\{\eta_k\}$, if (1) $\mathbb{E}[\eta_k] = \mu$ is finite and (2) the autocovariance $\gamma(j)$ is absolutely summable, then $$\frac{S_n - n\mu}{\sqrt{n \cdot \sigma^2}} \xrightarrow{d} N(0,1)$$ where $\sigma^2 = \gamma(0) + 2\sum_{j=1}^\infty \gamma(j)$. Computing: $$\sigma^2 = \frac{3}{16} + 2\left(-\frac{1}{16}\right) = \frac{1}{16}$$ Hence $\sigma = \frac{1}{4}$. Step 6. Conclusion. There exist constants $\mu = \frac{1}{4}$ and $\sigma = \frac{1}{4}$ such that $$\frac{S_n - n\mu}{\sigma \sqrt{n}} \xrightarrow{d} N(0,1).$$

QED.