I was reading a paper about variance reduction techniques, where they want to create a valid correlation matrix of size $n \times n$, but with the same pairwise-correlation coefficient, therefore, every off-diagonal entries are set to the same number $\alpha$:
$$ R_n = (r_{ij}) =\begin{pmatrix}
1 & \alpha & \cdots & \alpha \\
\alpha & 1 & \cdots & \alpha \\
\vdots & \vdots & \ddots & \vdots \\
\alpha & \alpha & \cdots & 1
\end{pmatrix}$$
Then, they say that this matrix must be positive-definite to be a valid correlation matrix which I understand perfeclty, but they say that this matrix is therefore constrained by the following relation:
$$0 \leq \sum_{i=1}^{n} \sum_{j=1}^{n} r_{ij} \leq n^2 $$
This relation, however, is totally new for me and I couldn't find where it came from. So my question is: Is this relation correct? Is there some kind of proof for it? Any help would be greatly appreaciated.
PS: English is not my first language, so apologies for my grammar.
Best Answer
The matrix will be positive semi-definite if and only if $-1/(n-1) \le \alpha \le 1$, as shown in answers at Bound for the correlation of three random variables .
The upper bound of $n^2$ in $0 \leq \sum_{i=1}^{n} \sum_{j=1}^{n} r_{ij} \leq n^2 $ is achieved using $\alpha = 1.$
The lower bound of $0$ is achieved using $\alpha = -1/(n-1)$, because there are $n(n-1)$ occurrences of $\alpha$ in the double sum, together with n occurrences of $1$ (the diagonal elements). Therefore the double sum = $n(n-1)*(-1)/(n-1) + n*1 = 0$.
So what you state from the paper is consistent with the link I provided.