# Random variables with arbitrary positive correlation but without arbitrary negative correlation. Then $Cor(X_i,X_j)<\frac{-1}{n-1}$ not possible

Let $$X_i \in L_2$$ be a sequence of pairwise correlated random variables. The random variables can have arbitrary positive correlation but can't have arbitrary negative correlation.

How can I show that for $$(X_1,\ldots,X_n)$$ and $$\forall i,j \in \{1,\ldots,n\}, i\neq j$$

$$\mathrm{Cor}(X_i,X_j)<\frac{-1}{n-1}$$ is not possible

• Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ? – dallonsi Dec 11 '18 at 15:27
• @dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality. – Marcus M Dec 11 '18 at 17:21
• yes Marcus, thanks. My mistake – dallonsi Dec 13 '18 at 13:11

The idea is to realize that covariance matrices are positive semi-definite. Define variables $$Y_j = X_j/ \sqrt{\operatorname{Var}(X_j)}.$$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$\operatorname{Cor}(X_i,X_j) = \operatorname{Cov}(Y_i,Y_j) < -\frac{1}{n-1}.$$
Let $$\Sigma$$ be the covariance matrix of $$(Y_1,\ldots,Y_n)$$. Then $$\Sigma$$ has $$1$$'s on the diagonal, and each off-diagonal entry is less than $$-\frac{1}{n-1}$$. If we define $$x = (1,1,\ldots,1)^T$$, then we see that $$x^T \Sigma x < 0$$, which contradicts the fact that covariance matrices are positive semi-definite.