On the expected distance of two von Mises distributed random variables












2












$begingroup$


About this thread



I opened a couple of threads here and at MathOverflow regarding the following question, but always put in a different context. Since there has been no satisfactory answer yet, I decided to collect all formulations and partial results of this problem in one single thread, hoping that this will attract more people and give us all a broader picture.





The Problem



Let $X, Y$ be two independent, centered von Mises distributed random variables on the circle, i.e. that they can be described via the density function



$$f_X(t mid kappa) = frac{e^{kappa cos(t)}}{2pi I_0(kappa)} cdot 1_{[-pi, pi]}(t) ; ,$$



where



$$I_{alpha}(z)
:= sum_{m=0}^{infty}frac{left(frac{z}{2}right)^{2m+alpha}}{m! Gamma(m+1+alpha)}
= frac{1}{2pi} int_{-pi}^{pi} e^{ialpha tau + z sin{tau}} dtau$$



denotes the modified Bessel functions of the first kind.
Let further



$$Delta := min{big{ |X_1 - X_2|, , 2pi - |X_1 - X_2| big}} = pi - big||X_1 - X_2| - pi big|$$



be their geodesic distance. The pdf for $Delta$ can be explicitly calculated and yields



$$f_{Delta}(t mid kappa) = frac{I_0 left( 2kappa cos{frac{t}{2}} right)}{pi I^2_0(kappa)} cdot 1_{[0, pi]}(t); .$$



The question now is:




Can we find a closed expression for
$$mathbb{E}[Delta]
= frac{1}{pi I^2_0(kappa)} int_0^{pi} t I_0 left( 2kappa cos{frac{t}{2}} right) dt
= frac{4}{pi I^2_0(kappa)} int_0^{pi/2} t I_0 left( 2kappa cos{t} right)dt quad ?$$






What was discussed so far



Tackling the integral in the series representation



Plugging in the series expansion representation of $I_0$ leads to



$$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = sum_{m=0}^{infty} left(frac{kappa^m}{m!}right)^2 int_0^{pi} tcos^{2m}{left(frac{t}{2}right)} dt,$$



which seemed discouraging first, but Robert Israel was able to deduce this beautiful identity



$$ int_0^pi t I_0(2kappa cos(t/2)) ; dt = frac{pi^2}{2} I_0(kappa)^2 - 4 sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2} ; .$$



This led to a follow-up thread, aiming to find a closed expression for the sum appearing on the right side of the equation.



Tackling the integral in the integral representation



Plugging in the integral representation of $I_0$ leads to



$$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = 4 int_{-pi}^{pi} int_0^{pi/2} t e^{2kappa cos{t}sin{tau}} dt , dtau ;$$



which was the actual trigger for this umbrella-thread because I didn't want to open yet another thread each time a new approach seemed somewhat promising.



Using the substitution $u = cos{t}$ and integration by parts, one can conclude that



$$int_0^{pi/2} t e^{alphacos{t}} dt = frac{pi^2}{8} + frac{alpha}{2} int_0^1 e^{alpha u} arccos^2{u} ; du ; ,$$



which looks equally hard to solve, but with Gradshteyn and Ryzhik, 4.551 one can deduce that



$$int_0^1 e^{alpha u} arccos{u} ; du = frac{pi}{2alpha} big( I_0(alpha) + boldsymbol{L_0}(alpha) -alphabig) ; ,$$



where $boldsymbol{L_0}$ is the modified Struve function, which makes the integral doesn't look that impossible anymore.



Tackling the integral directly



Again Gradshteyn and Ryzhik mention in 6.519.1 that



$$int_0^{pi/2} J_{2r}(2z cos{t}) dt = frac{pi}{2} J_r^2(z) ; ,$$
where $J_v(z) = i^z I_v(-iz)$, for $v in mathbb{N}$. Which made me question if that information is already enough to solve the sought integral.



Tackling the integral with a Fourier transform



One can exploit the fact that



$$mathbb{E}[Delta] = -ivarphi'_{Delta}(0)
= lim_{omega rightarrow 0} frac{varphi_{Delta}(omega) - varphi_{Delta}(-omega)}{2iomega}
= lim_{omega rightarrow 0} frac{mathcal{Im}left(varphi_{Delta}(omega)right)}{omega} ,$$



where



$$varphi_{Delta}(omega) := int_{-infty}^{infty} e^{itomega}f_{Delta}(t) dt ,$$



is the characteristic function of $Delta$.

Note, however that by pulling the limit into the integral (e.g. by dominated convergence) we end up with the same integral as above. Hence, this approach can only be useful if one manages to exploit some clever properties of the Fourier transform.





What do we gain from that?



Besides solving the original problem which I think is interesting enough in its own right, I feel like the question has gotten to a point where solving it would result in a bunch of identities that seem to be missing in the literature.

I am not intending to publish any of this (my research doesn't lie in pure math anyways) although citing StackExchange is actually possible since a couple of years now. I would rather want to see these identities to be added in appropriate books for integral tables and series identities and somehow give credit where credit is due.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    About this thread



    I opened a couple of threads here and at MathOverflow regarding the following question, but always put in a different context. Since there has been no satisfactory answer yet, I decided to collect all formulations and partial results of this problem in one single thread, hoping that this will attract more people and give us all a broader picture.





    The Problem



    Let $X, Y$ be two independent, centered von Mises distributed random variables on the circle, i.e. that they can be described via the density function



    $$f_X(t mid kappa) = frac{e^{kappa cos(t)}}{2pi I_0(kappa)} cdot 1_{[-pi, pi]}(t) ; ,$$



    where



    $$I_{alpha}(z)
    := sum_{m=0}^{infty}frac{left(frac{z}{2}right)^{2m+alpha}}{m! Gamma(m+1+alpha)}
    = frac{1}{2pi} int_{-pi}^{pi} e^{ialpha tau + z sin{tau}} dtau$$



    denotes the modified Bessel functions of the first kind.
    Let further



    $$Delta := min{big{ |X_1 - X_2|, , 2pi - |X_1 - X_2| big}} = pi - big||X_1 - X_2| - pi big|$$



    be their geodesic distance. The pdf for $Delta$ can be explicitly calculated and yields



    $$f_{Delta}(t mid kappa) = frac{I_0 left( 2kappa cos{frac{t}{2}} right)}{pi I^2_0(kappa)} cdot 1_{[0, pi]}(t); .$$



    The question now is:




    Can we find a closed expression for
    $$mathbb{E}[Delta]
    = frac{1}{pi I^2_0(kappa)} int_0^{pi} t I_0 left( 2kappa cos{frac{t}{2}} right) dt
    = frac{4}{pi I^2_0(kappa)} int_0^{pi/2} t I_0 left( 2kappa cos{t} right)dt quad ?$$






    What was discussed so far



    Tackling the integral in the series representation



    Plugging in the series expansion representation of $I_0$ leads to



    $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = sum_{m=0}^{infty} left(frac{kappa^m}{m!}right)^2 int_0^{pi} tcos^{2m}{left(frac{t}{2}right)} dt,$$



    which seemed discouraging first, but Robert Israel was able to deduce this beautiful identity



    $$ int_0^pi t I_0(2kappa cos(t/2)) ; dt = frac{pi^2}{2} I_0(kappa)^2 - 4 sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2} ; .$$



    This led to a follow-up thread, aiming to find a closed expression for the sum appearing on the right side of the equation.



    Tackling the integral in the integral representation



    Plugging in the integral representation of $I_0$ leads to



    $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = 4 int_{-pi}^{pi} int_0^{pi/2} t e^{2kappa cos{t}sin{tau}} dt , dtau ;$$



    which was the actual trigger for this umbrella-thread because I didn't want to open yet another thread each time a new approach seemed somewhat promising.



    Using the substitution $u = cos{t}$ and integration by parts, one can conclude that



    $$int_0^{pi/2} t e^{alphacos{t}} dt = frac{pi^2}{8} + frac{alpha}{2} int_0^1 e^{alpha u} arccos^2{u} ; du ; ,$$



    which looks equally hard to solve, but with Gradshteyn and Ryzhik, 4.551 one can deduce that



    $$int_0^1 e^{alpha u} arccos{u} ; du = frac{pi}{2alpha} big( I_0(alpha) + boldsymbol{L_0}(alpha) -alphabig) ; ,$$



    where $boldsymbol{L_0}$ is the modified Struve function, which makes the integral doesn't look that impossible anymore.



    Tackling the integral directly



    Again Gradshteyn and Ryzhik mention in 6.519.1 that



    $$int_0^{pi/2} J_{2r}(2z cos{t}) dt = frac{pi}{2} J_r^2(z) ; ,$$
    where $J_v(z) = i^z I_v(-iz)$, for $v in mathbb{N}$. Which made me question if that information is already enough to solve the sought integral.



    Tackling the integral with a Fourier transform



    One can exploit the fact that



    $$mathbb{E}[Delta] = -ivarphi'_{Delta}(0)
    = lim_{omega rightarrow 0} frac{varphi_{Delta}(omega) - varphi_{Delta}(-omega)}{2iomega}
    = lim_{omega rightarrow 0} frac{mathcal{Im}left(varphi_{Delta}(omega)right)}{omega} ,$$



    where



    $$varphi_{Delta}(omega) := int_{-infty}^{infty} e^{itomega}f_{Delta}(t) dt ,$$



    is the characteristic function of $Delta$.

    Note, however that by pulling the limit into the integral (e.g. by dominated convergence) we end up with the same integral as above. Hence, this approach can only be useful if one manages to exploit some clever properties of the Fourier transform.





    What do we gain from that?



    Besides solving the original problem which I think is interesting enough in its own right, I feel like the question has gotten to a point where solving it would result in a bunch of identities that seem to be missing in the literature.

    I am not intending to publish any of this (my research doesn't lie in pure math anyways) although citing StackExchange is actually possible since a couple of years now. I would rather want to see these identities to be added in appropriate books for integral tables and series identities and somehow give credit where credit is due.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      About this thread



      I opened a couple of threads here and at MathOverflow regarding the following question, but always put in a different context. Since there has been no satisfactory answer yet, I decided to collect all formulations and partial results of this problem in one single thread, hoping that this will attract more people and give us all a broader picture.





      The Problem



      Let $X, Y$ be two independent, centered von Mises distributed random variables on the circle, i.e. that they can be described via the density function



      $$f_X(t mid kappa) = frac{e^{kappa cos(t)}}{2pi I_0(kappa)} cdot 1_{[-pi, pi]}(t) ; ,$$



      where



      $$I_{alpha}(z)
      := sum_{m=0}^{infty}frac{left(frac{z}{2}right)^{2m+alpha}}{m! Gamma(m+1+alpha)}
      = frac{1}{2pi} int_{-pi}^{pi} e^{ialpha tau + z sin{tau}} dtau$$



      denotes the modified Bessel functions of the first kind.
      Let further



      $$Delta := min{big{ |X_1 - X_2|, , 2pi - |X_1 - X_2| big}} = pi - big||X_1 - X_2| - pi big|$$



      be their geodesic distance. The pdf for $Delta$ can be explicitly calculated and yields



      $$f_{Delta}(t mid kappa) = frac{I_0 left( 2kappa cos{frac{t}{2}} right)}{pi I^2_0(kappa)} cdot 1_{[0, pi]}(t); .$$



      The question now is:




      Can we find a closed expression for
      $$mathbb{E}[Delta]
      = frac{1}{pi I^2_0(kappa)} int_0^{pi} t I_0 left( 2kappa cos{frac{t}{2}} right) dt
      = frac{4}{pi I^2_0(kappa)} int_0^{pi/2} t I_0 left( 2kappa cos{t} right)dt quad ?$$






      What was discussed so far



      Tackling the integral in the series representation



      Plugging in the series expansion representation of $I_0$ leads to



      $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = sum_{m=0}^{infty} left(frac{kappa^m}{m!}right)^2 int_0^{pi} tcos^{2m}{left(frac{t}{2}right)} dt,$$



      which seemed discouraging first, but Robert Israel was able to deduce this beautiful identity



      $$ int_0^pi t I_0(2kappa cos(t/2)) ; dt = frac{pi^2}{2} I_0(kappa)^2 - 4 sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2} ; .$$



      This led to a follow-up thread, aiming to find a closed expression for the sum appearing on the right side of the equation.



      Tackling the integral in the integral representation



      Plugging in the integral representation of $I_0$ leads to



      $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = 4 int_{-pi}^{pi} int_0^{pi/2} t e^{2kappa cos{t}sin{tau}} dt , dtau ;$$



      which was the actual trigger for this umbrella-thread because I didn't want to open yet another thread each time a new approach seemed somewhat promising.



      Using the substitution $u = cos{t}$ and integration by parts, one can conclude that



      $$int_0^{pi/2} t e^{alphacos{t}} dt = frac{pi^2}{8} + frac{alpha}{2} int_0^1 e^{alpha u} arccos^2{u} ; du ; ,$$



      which looks equally hard to solve, but with Gradshteyn and Ryzhik, 4.551 one can deduce that



      $$int_0^1 e^{alpha u} arccos{u} ; du = frac{pi}{2alpha} big( I_0(alpha) + boldsymbol{L_0}(alpha) -alphabig) ; ,$$



      where $boldsymbol{L_0}$ is the modified Struve function, which makes the integral doesn't look that impossible anymore.



      Tackling the integral directly



      Again Gradshteyn and Ryzhik mention in 6.519.1 that



      $$int_0^{pi/2} J_{2r}(2z cos{t}) dt = frac{pi}{2} J_r^2(z) ; ,$$
      where $J_v(z) = i^z I_v(-iz)$, for $v in mathbb{N}$. Which made me question if that information is already enough to solve the sought integral.



      Tackling the integral with a Fourier transform



      One can exploit the fact that



      $$mathbb{E}[Delta] = -ivarphi'_{Delta}(0)
      = lim_{omega rightarrow 0} frac{varphi_{Delta}(omega) - varphi_{Delta}(-omega)}{2iomega}
      = lim_{omega rightarrow 0} frac{mathcal{Im}left(varphi_{Delta}(omega)right)}{omega} ,$$



      where



      $$varphi_{Delta}(omega) := int_{-infty}^{infty} e^{itomega}f_{Delta}(t) dt ,$$



      is the characteristic function of $Delta$.

      Note, however that by pulling the limit into the integral (e.g. by dominated convergence) we end up with the same integral as above. Hence, this approach can only be useful if one manages to exploit some clever properties of the Fourier transform.





      What do we gain from that?



      Besides solving the original problem which I think is interesting enough in its own right, I feel like the question has gotten to a point where solving it would result in a bunch of identities that seem to be missing in the literature.

      I am not intending to publish any of this (my research doesn't lie in pure math anyways) although citing StackExchange is actually possible since a couple of years now. I would rather want to see these identities to be added in appropriate books for integral tables and series identities and somehow give credit where credit is due.










      share|cite|improve this question











      $endgroup$




      About this thread



      I opened a couple of threads here and at MathOverflow regarding the following question, but always put in a different context. Since there has been no satisfactory answer yet, I decided to collect all formulations and partial results of this problem in one single thread, hoping that this will attract more people and give us all a broader picture.





      The Problem



      Let $X, Y$ be two independent, centered von Mises distributed random variables on the circle, i.e. that they can be described via the density function



      $$f_X(t mid kappa) = frac{e^{kappa cos(t)}}{2pi I_0(kappa)} cdot 1_{[-pi, pi]}(t) ; ,$$



      where



      $$I_{alpha}(z)
      := sum_{m=0}^{infty}frac{left(frac{z}{2}right)^{2m+alpha}}{m! Gamma(m+1+alpha)}
      = frac{1}{2pi} int_{-pi}^{pi} e^{ialpha tau + z sin{tau}} dtau$$



      denotes the modified Bessel functions of the first kind.
      Let further



      $$Delta := min{big{ |X_1 - X_2|, , 2pi - |X_1 - X_2| big}} = pi - big||X_1 - X_2| - pi big|$$



      be their geodesic distance. The pdf for $Delta$ can be explicitly calculated and yields



      $$f_{Delta}(t mid kappa) = frac{I_0 left( 2kappa cos{frac{t}{2}} right)}{pi I^2_0(kappa)} cdot 1_{[0, pi]}(t); .$$



      The question now is:




      Can we find a closed expression for
      $$mathbb{E}[Delta]
      = frac{1}{pi I^2_0(kappa)} int_0^{pi} t I_0 left( 2kappa cos{frac{t}{2}} right) dt
      = frac{4}{pi I^2_0(kappa)} int_0^{pi/2} t I_0 left( 2kappa cos{t} right)dt quad ?$$






      What was discussed so far



      Tackling the integral in the series representation



      Plugging in the series expansion representation of $I_0$ leads to



      $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = sum_{m=0}^{infty} left(frac{kappa^m}{m!}right)^2 int_0^{pi} tcos^{2m}{left(frac{t}{2}right)} dt,$$



      which seemed discouraging first, but Robert Israel was able to deduce this beautiful identity



      $$ int_0^pi t I_0(2kappa cos(t/2)) ; dt = frac{pi^2}{2} I_0(kappa)^2 - 4 sum_{r=0}^infty frac{I_{2r+1}(kappa)^2}{(2r+1)^2} ; .$$



      This led to a follow-up thread, aiming to find a closed expression for the sum appearing on the right side of the equation.



      Tackling the integral in the integral representation



      Plugging in the integral representation of $I_0$ leads to



      $$int_0^{pi} t cdot I_0 left( 2kappa cos{frac{t}{2}} right) dt = 4 int_{-pi}^{pi} int_0^{pi/2} t e^{2kappa cos{t}sin{tau}} dt , dtau ;$$



      which was the actual trigger for this umbrella-thread because I didn't want to open yet another thread each time a new approach seemed somewhat promising.



      Using the substitution $u = cos{t}$ and integration by parts, one can conclude that



      $$int_0^{pi/2} t e^{alphacos{t}} dt = frac{pi^2}{8} + frac{alpha}{2} int_0^1 e^{alpha u} arccos^2{u} ; du ; ,$$



      which looks equally hard to solve, but with Gradshteyn and Ryzhik, 4.551 one can deduce that



      $$int_0^1 e^{alpha u} arccos{u} ; du = frac{pi}{2alpha} big( I_0(alpha) + boldsymbol{L_0}(alpha) -alphabig) ; ,$$



      where $boldsymbol{L_0}$ is the modified Struve function, which makes the integral doesn't look that impossible anymore.



      Tackling the integral directly



      Again Gradshteyn and Ryzhik mention in 6.519.1 that



      $$int_0^{pi/2} J_{2r}(2z cos{t}) dt = frac{pi}{2} J_r^2(z) ; ,$$
      where $J_v(z) = i^z I_v(-iz)$, for $v in mathbb{N}$. Which made me question if that information is already enough to solve the sought integral.



      Tackling the integral with a Fourier transform



      One can exploit the fact that



      $$mathbb{E}[Delta] = -ivarphi'_{Delta}(0)
      = lim_{omega rightarrow 0} frac{varphi_{Delta}(omega) - varphi_{Delta}(-omega)}{2iomega}
      = lim_{omega rightarrow 0} frac{mathcal{Im}left(varphi_{Delta}(omega)right)}{omega} ,$$



      where



      $$varphi_{Delta}(omega) := int_{-infty}^{infty} e^{itomega}f_{Delta}(t) dt ,$$



      is the characteristic function of $Delta$.

      Note, however that by pulling the limit into the integral (e.g. by dominated convergence) we end up with the same integral as above. Hence, this approach can only be useful if one manages to exploit some clever properties of the Fourier transform.





      What do we gain from that?



      Besides solving the original problem which I think is interesting enough in its own right, I feel like the question has gotten to a point where solving it would result in a bunch of identities that seem to be missing in the literature.

      I am not intending to publish any of this (my research doesn't lie in pure math anyways) although citing StackExchange is actually possible since a couple of years now. I would rather want to see these identities to be added in appropriate books for integral tables and series identities and somehow give credit where credit is due.







      sequences-and-series probability-distributions definite-integrals bessel-functions






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      share|cite|improve this question













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      edited Jan 14 at 14:32









      Lee David Chung Lin

      4,50841342




      4,50841342










      asked Jan 14 at 10:37









      chickenNinja123chickenNinja123

      13813




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