1

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

Marco
  • 43
  • 1
    Do you want the ball and the gaussian to have the same dimension? – Kroki Feb 05 '24 at 19:08
  • Yes, the gaussian is the multidimensional standard gaussian on R^n, and the euclidean ball is defined as the set of points in R^n with euclidean norm less or equal than square root of n – Marco Feb 05 '24 at 19:12

1 Answers1

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.

Kroki
  • 13,619
  • In this way we define a function such that the image of Z is the uniform on the euclidean ball of radius 1, right? To have the uniforn on the ball of radius square root of n, it is enough to consider the same function multiplied by square root of n? @Kroki – Marco Feb 06 '24 at 15:25
  • Yes that's correct – Kroki Feb 06 '24 at 15:33
  • And it's possible to show that this function have a bounded lipschitz constant? @Kroki – Marco Feb 06 '24 at 15:52
  • You want to prove that $z \to f(z)$ is lipschitz? – Kroki Feb 06 '24 at 16:18
  • Yes, since I need that the map is lipschitz for the rest of the exercise I'm doing @Kroki – Marco Feb 06 '24 at 16:21
  • It is lipschitz if $n \ge 2$ – Kroki Feb 06 '24 at 16:34
  • Is that enough for you? Or you want also the function to be lipschitz for $n=1$? – Kroki Feb 06 '24 at 16:36
  • It's enough. How can I prove it? @Kroki – Marco Feb 06 '24 at 16:39
  • let me edit the answer to include the proof. – Kroki Feb 06 '24 at 16:40
  • I have another question: in the first part of your answer, you said that F(||Z||) is uniformly distributed on (0,1). How to show this? @Kroki – Marco Feb 06 '24 at 18:21
  • I am editing my answer and including the full proof. – Kroki Feb 06 '24 at 18:22
  • @Marco you can look at the answer now. – Kroki Feb 06 '24 at 19:54
  • thank you!! A last question: we have that, since Z is a multidimensional standard gaussian, its component are independent so ||Z||^2 is distributed as a chi-squared distribution, so ||Z|| is distributed as the square root of a chi-squared distribution. So we can apply the claim 2 since the cumulative distribution function of such a distribution is continuous, right? @Kroki – Marco Feb 07 '24 at 12:15
  • Yes the answer to your question. – Kroki Feb 07 '24 at 12:59
  • If you feel that the answer answers you question don't hesitate to upvote it or to accept it. – Kroki Feb 07 '24 at 13:00