📂Probability Distribution

Von Mises Distribution

Definition ^{1 2}

The von Mises Distribution is a continuous probability distribution with the probability density function given for Mean Direction $\mu \in \mathbb{R}$ and Concentration $\kappa > 0$ as follows: $$ f(x) = {{ 1 } \over {2 \pi I_{0} \left( \kappa \right) }} \exp \left( \kappa \cos \left( x - \mu \right) \right) \qquad , x \in \mathbb{R} \pmod{2 \pi} $$

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• $I_{\nu}$ is a modified Bessel function of the first kind of order $\nu$, and the reason for using such complex functions is explained in the post Why Modified Bessel Functions of the First Kind Appear in Directional Statistics.

Description

The von Mises Distribution represents the simplest distribution encountered in Directional Statistics, modeling data sampled on the circumference of a circle $S^{1}$. It is also referred to as the Circular Normal Distribution or Tikhonov Distribution.

The probability of sampling from the exponential function $\exp$ is higher as it approaches $\infty$ and lower as it approaches $-\infty$, naturally determined by $\cos \left( x - \mu \right)$. $x \approx \mu$, places closer to the mean direction, become $\cos \approx 1$, thus being sampled more frequently, whereas the opposite direction has a very low probability.

The Concentration $\kappa$ feels the opposite of dispersion, with higher values increasing the probability of the mean direction.

Generalizations of the von Mises distribution include the dimension-increased von Mises-Fisher distribution, the expansion to a torus leading to the Bivariate von Mises distribution, the von Mises-Bingham distribution² using eight parameters, and the Kent distribution using only five parameters.

Theorem

The following summarizes why it is appropriate to call the von Mises distribution a Circular Normal Distribution. The assumption that $\kappa$ is sufficiently large means that the probability is concentrated near $\mu$, and by only sampling around $\mu$ and not using a wide range of $S^{1}$, it closely approximates a normal distribution along its tangent. This is also referred to as LAN(Local Asymptotic Normality).

Circular Normal Distribution

For sufficiently large $\kappa = \sigma^{-2}$, $f(x)$ approximates the probability density function of a normal distribution. $$ f(x) \approx {{ 1 } \over { \sigma \sqrt{2 \pi} }} \exp \left[ {{ - \left( x - \mu \right)^{2} } \over { 2 \sigma^{2} }} \right] $$

Proof

Taylor expansion of the cosine function: $$ \cos x = \frac { 1 }{ 0! }-\frac { { x } ^{ 2 } }{ 2! }+\frac { { x } ^{ 4 } }{ 4! }-\frac { { x } ^{ 6 } }{ 6! }+ \cdots $$

Assuming $\kappa$ is sufficiently large, discarding the third and subsequent terms in the Taylor expansion of the cosine around $\mu$ yields the following.

$$ \begin{align*} f(x) = & {{ 1 } \over {2 \pi I_{0} \left( \kappa \right) }} \exp \left( \kappa \cos \left( x - \mu \right) \right) \\ \approx& {{ 1 } \over {2 \pi I_{0} \left( \kappa \right) }} \exp \left( \kappa \left[ 1 - {{ \left( x - \mu \right)^{2} } \over { 2 }} \right] \right) \\ =& {{ 1 } \over {2 \pi I_{0} \left( \kappa \right) }} e^{\kappa} \exp \left( - {{ \left( x - \mu \right)^{2} } \over { 2 \sigma^{2} }}\right) \end{align*} $$

Meanwhile, since $\pi = 3.141592 \cdots$ can be considered as $\kappa = 1$ and is much greater than $z_{0.99} = 2.58 \cdots$ of the standard normal distribution under the assumption that $\kappa$ is sufficiently large, $I_{0} (\kappa)$ also $$ \begin{align*} 2\pi I_{0} (\kappa) =& \int_{-\pi}^{\pi} \exp \left( \kappa \cos \left( x - \mu \right) \right) dx \\ =& \int_{-\pi}^{\pi} \exp \left( \kappa \cos t \right) dt \\ \approx& \int_{-\infty}^{\infty} \exp \left( \kappa - {{ t^{2} } \over { 2 \sigma^{2} }}\right) dt \\ = & e^{\kappa} \int_{-\infty}^{\infty} \exp \left( - {{ t^{2} } \over { 2 \sigma^{2} }}\right) dt \\ = & \sigma \sqrt{2 \pi} e^{\kappa} \int_{-\infty}^{\infty} {{ 1 } \over { \sigma \sqrt{2 \pi} }} \exp \left( - {{ t^{2} } \over { 2 \sigma^{2} }}\right) dt \\ = & \sigma \sqrt{2 \pi} e^{\kappa} \end{align*} $$ leads to the following approximation. $$ f(x) \approx {{ 1 } \over { \sigma \sqrt{2 \pi} }} \exp \left[ {{ - \left( x - \mu \right)^{2} } \over { 2 \sigma^{2} }} \right] $$

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