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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.

Steffen Moritz
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Aulin Cadeac
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    What non-closed-form solutions have you found? And since two perfectly correlated variables are almost surely equal, you're really asking for the probability function of a *single* Normal variable--so what is the obstacle to finding that? Is it that you are concerned about mathematically describing a singular multivariate distribution (which indeed can be complicated)? If so, please see the related post http://stats.stackexchange.com/questions/63817 for some ideas. – whuber Sep 07 '16 at 12:58
  • This question is nearly identical to [this one](http://stats.stackexchange.com/q/230860/127790) which I had attempted (badly) to answer, as well as a few other ones. All of these either have no accepted answer, or the question is not quite the same. It might be worthwhile to get an authoritative answer here, for future reference. (A hint of one for the more general case is currently at [Wikipedia](https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Degenerate_case).) – GeoMatt22 Sep 07 '16 at 13:54
  • Joint density does not exist when $\rho=1$. – StubbornAtom Nov 30 '19 at 12:51

1 Answers1

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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.

Joseph Santarcangelo
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