Multivariate normal joint density of two correlated ($\rho$ = 1) variables

Question

I cannot find a close form solution to calculate the normal joint probability of two variables assuming they are fully correlated ($\rho$=1).

Thanks in advance.

Answer 1

Unfortunately when two variables are fully correlated i.e $\rho=1$ the covariance matrix ${\displaystyle {\boldsymbol {\Sigma }}}$ is not invertible . This is a Degenerate case.

The formula for normal joint probability is given by:

\begin{align} f_{\mathbf x}(x_1,\ldots,x_k) = \frac{1}{\sqrt{(2\pi)^{k}|\boldsymbol\Sigma|}} \exp\left(-\frac{1}{2}({\mathbf x}-{\boldsymbol\mu})^\mathrm{T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu}) \right). \end{align}

The covariance matrix for the two variable case is given by: $$\Sigma = \begin{pmatrix} \sigma_X^2 & \rho \sigma_X \sigma_Y \\ \rho \sigma_X \sigma_Y & \sigma_Y^2\end{pmatrix}.$$

Setting $\rho=1$ we get:

$$\Sigma = \begin{pmatrix} \sigma_X^2 & \sigma_X \sigma_Y \\ \sigma_X \sigma_Y & \sigma_Y^2\end{pmatrix},$$

the determinant of this matrix is zero:
$$det(\Sigma)=\sigma_X^2\sigma_Y^2-\sigma_X \sigma_Y\sigma_X \sigma_Y=0,$$

therefor not invertible

You can also see this with the formula for normal joint probability of two variables $x$ and $y$ is given by:

\begin{align} f(x,y)= \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right] \right)\\ \end{align}

if $\rho=1$ we get division by zero.