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.

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#### Best Answer

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{(2pi)^{k}|boldsymbolSigma|}} expleft(-frac{1}{2}({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu}) right). end{align}

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

Setting $rho=1$ we get:

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

the determinant of this matrix is zero:

$$det(Sigma)=sigma_X^2sigma_Y^2-sigma_X sigma_Ysigma_X sigma_Y=0,$$

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}} expleft( -frac{1}{2(1-rho^2)}left[ frac{(x-mu_X)^2}{sigma_X^2} + frac{(y-mu_Y)^2}{sigma_Y^2} – frac{2rho(x-mu_X)(y-mu_Y)}{sigma_X sigma_Y} right] right)\ end{align}

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

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