Existence of Multivariate normal distribution $(X_1,.....,X_n)$ with $Corr(X_i,X_j) = \rho\;$

Question

Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 \;,\; 1\le i \le n\; ;} $$ $${Var(X_i) = 1\;,\; 1\le i \le n\; ;} $$ $${Corr(X_i,X_j) = \rho\;;\;1\le i\;\neq j \le n\; ;}$$ for $ some\; 0\le \rho < 1$.
Show that for any ${\rho\in [0,1),(X_1,.....,X_n)}$ as above exist, but not necessarily for $\rho\in (-1,0]$.

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I thought of doing it using density function of multivariate normal
That is,if density of $(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) \in \mathbb{R^n}$
$${{g(x)=\frac{1}{{(2\pi)^\frac n2}\sqrt{\det\Sigma}}e^{-\frac12 x^T\Sigma^{-1}x},1(x\in \mathbb{R^n})\;;}}$$
where $\Sigma = \begin{bmatrix} 1 & \rho &\cdots&\cdots &\rho \\ \rho & 1& \ddots &&\rho \\ \vdots&\ddots&\ddots&\ddots &\vdots\\ \rho&&&1&\rho\\ \rho &\cdots&&\rho&1\\ \end{bmatrix} _{n\times n}$
${\det\Sigma = (1+(n-1)\rho){(1-\rho)}^{n-1}} $
I further thought that as $\det\Sigma$ need to be $\ge 0$ we can come up with some conditions that $\rho$ need to satisfy.But I didn't get anything useful. Maybe my thought process was entirely wrong.
Anyway I would be thankful for any help

Answer 1

Here's the idea: let $e$ be the n-dim standard normal, then a linear transformation of $e$, say $X=\mu+Ae$ is again multivariate normal (prove it by using, say, characteristic functions for multivariate distributions) whose covariance is clearly $AA^T$. Now what $A$ can satisfy your requirement?

To elaborate: for existence for rho in [0, 1), rewrite the required correlation matrix as $$C=\rho 11^T + (1-\rho)I$$ And therefore the required covariance matrix is just $$\Sigma=vCv^T,\quad\text{where}\,v=[\sigma(X_1)^2,\cdots \sigma(X_n)^2]^T.$$ which is positive definite and admits a non-singular Cholesky decomposition.

For $\rho\in(-1,0]$, $C$ isn't even positive semidefinite (proof easy), and therefore cannot be a correlation matrix.

Edit: sorry the last statement is wrong. Whether or not $C$ is positive semidefinite seems to depend on how large $\rho$ is. Actually, your required distribution exists if and only if the correlation matrix $C$ is semi-positive definite (in which case a Cholesky decomposition is possible). I'm talking only if because correlation matrix must be PSD.

For $\rho<0$, when does $C$ fail to be PSD? When there exists a non zero $x\in\Bbb R^n$ such that $$x^TCx = \rho (1^T x)^2 + (1-\rho)\|x\|^2 < 0$$ Or $$|1^Tx| / \|x\|> \left(\frac{1 + |\rho|}{|\rho|}\right)^{1/2}.$$ But since $x\mapsto 1^Tx$ is a bounded linear map, this cannot happend with a too small $|\rho|$ (in which case RHS would be too large). Indeed, the above inequality has solution if and only if the matrix norm (spectrual radius, namely) $\|(x\mapsto 1^Tx)\| > \left(\frac{1 + |\rho|}{|\rho|}\right)^{1/2}$.

Answer 2

The $\Sigma$ you desire can be written as $(1-\rho)I + \rho J$ where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Can you compute the eigenvalues of $\Sigma$ now, and state a condition for $\Sigma\succeq 0$?

The eigenvectors of $J$ are the all ones vector (with eigenvalue $n$) and the vectors orthogonal to the all ones vector (with eigenvalue $0$). Thus the eigenvalues of $\Sigma$ are $1-\rho + n\rho$ and $1 - \rho$. So if $\rho \in [0, 1)$, these eigenvalues are nonnegative so $\Sigma \succeq 0$. If $\rho \in (-1, 0]$, then you can find an $n$ such that $1 - \rho + n \rho < 0$ so $\Sigma \not \succeq 0$.