Constructing a bivariate normal from three univariate normals

I'm trying to construct correlated bivariate normal random variables from three univariate normal random variables. I realize there is a formula for constructing a bivariate normal random variable from two univariate random normal variables, but I have reasons for wanting to adjust two previously sampled variables by a third in order to give them a correlation $$\rho$$.

Based on the approaches for constructing bivariate normals from two univariate normals, I came up with the following approach and need help verifying its correctness.

First, imagine that we have three univariate normal variables. For simplicity here, we just assume they all have $$\sigma=1$$.

$$X_0 \sim Normal(0, 1)$$ $$Y_0 \sim Normal(0, 1)$$ $$Z \sim Normal(0, 1)$$

Given these three univariate random variables, I construct two new random variables using the following linear combinations:

$$X = |\rho| * Z + \sqrt{1-\rho^2} * X_0$$ $$Y = \rho * Z + \sqrt{1-\rho^2} * Y_0$$

where $$\rho \in [-1, 1]$$ represents the correlation coefficient between the two univariate normals.

Can someone help me formally verify that $$X$$ and $$Y$$ are now correlated random variables with correlation $$\rho$$?

I've convinced my self through empirical simulation. Here are plots of values sampled from $$X$$ and $$Y$$ for the cases where $$\rho=0$$, $$\rho=1$$, and $$\rho=-1$$.

Plot of X vs. Y for rho of 0

Plot of X vs. Y for rho of 1

Plot of X vs. Y for rho of -1

These plots are exactly as I would expect, but it would be nice to have a formal proof based on my construction. Thanks in advance!

• Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent. – Chris MacLellan Jan 5 at 23:24

I presume $$X_0, Y_0$$ and $$Z$$ are assumed to be independent? Well, if we define $$\mathbf U := \begin{bmatrix} X_0 \\ Y_0 \\Z \end{bmatrix}, \ \ \ \ \ \ \mathbf V := \begin{bmatrix} X \\ Y\end{bmatrix},$$

then

$$\mathbf V = \mathbf M \mathbf U,$$

where

$$\mathbf M := \begin{bmatrix} \sqrt{1- \rho^2} & 0 & | \rho| \\ 0 & \sqrt{1- \rho^2} & \rho \end{bmatrix}$$

Since $$X_0, Y_0$$ and $$Z$$ are independent with unit variance, the covariance matrix for $$\mathbf U$$ is the $$3 \times 3$$ identity matrix: $$\mathbb E[\mathbf U \mathbf U^T] = \mathbf I.$$

Hence the covariance matrix for $$\mathbf V$$ is given by $$\mathbb E [\mathbf V \mathbf V^T ] = \mathbf M \ \mathbb E[\mathbf U \mathbf U^T ] \ \mathbf M^T = \mathbf M\mathbf M^T=\begin{bmatrix} 1 & \rho | \rho| \\ \rho | \rho| & 1\end{bmatrix}$$

Thus the correlation coefficient between $$X$$ and $$Y$$ is $$\rho_{X,Y} = \frac{\mathbb E[XY]}{\sqrt{\mathbb E[X^2] \ \mathbb E[{Y^2]}}} = \frac{\rho| \rho|}{\sqrt{1 \times 1}} = \rho|\rho|.$$ [I used the fact that $$\mathbb E[X] = \mathbb E[Y] = 0$$ here.]

So I'm afraid the correlation coefficient is not $$\rho$$. It's $${\rm sign}(\rho) \times |\rho|^2$$.

But that's easily fixed. If you redefine $$\mathbf M$$ as $$\mathbf M := \begin{bmatrix} \sqrt{1- |\rho |} & 0 & {\rm sign}(\rho)\sqrt{ |\rho |} \\ 0 & \sqrt{1- |\rho | } & \sqrt{|\rho |} \end{bmatrix},$$ then it should work out.

Finally, I'll point out that the $$X$$ and $$Y$$ that you've constructed are guaranteed to be Gaussian, since $$\mathbf V$$ is related to the Gaussian vector $$\mathbf U$$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $$\mathbb E[\mathbf V \mathbf V^T]$$.