CrossValidated user @whuber has earlier answered at: https://stats.stackexchange.com/a/158008/160034 with a neat and quick method that generates a norm-constrained multivariate Gaussian. It is just three steps:

  1. Generate $X \sim \mathcal{N}(0,\mathbb{I}_n)$. ($n$ standard normal elements)

  2. Generate $P$ as the square root of a $\chi^2(d)$ distribution truncated at $(a/\sigma)^2$. (May use CDF-inversion where $a$ is the norm-constraint, $\sigma$ is the usual std deviation of the normal distribution in question)

  3. Let $Y = \sigma P\, X/||X||$ ==> the desired norm-constrained multivariate Gaussian.

My question is related to that, and goes larger than a comment as below. (hence this separate question)

@whuber's answer is really exciting and appeals to intuition. But I do not find it fully convincing. I rephrase that the core of @whuber's idea is the decomposition

\begin{align} X &= \left(\frac{X}{||X||}\right) \cdot \Big( ||X|| \Big) \\[2mm] &= A \cdot B \end{align}

and then to generate the two product terms separately. I have apprehensions about such an approach as I fail to find answers to below questions.

Question-1: Are $A$ and $B$ independent?

$ ||X||^2 = P^2 = \rho $ is certainly $\chi^2$ distributed under unconstrained norm, and CDF-inversion method can be efficiently used to generate the constrained version.

But after generating $P$, is it independent of $\frac{X}{||X||}$ ?

Only hint is that it happens so in 2D case and such a concept is extensively used as basis for polar-coordinate system. For higher dimensions, I am unaware of any extension.

Question-2: Does the final product vector follow the Gaussian distribution with norm constraint?

Okay, lets say $A$ and $B$ are independent. How can we be sure that the product $AB$ generated as per @whuber's suggestion still follows a multivariate Gaussian?

At the very outset, the vectors are simply complicated norm-normalized standard Gaussian vectors (whose distribution is unknown) scaled by another random constant distributed as truncated-$\chi$. Is the result really a much simpler norm-constrained multivariate Gaussian?


For Question 1 (independence): On the generation side, this follows directly from the fact that step 2 uses no information from step 1. On the other side, this follows because you have a spherically-symmetric multivariate Gaussian: if knowing the angle (i.e. orientation, unit vector) gave any information on the radius, it would not be spherically symmetric, no?

For Question 2, I am not sure what your uncertainty is. If we accept that the two components are independent, then the only remaining question is if they have the correct marginals. For the orientation, it has a "spherical Gaussian orientation" marginal by construction. Truncating the norm will not change this, since the Gaussian orientation and norm are independent. The marginal for the un-truncated norm will follow a $\chi_n$ distribution. Does it make sense that the product of these would give an un-truncated Gaussian?


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