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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?

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    $\begingroup$ The title could be improved, I tried. $\endgroup$
    – Bob Jansen
    Commented Apr 19 at 14:45

1 Answer 1

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In your example,

$$ \begin{align} V(\sum X_i) &= \sum V(x_i)+2\sum_{i=1}^n\sum_{j=i+1}^n V(X_i,X_j)\\ &=n+2\sum_i^n\sum_{j=i+1}^n\rho\\ &=n+2\rho \frac{n(n-1)}{2}\\ &=n+n(n-1)\rho \end{align} $$ 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 $\lambda_2=1+(n-1)\rho$.

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

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