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From this post, Confidence regions on bivariate normal distributions using $\hat{\Sigma}_{MLE}$ or $\mathbf{S}$, the equation

$(\mathbf{x} - \boldsymbol{\mu})^{T} \mathbf{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu}) \leq \chi_{p, \alpha}^2$

can be described by the parametric curve

$ \mathbf{x} = \boldsymbol{\mu} + \sqrt{\chi_{p, \alpha}^2} \mathbf{L} \begin{bmatrix} \cos(\theta)\\ \sin(\theta) \end{bmatrix} $

for $ 0 < \theta < 2 \pi $

Is it possible to generalize to larger dimensions?

For example, for three dimensions

$ \mathbf{x} = \boldsymbol{\mu} + \sqrt{\chi_{p, \alpha}^2} \mathbf{L} \begin{bmatrix} \cos(\theta) \cos(\phi)\\ \cos(\theta) \sin(\phi)\\ \sin(\theta) \end{bmatrix} $

for $ 0 < \theta < \pi $ and $ 0 < \phi < 2 \pi $

And so on?

Javier Morlet
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  • Welcome to CV, Javier. Of course it's possible! But how do you want to generalize? In $d$ dimensions the "curve" becomes a hypersurface. A good way to view this hypersurface is that it's a sphere in Mahalanobis coordinates. There are many ways to parameterize it, if you need to -- but it's rare to need to. Usually all you need is to determine whether a point lies inside or outside this sphere, and that's easily done. – whuber Mar 14 '23 at 22:40
  • Thank you. I want to sample or generate points at the boundary of the hypersurface. So I mean if its possible to generalize the parametrization – Javier Morlet Mar 15 '23 at 15:12
  • Please make that your question, then. The reason is that parameterization is only a means to that end and is both unnecessary and overly complicated. The simple answer is that to sample the sphere you can choose one point and then rotate it (randomly or deterministically). A random rotation is effected by a random orthogonal transformation, which is incredibly easy to obtain, but orthogonal transformations do not parameterize the sphere. – whuber Mar 15 '23 at 16:40

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