Consider a $d$-dimensional Gaussian random vector $\mathbf{Z}=[Z_i]_i$ with mean $\boldsymbol{\mu}$ and covariance matrix $\boldsymbol{\Sigma}$. What would be the more efficient method(s) to simulate $\mathbf{Z}$ conditional on the sum of squares of its components $Y := Z_1^2 + Z_2^2 + \dots + Z_d^2$? The independent case with $\boldsymbol{\Sigma}$ diagonal can be of interest.

My first simple attempt is via importance sampling using as importance distribution a Von Mises distribution on the sphere $\{\mathbf{z}: \: \|\mathbf{z}\|^2 = Y\}$ with pdf $g(\mathbf{z}) \propto \exp\{ \boldsymbol{\theta}^\top \mathbf{z}\}$, choosing $\boldsymbol{\theta} = \boldsymbol{\Sigma}^{-1}\boldsymbol{\mu}$. This question relates to my question about the chi-square process.

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    $\begingroup$ Importance sampling does not allow you to simulate a given distribution but only and at best to approximate moments from this distribution. To simulate exactly from your target, you should seek an accept-reject method, using a proposal that is supported by the sphere. $\endgroup$
    – Xi'an
    Jun 27, 2017 at 12:46
  • $\begingroup$ @Xi'an. Oh yes, thank you. I had in mind using particles for the specific filtering problem, hence using resampling. A concern is the ratio of the two pdf, which can be computed exactly with for the Von Mises'. $\endgroup$
    – Yves
    Jun 27, 2017 at 14:56
  • $\begingroup$ But can the ratio be bounded? This is the important issue for exact simulation. Actually, given the von Mises density, it is likely to be bounded. $\endgroup$
    – Xi'an
    Jun 27, 2017 at 15:43
  • $\begingroup$ I think that when $\boldsymbol{\theta}$ is chosen as I proposed the ratio of pdf target / proposal is $\propto \exp\{-\mathbf{z}^\top \boldsymbol{\Sigma}^{-1} \mathbf{z} / 2\}$ so the maximum and minimum relate to those of the Rayleigh quotient hence to the eigenvalues of $\boldsymbol{\Sigma}$. When $\boldsymbol{\Sigma}$ is scalar the ratio is constant hence must be $1$. $\endgroup$
    – Yves
    Jun 27, 2017 at 17:04
  • $\begingroup$ I'd be interesting in seeing this even for the cases with $d=2$; $\mu=(1,1)$ or $\mu=(0,1)$; $\Sigma=\pmatrix{1 & 0\\0 & 1}$ $\endgroup$
    – Matt F.
    May 9, 2021 at 2:42


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