# Joint density of two correlated normal random variables

Let $X \text{~}N(\mu_x, \sigma_x^2)$ and $Y \text{~}N(\mu_y, \sigma_y^2)$ be two correlated normal random variables and $U = \frac{X - \mu_x}{\sigma_x}$ and $U = \frac{Y - \mu_x}{\sigma_x}$ be two normalized random variables with $\text{Cov(U,V)} = \rho = \frac{\sigma_{x,y}^2}{\sigma_x^2\sigma_y^2}$ where $\sigma_{xy}^2$ is the covariance of $X$ and $Y$.

The book I'm reading states that the joint density derived of $U$ and $V$ is $$f_{UV}(u,v) = \frac{1}{2\pi \sqrt{1-\rho^2}}\text{exp}(-\frac{u^2 - 2\rho u v + v^2}{2(1-p^2)})$$

How is this joint density derived? And how can I convert this density from $UV$ to $XY$?

The multivariate normal distribution of a vector $x\in \mathbf{R}^n$ is defined as $$f(x) = \frac{1}{(2\pi)^n |\Sigma|} e^{-1/2 \cdot (x-\mu)^T\Sigma^{-1}(x-\mu)}$$ where $\Sigma\in \mathbf{R}^{n\times n}$ is the covariance matrix where $\Sigma_{ij} = \mathbf{Cov}(x_i,x_j)$ (note the special case when $i=j$ is just $\mathbf{var}(x_i)$), $|\Sigma|$ is the determinant of the matrix, and $\mu\in\mathbf{R}^n$ is the mean.

With all this out of the way, the answer to your second question about how to represent the distribution in $XY$ rather than $UV$ is already answered by the definition of the normal distribution. Specifically, $$\mu = [\mu_x, \mu_y]^T$$ $$\Sigma = \begin{bmatrix}\sigma_x^2 & \sigma_{xy}^2\\ \sigma_{yx}^2 & \sigma_y^2\end{bmatrix}$$ $$|\Sigma| = \sigma_x^2\sigma_y^2 - \sigma_{xy}^2 \cdot \sigma_{yx}^2$$ Plug in $x = [X,Y]^T$ and that is distribution.

For your first question, notice that we can relate $UV$ and $XY$ by a linear (technically affine) transformation $$\begin{bmatrix} U\\ V \end{bmatrix} = \begin{bmatrix} 1/\sigma_x & 0\\ 0 & 1/\sigma_y \end{bmatrix} \begin{bmatrix} X\\ Y \end{bmatrix} + \begin{bmatrix} -\mu_x /\sigma_x\\ -\mu_y / \sigma_y \end{bmatrix}$$ More succinctly, we have $u = Ax + b$. It just so happens that a linear combination (plus a possible constant) of Gaussian random variables, is in fact Gaussian (this is not obvious). And since we know the Gaussian PDF is defined by the two parameters $\Sigma$ and $\mu$, if we can find those, then we are done.

It should be clear from linearity of expectation that the mean of $u\in \mathbf{R}^2$ is just $A\mu +b$. For the covariance we have the identity $\mathbf{cov}(Az + b) = A\mathbf{cov}(z)A^T$ (this also more or less falls out from linearity of expectation and you can find numerous proofs online)

Doing everything out we find that the distribution of $UV$ is Gaussian with mean vector $$\mu_{UV} = \begin{bmatrix} 1/\sigma_x & 0\\ 0 & 1/\sigma_y \end{bmatrix} \begin{bmatrix} \mu_x\\ \mu_y \end{bmatrix} + \begin{bmatrix} -\mu_x /\sigma_x\\ -\mu_y / \sigma_y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$ $$\Sigma_{UV} = \begin{bmatrix} 1/\sigma_x & 0\\ 0 & 1/\sigma_y \end{bmatrix} \begin{bmatrix}\sigma_x^2 & \sigma_{xy}^2\\ \sigma_{yx}^2 & \sigma_y^2 \end{bmatrix} \begin{bmatrix} 1/\sigma_x & 0\\ 0 & 1/\sigma_y \end{bmatrix} = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}$$

You can go ahead and compute the inverse and determinant and see that you get the joint PDF you have described

• Can you expand on what you mean by "plug in $x=[X,Y]^T$ and that is the distribution"? By substituting in $(x,y)$ for $(u,v)$ in $f_{UV}(u,v)$ I don't see why that would be the distribution. – Oliver G Apr 28 '18 at 13:41
• The distribution is given by $f(x)$ (see my answer above) where $x$ is a column vector $[X,Y]^T$ (i.e. $x$ is not the same $x$ you are using). You know all the parameters so you just substitute in all the values. $f_{UV}(u,v)$ does not come into play – Casey Apr 28 '18 at 18:21
• But by substituting in all the values then you're just writing $f_{XY}(x,y)$ and not inverting $f_{UV}(u,v)$ to get $f_{XY}(x,y)$. Given the information in the box of my question how do you invert $f_{UV}(u,v)$ to get $f_{XY}(x,y)$? – Oliver G Apr 28 '18 at 20:27
• Exactly, there is no need to transform $f_{UV}(u,v)$ to $f_{XY}(x,y)$. The problem states the properties of the joint density of $X$ and $Y$ and that is the end of the story. Going from $UV$ to $XY$ is circular. In order to get $UV$ you have to know the distribution of $XY$ in the first place. If you insist on doing it, purely as an exercise, then you can apply everything above to $x = A^{-1}u -A^{-1}b$ – Casey Apr 29 '18 at 2:44