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I am trying to find a function $\phi$ such that for $Z \sim \mathcal{N}(0, I_n)$, which denotes the multidimensional gaussian distribution, we have that $\phi(Z) \sim Uniform(B_n^1)$, where $B_n^1$ is the unit ball.

For the unit cube, $\phi$ simply maps each coordinate $z_j$ to $\Phi(z_j)$. For the unit ball though, it's clear that the direction will be uniformly distributed, but I need to normalize $Z$ to weight the larger radii since the density of the uniform distribution on the unit ball as a function of the distance from the origin, r, should go as $r^n$.

JohnKnoxV
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1 Answers1

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Passing through the direction as you suggested, you can take $$\phi(Z)=\frac{Z}{\|Z\|^2}F(\|Z\|^2; n)^{1/n}$$

where $F(x; n)$ is the cumulative distribution function for $\chi^2(n)$ (sum of squares of $n$ independent normal variables) as described here: https://en.wikipedia.org/wiki/Chi-squared_distribution#Cumulative_distribution_function

Since the volume of the sphere of radius $r$ is proportional to $r^n,$ you need the distance from the origin $R$ to satisfy $\mathbb P[R^n\leq r^n]=r^n,$ i.e. it is necessary and sufficient that $R^n$ is uniformly distributed in $[0,1].$ This can be achieved by taking the uniformly distributed variable $F(\|Z\|^2; n)$ and raising it to the power $1/n.$

Dap
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  • Thanks, Dap. Can you explain why this is the necessary transform? – JohnKnoxV Oct 24 '17 at 14:44
  • @JKV: actually I think I had the power the wrong way round. I've added an explanation of why I think taking an $n$'th root is correct. – Dap Oct 24 '17 at 18:19