Jointy normal correlated random variables Suppose that we have a collection of standard normal, independent random variables: $V, \epsilon_i \sim N(0,1)$ for $i=1, \cdots n$ and let $|\rho| <1$. 
It is easy to see that the random variables $X_i$ defined by 
\begin{equation}
X_i = \sqrt\rho \ V + \sqrt{1-\rho}\ \epsilon_i
\end{equation}
compose a random vector $(X_1, \cdots, X_n)$ which has a multivariate normal distribution, with $corr(X_i, X_j) = \rho$, whenever $i\neq j$. The proof of this claim is pretty much straightforward: take any constant vector $a=(a_1, \cdots, a_n)$ and consider the linear combination $a^TX$: as the summands are all independent normal random variables, the sum will be also a normal random variable, hence $X$ has the multivariate normal distribution.
Question: is the converse of the above statement true? 
That is, suppose that we have a multivariate normal random vector $X = (X_1, \cdots, X_n)$, such that for all $i\neq j$ we have $X_i$ standard normal and  $corr(X_i, X_j) = \rho$. Can we prove the existence of standard normal, independent random variables $V, \epsilon_i \sim N(0,1)$ for which we have
\begin{equation}
X_i = \sqrt\rho \ V + \sqrt{1-\rho}\ \epsilon_i ?
\end{equation}
I am not sure about it. It is certainy true when $\rho = 0$, but I have tried to look for a counterexample, but could not find any, so I suspect that it could be true, but I don't see how to prove it in the general case.
Thanks in advance for your answers.
 A: This construction is known, but I cannot provide references for it, and so I had to prove this statement anew. 
Let $U$ be a standard normal random variable that does not depend on $X=(X_1,\ldots,X_n)$, where $X$ is some multivariate normal random vector, such that for all $i\neq j$ we have $corr(Xi,Xj)=\rho$, $E(X_i)=0$ and $\text{Var}(X_i)=1$ for all $i=1,\ldots,n$. 
Define constants
$$
a=\frac{\sqrt{\rho}}{1+(n-1)\rho}, \quad b=\frac{\sqrt{1-\rho}}{\sqrt{1+(n-1)\rho}}.
$$
Now set 
$$
V=a\sum_{i=1}^n X_i + bU, \quad \epsilon_i=\frac{X_i}{\sqrt{1-\rho}} - V\frac{\sqrt{\rho}}{\sqrt{1-\rho}}, \quad i=1,\ldots, n.
$$
The equation 
$$
X_i = \sqrt\rho \ V + \sqrt{1-\rho}\ \epsilon_i
$$
follows immediately.
Let us prove that new vector $(V,\epsilon_1,\ldots,\epsilon_n)$ consists of independent standard normal random variables. Since this vector is obtained by linear transformation of multivariate normal random vector $(U,X_1,\ldots,X_n)$, it is also multivariate normal. It suffices to show that variances are equal to $1$ and covariances are zero.
$$\text{Var}(V) =\text{Var}\left(a\sum_{i=1}^n X_i + bU\right)=a^2\text{Var}\left(\sum_{i=1}^n X_i\right) + b^2\text{Var}(U)=a^2\left(n+n(n-1)\rho\right)+b^2=1.
$$
Before calculation of variance of $\epsilon_1$, find covariance of $\epsilon_1$ and $V$: 
$$\tag{1}\label{1}
\text{cov}(\epsilon_1, V) =\text{cov}\left(\frac{X_1}{\sqrt{1-\rho}} - V\frac{\sqrt{\rho}}{\sqrt{1-\rho}}, V\right)=\frac{1}{\sqrt{1-\rho}}\left[\text{cov}(X_1,V)-\sqrt{\rho}\text{Var}(V)\right]=\frac{1}{\sqrt{1-\rho}}\left[\text{cov}(X_1,V)-\sqrt{\rho}\right]
$$
Here
$$
\text{cov}(X_1,V) = \text{cov}\left(X_1,a\sum_{i=1}^n X_i + bU\right)=
a\bigl(\text{Var}(X_1)+(n-1)\rho\bigr)=a\bigl(1+(n-1)\rho\bigr)=\sqrt{\rho}.
$$
Substitute this result into (\ref{1}) and obtain $\text{cov}(\epsilon_1, V) =0$. And the same for $\text{cov}(\epsilon_i, V)$ for $i=2,\ldots,n$.
Next find
$$
\text{Var}(\epsilon_1) = \text{Var}\left(\frac{X_i}{\sqrt{1-\rho}} - V\frac{\sqrt{\rho}}{\sqrt{1-\rho}}\right)=\frac{1}{1-\rho}\left[\text{Var}\left(X_1\right)+\rho\text{Var}(V) -2\sqrt{\rho}\text{cov}\left(X_i, V\right)\right]=1.
$$
And finally find covariance of $\epsilon_i$, $\epsilon_j$ ($i\neq j$):
$$
\text{cov}(\epsilon_1, \epsilon_2)=\text{cov}\left(\frac{X_1}{\sqrt{1-\rho}} - V\frac{\sqrt{\rho}}{\sqrt{1-\rho}}, \frac{X_2}{\sqrt{1-\rho}} - V\frac{\sqrt{\rho}}{\sqrt{1-\rho}}\right)=\frac{1}{1-\rho}\bigl[\,\underbrace{\text{cov}(X_1,X_2)}_{\rho}-2\sqrt{\rho}\,\underbrace{\text{cov}(X_1,V)}_{\sqrt{\rho}}+\rho\underbrace{\text{Var}(V)}_{1}\bigr]=0
$$
By symmetry, $\text{cov}(\epsilon_i, \epsilon_j)=0$ for all $i\neq j$.
