# Admissible values for non diagonal elements of correlation matrix

Preparing for possible job market interview questions I was reading some questions on the site.

Regarding this question with its solution interview questions

Question
Let $$\mathbf{C}$$ be a $$n\times n$$ covariance matrix such that all diagonal elements are equal to 1, and the non-diagonal ones to $$\rho$$ with $$-1\leq\rho\leq1$$. Which range of values is admissible for $$\rho$$?

Solution 1
Let $$X_1,\dots,X_n$$ be a sequence of independent random variables with unit variance and pairwise correlation $$\rho$$ for any $$i\not= j$$. Let $$Y:=\sum_iX_i$$ then: \begin{align} \notag V\left(Y\right) &=\sum_{i=1}^nV\left(X_i\right)+\sum_{i\not=j}Cov(X_i,X_j) \\ &=n+n(n-1)\rho \end{align} The variance of $$Y$$ is positive, therefore: \begin{align} n+n(n-1)\rho\geq0 \quad\Leftrightarrow\quad \boxed{\rho\geq\frac{1}{1-n}} \end{align}

I do not understand why $$V(Y)$$ is equal to $$n+n(n-1)\rho$$. Shouldn't be equal to $$n+\frac{n(n-1)}{2}\rho$$ since $$\sum_{i\not=j}Cov(X_i,X_j) = \frac{n(n-1)}{2}\rho$$ ?

This would change the final solution to

\begin{align} n+\frac{n(n-1)}{2}\rho\geq0 \quad\Leftrightarrow\quad \boxed{\rho\geq\frac{2}{1-n}} \end{align}

What Am I missing?

• The title could be improved, I tried. Commented Apr 19 at 14:45

\begin{align} V(\sum X_i) &= \sum V(x_i)+2\sum_{i=1}^n\sum_{j=i+1}^n V(X_i,X_j)\\ &=n+2\sum_i^n\sum_{j=i+1}^n\rho\\ &=n+2\rho \frac{n(n-1)}{2}\\ &=n+n(n-1)\rho \end{align} as was the original.

IMHO, a better and cleaner way to obtain this result is by imposing the condition that the corresponding covariance matrix is positive (semi)definite, i.e.

$$\lambda_{min}(\Sigma)\geq 0$$

For a $$n\times n$$ matrix of the type

$$\Sigma=(1-\rho)\mathbf{I}+\rho\mathbf{1}\mathbf{1}^T$$

where $$\mathbf{I}$$ the identity matrix and $$\mathbf{1}$$ a vector of ones, the eigenvalues are $$1-\rho$$ and $$(n-1)\rho+1$$. Thus we require

$$(n-1)\rho+1\geq0 \Rightarrow \boxed{\rho\geq\frac{1}{1-n}}$$

#### Derivation

Using the matrix determinant lemma, we can solve for the eigenvalues as:

\begin{align} 0&=\mathrm{det}\left(\Sigma-\lambda\mathbf{I}\right)\\ &=\mathrm{det}\left((1-\rho)\mathbf{I}+\rho\mathbf{11^T}-\lambda\mathbf{I}\right)\\ &=\mathrm{det}\left((1-\rho-\lambda)\mathbf{I}+\rho\mathbf{11^T}\right)\\ &=\mathrm{det}\left((1-\rho-\lambda)\mathbf{I}\left(\mathbf{I}+\frac{\rho}{1-\rho-\lambda}\mathbf{11^T}\right)\right)\\ &=\mathrm{det}\left((1-\rho-\lambda)\mathbf{I}\right)\mathrm{det}\left(\mathbf{I}+\frac{\rho}{1-\rho-\lambda}\mathbf{11^T}\right)\\ &=\left(1-\rho-\lambda\right)^{n}\left(1+\frac{n\rho}{1-\rho-\lambda}\right)\\ &=\left(1-\rho-\lambda\right)^{n}+n\rho\left(1-\rho-\lambda\right)^{n-1}\\ &=\left(1-\rho-\lambda\right)^{n-1}\left(1+(n-1)\rho-\lambda\right) \end{align}

and hence $$\lambda_1=1-\rho$$ and $$\lambda_2=1+(n-1)\rho$$.

• Yes, I also thought the positive definiteness was the way to go with the same disection of Covariance matrrix as you made.
– Attack68
Commented Apr 19 at 14:32