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Background:

In psychophysics or the study of ant navigation it's important to represent random variables on a circle. The most popular distribution for doing so is the von-Mises distribution (the wrapped Gaussian is too messy to work with). This is the von-Mises distribution for a random variable on a circle, $\theta \in [-\pi,\pi)$: \begin{align} p(\theta;\mu,\kappa) = \frac{e^{\kappa cos(\theta - \mu)}}{\int_{-\pi}^\pi e^{\kappa cos(\theta)} \text{d}\theta} = \frac{e^{\kappa cos(\theta - \mu)}}{2 I_0(\kappa)} , \end{align} where $\kappa$ is the concentration parameter (have the inverse variance $\kappa \sim \sigma^{-2}$ in mind), $\mu$ the mean and the normalisation can be expressed in terms of the modified Bessel function of the first kind, $I_0(\kappa)$.

Problem statement:

Wikipedia (https://en.wikipedia.org/wiki/Von_Mises_distribution) claims that \begin{align} \lim_{\kappa \rightarrow \infty} p(\theta;\mu,\kappa) = \text{Normal}(\theta; \mu, \kappa^{-1/2}). \end{align}

(The limit is not a rigorous mathematical limit but more a physicist-way of saying: expand the expression and throw higher-order terms away). It makes sense: If the distribution is highly concentrated, we can forget about the circularity and just approximate the von-Mises as normal. Mathematically it makes intuitive sense, too. The argument of the exponential ranges between $\pm \kappa$, being 0 always at $\theta = \pm \pi/2$. At the same time, the normalisation constant grows large such that the whole distribution is essentially divided by a very large number and tends to zero everywhere except around it's mode. Around the mode, the cosine can be approximated by $\cos(\epsilon) = 1- \epsilon^2/2$, yielding a nice normal distribution. Despite this, I find it incredibly hard to show this! Can anyone help me with this technical challenge?

My (unsuccessful) approach:

I've tried to use Taylor expansion for the cosine and the following series representations (http://mhtlab.uwaterloo.ca/courses/me755/web_chap4.pdf) of the Bessel function for large $\kappa$: \begin{align} I_0(\kappa) = \sum_{n}^\infty \frac{(\kappa/2)^{2n}}{n!\Gamma(n+1)} \approx \frac{e^\kappa}{\sqrt{2\pi \kappa}}(1 + \frac{1}{8\kappa} + \mathcal{O}(\kappa^{-2})), \end{align} however, it didn't get me anywhere. Does anyone have any ideas how to attack this problem? Any help is much appreciated!

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  • $\begingroup$ How can the limit as $\kappa\to\infty$ still be a function of $\kappa$? $\endgroup$ – Augustin Mar 3 '16 at 14:05
  • $\begingroup$ The meaning of this limit is 'take $\kappa$ as large as you want and consider the lowest non-trivial terms'. I guess coming from Physics, I'm not super-rigorous about using the limes. I'll edit the post. $\endgroup$ – JanJeg Mar 3 '16 at 14:41
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I'm not entirely sure how rigorous this is so check over my work carefully.

From the pdf $\frac{e^{\kappa\cos(x-\mu)}}{2\pi I_0(\kappa)}$

You already got that $I_0(\kappa)= \frac{e^{\kappa}}{\sqrt{2 \pi \kappa}}(1+\frac{1}{8\kappa})$

In the limit only angles close to the mean will be significant therefore you can assume that $x-\mu$ is small.

For small $z$ we have $cos(z)=1-\frac{z^2}{2}$

Substitute the approximations for $cos(x)$ and $I(\kappa)$ into the pdf to get

$ \frac{1}{2\pi I_0(\kappa)} e^{\kappa \left(1-\frac{(x-\mu)^2}{2}\right)}$

$ \frac{1}{2\pi \frac{e^{\kappa}}{\sqrt{2 \pi \kappa}}(1+\frac{1}{8\kappa})} e^{\kappa \left(1-\frac{(x-\mu)^2}{2}\right)}$

Expanding out the exponent in the numerator and canceling $e^{\kappa}$ terms gives:

$ \frac{1}{2\pi \frac{1}{\sqrt{2 \pi \kappa}}(1+\frac{1}{8\kappa})} e^{ \left(\frac{\kappa(x-\mu)^2}{2}\right)}$

Neglecting the $\frac{1}{8\kappa}$ term in the denominator and tidying up the constants gives you

$ \frac{1}{\sqrt{\frac{2 \pi}{\kappa}}} e^{ \left(\frac{\kappa(x-\mu)^2}{2}\right)}$

Which is a normal distribution

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  • $\begingroup$ Thanks for the effort. But why do you assume that only small angles are important? Why can you just neglect the entire $\cos(x)$ series except for the first two terms? I would like to see that the denominator basically 'kills' the terms within the cosine series. The argument of the cosine could be as big a $/pi$ making your approximation at least questionable, I think. $\endgroup$ – JanJeg Mar 3 '16 at 15:53
  • $\begingroup$ @JanJeg I admit that was a weak part of the proof. To show that you can neglect higher order terms in the limit you can show that exp(-\kappa(x^2+[other terms])) - exp(-\kappa(x^2)) = 0 in the limit $\endgroup$ – Hugh Mar 3 '16 at 18:24
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Thanks Hugh, you got me onto the right path!

Comparing two values $x_d,x_D$ at distances $d$ and $D$ from the mean and $d < D$, the ratio of pdf's approaches zero quickly as I let $\kappa$ grow:

\begin{align} \frac{p(x_D)}{p(x_d)} = e^{\kappa (\cos(D) - \cos(d)} \equiv e^{- \kappa \epsilon }, \end{align} where $\epsilon := \cos(d) - \cos(D) > 0$. Suppose $\cos(y) \approx 1 - \frac{y^2}{2}$ is a valid approximation for $y=d$ but not for $y=D$. Then, I choose $\kappa \sim \epsilon^{-1}$ to ensure that the approximated pdf is only significantly different from zero where $d$ is small enough to taylor-expand the cosine to the quadratic order, as you suggested.

I am aware of the fact that this might not be a rigorous mathematical proof but it suffices for my application. Thanks again for your help!

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