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It is well known that convolution of two ball measures (i.e. a uniform measure over a ball) in $\mathbb{R}^{n}$ is absolutely continuous with respect to the Lebesgue measure. My question is - how to compute its density? I am mostly interested in the density when the two radii are not the same. It is evident it is spherical density.

Edit - I managed to compute the density of two spheres, which ends up to be polynomial by easy considerations. It seems that the answer for the ball averages might involve hypergeometric functions to some extant. Any help is appreciated.

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    $\begingroup$ What is a ball measure? Is it a uniform distribution on a ball? $\endgroup$
    – Gerald Edgar
    Commented Dec 30, 2020 at 23:51
  • $\begingroup$ @GeraldEdgar - yes. $\endgroup$
    – A K
    Commented Dec 31, 2020 at 5:51

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Equivalently, you ask what is the volume of the intersection of two balls (with, in general, different centers and different radii.) This intersection is partitioned by their radical plane onto two ball segments. For computing the volume of the ball segment, say $\{x=(x_1,\ldots,x_n):\|x\|\leqslant 1,-1\leqslant x_n\leqslant t\}$, you may integrate the volume of the section to get an integral $\int_{-1}^t (1-s^2)^{(n-1)/2}ds$ (times the volume of $(n-1)$-dimensional ball.) This integral is a partial case of the incomplete Beta function, but it also may be computed in elementary functions (denote $s=\cos \theta$, for example).

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