等相关：7 个变量的最小相关系数

考虑样本均值 $\overline{X}=\frac{1}{7}\sum_{i=1}^7 X_i$。

有

$$\mathrm{Var}(\overline{X})=\frac{1}{49}\left(\sum_{i=1}^7\mathrm{Var}(X_i)+\sum_{i\ne j}\mathrm{Cov}(X_i,X_j)\right) =\frac{1}{49}(7+42\rho)\ge 0.$$

因此 $7+42\rho\ge 0\Rightarrow \rho\ge -\frac{1}{6}$。

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Original Explanation

We are going to consider $\text{Var}(\overline{X})$ here, where $\overline{X} = \frac{X_1 + \dots + X_7}{7}$. By plugging this in and using properties of variance, we see that

$$\text{Var}(\overline{X}) = \frac{1}{49} \text{Var}(X_1 + \dots + X_7) = \frac{1}{49}\left[\sum_{i=1}^7 \text{Var}(X_i) + \sum_{i \neq j} \text{Cov}(X_i,X_j)\right]$$

Since we know the mean and variance of each of the random variables, we know that the first sum is just $7$. Similarly, we know that $\text{Cov}(X_i,X_j) = \rho(1)(1) = \rho$ and that there are $7 \cdot 6 = 42$ terms in that sum. Therefore, the second sum is just $42\rho$. Thus,

$$\text{Var}(\overline{X}) = \frac{7 + 42\rho}{49}$$

Our condition is that $\text{Var}(\overline{X}) \geq 0$, as the variance of any random variable must be non-negative. Thus, we can disregard the denominator and find that

$$7 + 42\rho \geq 0 \iff \rho \geq -\frac{1}{6}$$

Therefore, our answer is $-\frac{1}{6}$.