禁用中心极限定理时，具有特定协方差结构的样本均值的极限分布求解问询

Hey there! Let's work through this problem step by step since you can't use the Central Limit Theorem. First, let's recap the given conditions to make sure we're aligned:

Let $X_1,\dots, X_n$ be identically distributed with mean $E[X_1]=\mu$ and $\operatorname{Var}[X_1]=\sigma^2$. Assume that $\operatorname{Cov}(X_k, X_{k+1})\neq 0$ for $k=1,\dots, n-1$ but $\operatorname{Cov}(X_k, X_l)=0$ for $|k-l|\ge 2$.

Step 1: Calculate the expected value of $\bar{X}$ #

This part is straightforward, regardless of the covariance structure:
$$E[\bar{X}] = E\left[\frac{1}{n}\sum_{i=1}^n X_i\right] = \frac{1}{n}\sum_{i=1}^n E[X_i] = \frac{1}{n} \cdot n\mu = \mu$$

Step 2: Calculate the variance of $\bar{X}$ #

To find the limiting behavior, we need to look at how the variance of $\bar{X}$ behaves as $n$ grows. Start with the variance of the sum $\sum_{i=1}^n X_i$:
$$\operatorname{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \operatorname{Var}(X_i) + 2\sum_{1 \le i < j \le n} \operatorname{Cov}(X_i, X_j)$$

Break this down using the problem's covariance rules:

• Each $\operatorname{Var}(X_i) = \sigma^2$, so the first sum equals $n\sigma^2$.
• Only adjacent pairs ($j = i+1$) have non-zero covariance. Let's denote $\gamma = \operatorname{Cov}(X_k, X_{k+1})$ (since the $X_i$ are identically distributed, this value is consistent across all adjacent pairs). There are exactly $n-1$ such pairs, so the covariance sum becomes $2(n-1)\gamma$.

Combine these results:
$$\operatorname{Var}\left(\sum_{i=1}^n X_i\right) = n\sigma^2 + 2(n-1)\gamma$$

Now divide by $n^2$ to get the variance of $\bar{X}$:
$$\operatorname{Var}(\bar{X}) = \frac{n\sigma^2 + 2(n-1)\gamma}{n^2} = \frac{\sigma^2}{n} + \frac{2(n-1)\gamma}{n^2} = \frac{\sigma^2 + 2\gamma}{n} - \frac{2\gamma}{n^2}$$

Step 3: Analyze the limiting behavior #

As $n \to \infty$:

• The term $\frac{\sigma^2 + 2\gamma}{n}$ approaches 0, and $\frac{2\gamma}{n^2}$ also approaches 0. So $\operatorname{Var}(\bar{X}) \to 0$ as $n$ grows large.

We can use Chebyshev's Inequality here (since CLT is off-limits). For any $\epsilon > 0$, Chebyshev's Inequality states:
$$P\left(|\bar{X} - \mu| \ge \epsilon\right) \le \frac{\operatorname{Var}(\bar{X})}{\epsilon^2}$$

Since $\operatorname{Var}(\bar{X}) \to 0$, the right-hand side of this inequality tends to 0 as $n \to \infty$. This means $\bar{X}$ converges in probability to $\mu$.

Step 4: Define the limiting distribution #

When a random variable converges in probability to a constant, its limiting distribution is the degenerate distribution that assigns probability 1 to the constant $\mu$. In plain terms, as $n$ becomes extremely large, $\bar{X}$ is almost certainly equal to $\mu$.

To summarize: We didn't need CLT here because we're looking at convergence to a fixed constant, not a normal distribution. The key was showing the variance of $\bar{X}$ vanishes at infinity, then using Chebyshev's Inequality to confirm convergence in probability, which translates to a degenerate limiting distribution centered at $\mu$.

备注：内容来源于stack exchange，提问作者Hermi