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