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How do I compute the volume of the intersection of two $n$-dimensional ellipsoids?

Given an $n$-vector $c$ and a symmetric positive-definite $n\times n$ matrix $A$, define the ellipsoid $$E(c,A)=\{x|(A(x-c),x-c)<1\}$$ where $(\cdot,\cdot)$ if the dot product.

Then $$\mathrm{vol}(E(c,A))=\frac{u}{\det A}$$ where $u=\mathrm{vol}(E(0,1))$ is the volume of the unit sphere.

The question is: how do I compute the volume of the intersection $$\mathrm{vol}(E(c_1,A_1)\cap E(c_2,A_2))$$

I am more interested in being able to compute something relevant reasonably fast than in the exact correctness of the value. E.g., I would be happy to use the volumes of parallelepipeds (and their intersections) instead of ellipsoids.

EDIT: another acceptable alternative would be to define normal densities $f_1$ and $f_2$ (with mean $c_i$ and covariance $A_i$). What is $\int_{\Bbb{R}^n}f_1 f_2$? Something ugly, alas.

Rodrigo de Azevedo
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sds
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2 Answers2

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By linear coordinate transform you can simplify the problem to a unit n-sphere intersecting with an ellipse that is aligned with the axes (i.e. not rotated).

The simple case is where you have a intersection in one continuous area only (rather than one ellipse poking through the other, which is more complicated, and I will ignore this general case). In that simple case, the intersecting hyper-plane can be found similarly as in https://math.stackexchange.com/questions/162250/how-to-compute-the-volume-of-intersection-between-two-hyperspheres

You can compute the cap of the n-sphere as in that question. The remainder is the intersection of the ellipsoid with the hyperplane. I think it should be possible to apply another linear coordinate transform, and get another n-sphere cap for the second term.

I'd myself be interested in the details of this solution though.

j13r
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  • Are you sure the intersection curve will be planar? –  Jan 29 '14 at 00:12
  • @RahulNarain: true. http://mathoverflow.net/questions/66431/parametrization-of-the-intersection-of-an-ellipsoid-with-a-sphere could be useful for working towards a solution. – j13r Jan 29 '14 at 00:40
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It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".

sds
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