Csdrhrt

Suppose that $(X_1,.....,X_n)$ follows multivariate normal with
$${E(X_i) = 0 ;,; 1le i le n; ;} $$
$${Var(X_i) = 1;,; 1le i le n; ;} $$
$${Corr(X_i,X_j) = rho;;;1le i;neq j le n; ;}$$
for $ some; 0le rho < 1$.


Show that for any ${rhoin [0,1),(X_1,.....,X_n)}$ as
above exist, but not necessarily for $rhoin (-1,0]$.

I thought of doing it using density function of multivariate normal

That is,if density of
$(X_1,.....,X_n)$ is g(x) where $x=(x_1,...x_n) in mathbb{R^n}$
$${{g(x)=frac{1}{{(2pi)^frac n2}sqrt{detSigma}}e^{-frac12 x^TSigma^{-1}x},1(xin mathbb{R^n});;}}$$


where $Sigma =
begin{bmatrix}
1 & rho &cdots&cdots &rho \
rho & 1& ddots &&rho \
vdots&ddots&ddots&ddots &vdots\
rho&&&1&rho\
rho &cdots&&rho&1\
end{bmatrix}
_{ntimes n}$

${detSigma = (1+(n-1)rho){(1-rho)}^{n-1}}
$


I further thought that as $detSigma$ 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

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 $Sigmasucceq 0$?

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 $1-rho + nrho$ 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 $Sigma not succeq 0$.

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
$$Sigma=vCv^T,quadtext{where},v=[sigma(X_1)^2,cdots sigma(X_n)^2]^T.$$
which is positive definite and admits a non-singular Cholesky decomposition.

For $rhoin(-1,0]$, $C$ 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 $xinBbb R^n$ 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 $xmapsto 1^Tx$ 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) $|(xmapsto 1^Tx)| > left(frac{1 + |rho|}{|rho|}right)^{1/2}$.