Show that in this case $\rho \geq -\frac{1}{n-1}$

Question

The bounds of correlation coefficient $\rho$ is shown to be $\pm 1$ in class. In many situations the bounds are sharper, i.e. they stay away from $+1$ or $-1$. Consider the random variables $X_1,\dots,X_n$ such that $E(X_i)=\mu$, $Var(X_i)=\sigma^2$, and for each pair $i \neq j$, $Corr(X_i,X_j)=\rho$. Such equi-correlation structure is sometimes referred to as exchangeable or compound symmetry. Show that in this case $$\rho \geq -\frac{1}{n-1}$$

Answer 1

Using that $$\begin{align*}0\le\operatorname{Var}(\overline{X}_n)&=\frac{1}{n^2}\left(\sum_{i=1}^{n}\operatorname{Var}(X_i)+\sum_{i=1}^{n}\sum_{i\neq j=1}^{n}\operatorname{Cov}(X_i,X_j)\right)=\\\\&=\frac{1}{n^2}\left(nσ^2+\sum_{i=1}^{n}\sum_{i\neq j=1}^{n}\frac{\operatorname{Cov}(X_i,X_j)}{\sqrt{σ^2}\sqrt{σ^2}}\cdot σ^2\right)=\\\\&=\frac{1}{n^2}\left(nσ^2+\underbrace{\sum_{i=1}^{n}\sum_{i\neq j=1}^{n}}_{\begin{matrix}n(n-1)\\\text{ summands}\end{matrix}}\underbrace{\operatorname{Corr}(X_i,X_j)}_{=ρ}\cdot σ^2\right)=\\\\&=\frac{1}{n^2}\left(nσ^2+n(n-1)ρσ^2\right)=\frac{σ^2}{n}\left(1+(n-1)ρ\right)\end{align*}$$ and since $\frac{σ^2}{n}>0$ you obtain that $$1+(n-1)ρ\ge 0 \implies ρ\ge-\frac{1}{n-1}$$