Function that sent the gaussian distribution to the uniform on the ball

Question

How do we define a function from $\mathbb R^n$ to the Euclidean ball in such a way that, if we consider the multivariate standard Gaussian distribution $Z$, we have that the image of $Z$ is uniformly distributed on the Euclidean ball? Moreover, I want such a function to have a bounded Lipschitz constant I was thinking about the function that sent $Z$ to $\sqrt{n}\frac{Z}{\left\|Z\right\|}$, where $\left\|Z\right\|$ denotes the Euclidean norm of $Z$, however with this choice I get the uniform distribution on the sphere $S^{n-1}$, not on the whole ball

Answer 1

Let $F$ be the cumulative distribution function of $\left\|Z\right\|$ then , $$F'(x) = \alpha_n x^{n-1} e^{-\frac12x^2}\mathbf 1_{x > 0}.$$

Let $h : \mathbb R^n \to \mathbb B^n$, $$h(z) = \frac1{\left\|z\right\|}F(\left\|z\right\|) z$$

Claim 1: $h$ is Lipschitz

If $z_1, z_2\in \mathbb R^n$ such $\left\|z_2\right\| \le \left\|z_1\right\|$,

\begin{align} \left\|h\left(z_1\right) - h\left(z_2\right)\right\| &= \left\|\frac1{\left\|z_1\right\|}F\left(z_1\right)z_1 - \frac1{\left\|z_2\right\|}F\left(z_2\right)z_2\right\|\\ &=\left\| \frac{z_1\left\|z_2\right\|F\left(\left\|z_1\right\|\right) - z_2\left\|z_1\right\|F\left(\left\|z_2\right\|\right)}{\left\|z_1\right\|\left\|z_2\right\|} \right\|\\ &=\left\| \frac{z_1\left\|z_2\right\|F\left(\left\|z_1\right\|\right) - z_1\left\|z_2\right\|F\left(\left\|z_2\right\|\right) + z_1\left\|z_2\right\|F\left(\left\|z_2\right\|\right) - z_2\left\|z_1\right\|F\left(\left\|z_2\right\|\right)}{\left\|z_1\right\|\left\|z_2\right\|} \right\|\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left\|z_1 \left\|z_2\right\| - z_2\left\|z_1\right\|\right\|\\ &= \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left\|z_1 \left\|z_2\right\| - z_2 \left\|z_2\right\| + z_2 \left\|z_2\right\| - z_2\left\|z_1\right\|\right\|\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left(\left\|z_2\right\|\left\|z_1 - z_2\right\| + \left\|z_2\right\|\left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &= \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_2\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &\le \sup_{x\ge 0}\left|F'(x)\right|\left|\left\|z_2\right\| - \left\|z_1\right\|\right| + \frac{\sup\limits_{x\ge 0}\left|F'(x)\right|\left\|z_2\right\|}{\left\|z_2\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ & \le 3 \sup\limits_{x\ge 0} \left|F'(x)\right| \left\|z_1 - z_2\right\| \end{align}

Claim 2 $F\left(\left\|Z\right\|\right)\sim U\left([0,1]\right)$.

Proof: Look at this post

Claim 3: $\left\|Z\right\|$ and $\frac{Z}{\left\|Z\right\|}$ are independant.

Proof: Look at this post.

Claim 4: $h(Z) \sim U\left(\mathbb B^n\right)$

This is a direct consequence of the claims 2 and 3.