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Let $\mathcal{M}_1$ denote the set of probability measures on the unit ball in $\mathbb{R}^d$ (which comes with its Borel $\sigma$-field). Denote by $\sigma$ the uniform measure on the orthogonal group O(d).

Consider the operator $\mathcal{M}_1 \to \mathcal{M}_1$ which takes a measure $\pi \in \mathcal{M}_1$ to $\overline{\pi}$, which satisfies $$ \overline \pi (E) = \int_{O(d)} \pi(QE) \, \mathrm{d} \sigma(Q), $$ for any measurable subset $E$ of the unit ball. Note that we define $QE = \{Q x: x\in E\}$. Clearly the mapping $\pi \mapsto \overline \pi$ produces $O(d)$-invariant measures.

Is this mapping well known? It is similar to the concept of "symmetrization" of scalar random variables, in the sense that a random variable sampled from $\overline \pi$ has the same distribution as $Q \theta$ where $Q$ is uniform on $O(d)$ and $\theta$ is (independently) drawn from $\pi$.

Some concrete questions are: is it a surjection to the family of rotationally invariant probability measures on $\Theta$? Is it invertible on its image? What else is known about it?

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  • $\begingroup$ Isn’t Lebesgue measure the unique rotation-invariant measure up to scaling? If that’s so, the map $\bar\pi$ should be pretty easy to describe. $\endgroup$ Commented Jun 30, 2022 at 1:25
  • $\begingroup$ There are other measures, in general, which are rotation invariant, right? For instance, a (truncated) Gaussian with identity covariance. $\endgroup$
    – Drew Brady
    Commented Jul 1, 2022 at 15:50
  • $\begingroup$ I think that’s also Lebesgue? $\endgroup$ Commented Jul 1, 2022 at 16:10
  • $\begingroup$ I don't think so. For instance a Gaussian truncated to the unit ball places much more mass near the origin than the Lebesgue (i.e., uniform) measure does, no? $\endgroup$
    – Drew Brady
    Commented Jul 1, 2022 at 16:11
  • $\begingroup$ My mistake. I somehow thought you were talking about the unit sphere. But in that case, aren’t measures completely described by the radial measure? $\endgroup$ Commented Jul 2, 2022 at 19:45

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