Volume of the intersection of ellipsoids












11












$begingroup$


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



Given an $n$-vector $c$ and a symmetric positive-definite $ntimes n$ 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.






linear-algebra geometry probability-distributions normal-distribution volume







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$endgroup$

















    11












    $begingroup$


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



    Given an $n$-vector $c$ and a symmetric positive-definite $ntimes n$ 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.






    linear-algebra geometry probability-distributions normal-distribution volume







    share|cite|improve this question











    $endgroup$















      11












      11








      11


      3



      $begingroup$


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



      Given an $n$-vector $c$ and a symmetric positive-definite $ntimes n$ 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.






      linear-algebra geometry probability-distributions normal-distribution volume







      share|cite|improve this question











      $endgroup$




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



      Given an $n$-vector $c$ and a symmetric positive-definite $ntimes n$ 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.






      linear-algebra geometry probability-distributions normal-distribution volume




      linear-algebra geometry probability-distributions normal-distribution volume






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 10 '18 at 12:24







      sds

















      asked Mar 28 '13 at 16:36









      sdssds

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          2 Answers
          2






          active

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          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Are you sure the intersection curve will be planar?
            $endgroup$
            – Rahul
            Jan 29 '14 at 0:12










          • $begingroup$
            @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
            $endgroup$
            – j13r
            Jan 29 '14 at 0:40



















          0












          $begingroup$

          It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Are you sure the intersection curve will be planar?
              $endgroup$
              – Rahul
              Jan 29 '14 at 0:12










            • $begingroup$
              @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
              $endgroup$
              – j13r
              Jan 29 '14 at 0:40
















            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Are you sure the intersection curve will be planar?
              $endgroup$
              – Rahul
              Jan 29 '14 at 0:12










            • $begingroup$
              @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
              $endgroup$
              – j13r
              Jan 29 '14 at 0:40














            1












            1








            1





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 28 '14 at 23:58









            j13rj13r

            27219




            27219












            • $begingroup$
              Are you sure the intersection curve will be planar?
              $endgroup$
              – Rahul
              Jan 29 '14 at 0:12










            • $begingroup$
              @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
              $endgroup$
              – j13r
              Jan 29 '14 at 0:40


















            • $begingroup$
              Are you sure the intersection curve will be planar?
              $endgroup$
              – Rahul
              Jan 29 '14 at 0:12










            • $begingroup$
              @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
              $endgroup$
              – j13r
              Jan 29 '14 at 0:40
















            $begingroup$
            Are you sure the intersection curve will be planar?
            $endgroup$
            – Rahul
            Jan 29 '14 at 0:12




            $begingroup$
            Are you sure the intersection curve will be planar?
            $endgroup$
            – Rahul
            Jan 29 '14 at 0:12












            $begingroup$
            @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
            $endgroup$
            – j13r
            Jan 29 '14 at 0:40




            $begingroup$
            @RahulNarain: true. mathoverflow.net/questions/66431/… could be useful for working towards a solution.
            $endgroup$
            – j13r
            Jan 29 '14 at 0:40











            0












            $begingroup$

            It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".






                share|cite|improve this answer











                $endgroup$



                It appears that for my purposes the Hellinger distance between the corresponding normal distributions is a perfectly acceptable "computable alternative".







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 27 '18 at 22:36

























                answered Dec 7 '18 at 16:08









                sdssds

                3,5381129




                3,5381129






























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