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.


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 $i\neq j$.

  • $\begingroup$ Thank you very much, I will read it carefully and will come back to you as soon as I have understood it. If you happen to find a reference, please post. Thanks again! $\endgroup$ – RandomGuy May 1 '17 at 11:34
  • $\begingroup$ @RandomGuy, I've seen this fact in papers.ssrn.com/sol3/papers.cfm?abstract_id=946782 (page 3), but without proof. $\endgroup$ – NCh May 1 '17 at 14:24
  • $\begingroup$ @RandomGuy I found and corrected misprint in denominator of $a$: here should be "plus", not "minus". $\endgroup$ – NCh May 1 '17 at 14:33
  • $\begingroup$ I have checked the calculations, but I cannot see why $Var(V)=1$. I certainly have like you that $Var(V)=a^2(n+n(n-1)\rho)+b^2$, but the constants for a and b do not add up to 1 in my calculations, are you sure that they are the right ones? Maybe it is just some stupid computation mistake on my part, but I checked and checked again and I come up with something different from 1. Thanks. $\endgroup$ – RandomGuy May 3 '17 at 11:23
  • 1
    $\begingroup$ $a^2=\frac{\rho}{(1+(n-1)\rho)^2}$, $b^2=\frac{1-\rho}{1+(n-1)\rho}$, $$a^2(n+n(n-1)\rho)+b^2=na^2(1+(n-1)\rho)+b^2=n\frac{\rho}{(1+(n-1)\rho)^2}(1+(n-1)\rho) + \frac{1-\rho}{1+(n-1)\rho}=n\frac{\rho}{1+(n-1)\rho}+\frac{1-\rho}{1+(n-1)\rho}=\frac{n\rho+1-\rho}{1+(n-1)\rho}=1. $$ $\endgroup$ – NCh May 3 '17 at 14:37

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