Assuming one draws two points from a von Mises distribution on a circle, I am looking for the expected distance between two such points.

Given the pdf of a centered von Mises distribution $$f_X(t \mid \kappa) = \frac{e^{\kappa \cos(t)}}{2\pi I_0(\kappa)} \cdot 1_{[-\pi, \pi]}(t) \; ,$$ one can calculate the pdf that describes the distance of two points by $$f_{\Delta}(t \mid \kappa) = \frac{I_0 \left( 2\kappa \cos{\frac{t}{2}} \right)}{\pi I^2_0(\kappa)} \cdot 1_{[0, \pi]}(t)\; ,$$ where $\Delta := \min{\big\{ |X_1 - X_2|, \, 2\pi - |X_1 - X_2| \big\}} = \pi - \big||X_1 - X_2| - \pi \big|$. Now, the expected distance corresponds to the expression $$\mathbb{E}[\Delta] = \frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt ,$$ which I was not able to solve in a meaningful way even though there are some nice known integrals involving Bessel functions like $\int x I_0(x)\, dx = xI_1(x)$.

Trying to integrate it on its power series yields $$\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = \sum_{m=0}^{\infty} \left(\frac{\kappa^m}{m!}\right)^2 \int_0^{\pi} t\cos^{2m}{\left(\frac{t}{2}\right)} dt,$$ which seemed equally hard to solve.

Another desperate attempt to tackle this problem with integration by parts led to this question.


Is there a closed form for the integral $$\frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt \quad ?$$

  • $\begingroup$ this integral from $0$ to $\pi$ is as hard as asking for the indefinite integral of $t I_0(\cos t)$, which has no closed-form expression. $\endgroup$ Jan 3, 2019 at 18:47
  • $\begingroup$ I refuse to believe that this is as hard as the indefinite integral. Take the power series representation for example. WolframAlpha was able to calculate $\int_0^{pi} t \cos^{2m}{t/2}$ for each term exactly; it was just not able to come up with a general formula (which in my opinion has to exist, if it can calculate each term individually). So at least a representation of the integral as an infinite, integral-free series should be achievable which might then correspond to a known analytic function evaluated at a certain point. $\endgroup$ Jan 3, 2019 at 19:08

1 Answer 1


$$I_0(2\kappa \cos(t/2)) = \sum_{k=0}^\infty \cos(t/2)^{2k} \frac{\kappa^{2k}}{k!^2}$$ and it seems to me that $$\int_0^\pi \cos(t/2)^{2k} t \; dt = {2k \choose k} 2^{-2k-1} \pi^2 - \sum_{r=0}^{\lfloor (k-1)/2 \rfloor} {2k \choose k+2r+1} \frac{2^{2-2k}}{(2r+1)^2} $$ so that, unless I've made an error, $$ \eqalign{\int_0^\pi & I_0(2\kappa \cos(t/2)) t \; dt \cr & = \sum_{k=0}^\infty \frac{\kappa^{2k}}{k!^2} {2k \choose k} 2^{-2k-1} \pi^2 - \sum_{r=0}^\infty \sum_{k=2r+1}^\infty {2k \choose k+2r+1} \frac{\kappa^{2k}}{k!^2} \frac{2^{2-2k}}{(2r+1)^2}\cr &= \frac{\pi^2}{2} I_0(\kappa)^2 - 4 \sum_{r=0}^\infty \frac{I_{2r+1}(\kappa)^2}{(2r+1)^2} } $$

Not closed form, of course, but still...

  • $\begingroup$ Thank you Robert! Your approach seems to be very promising and numerical results show, that no error has been done in your index-battle. I opened a new threat to follow up on your thoughts. Once someone finds a closed form to this sum, I will mark your answer as accepted. $\endgroup$ Jan 5, 2019 at 20:10

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