# How to prove that von Mises distribution belongs to exponential family?

Can anyone help me prove this, I'm not able to simplify the distribution to find the sufficient statistics, log normalizer, etc.

• Could you write down the pdf in your post? Here is a MathJax tutorial for typesetting math. Commented May 18, 2021 at 11:56
• Apply the algorithm I give at stats.stackexchange.com/a/519715/919. It solves your problem immediately, with little work.
– whuber
Commented May 18, 2021 at 14:04

It is an exponential family distribution

An exponential family distribution with parameter vector $$\boldsymbol{\theta}$$ is one that has a log-density of the form: $$\log f(x \mid \boldsymbol{\theta})=\boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(x)+A(\boldsymbol{\theta})+B(x) .$$ As can be seen, in an exponential family the log-density the argument variable $$x$$ must be separable from the parameter vector through a sufficient statistic $$\mathbf{T}$$. For the Von-Mises distribution we have the log-density: \begin{align*} \log \operatorname{Von-Mises}(x \mid \mu, \kappa) &=\kappa \cos (x-\mu)-\log (2 \pi)-\log I_{0}(\kappa) \\ &= \kappa \cos(x)\cos(\mu) + \kappa \sin(x)\sin(\mu) -\log (2 \pi)-\log I_{0}(\kappa), \end{align*} where the support can be taken over any interval of length $$2 \pi$$. The term $$\kappa \cos (x-\mu)$$ can be manipulated into the form required for an exponential family, so it is an exponential family distribution with $$T(x) = (\cos(x), \sin(x))$$ and $$\eta(\theta) = \kappa (\cos(\mu), \sin(\mu))$$

#### It is an exponential family distribution

An exponential family distribution with parameter vector $$\boldsymbol{\theta}$$ is one that has a log-density of the form:

$$\log f(x|\boldsymbol{\theta}) = \boldsymbol{\eta}(\boldsymbol{\theta}) \cdot \mathbf{T}(x) + A(\boldsymbol{\theta}) + B(x).$$

As can be seen, in an exponential family the log-density the argument variable $$x$$ must be separable from the parameter vector through a sufficient statistic $$\mathbf{T}$$. For the Von-Mises distribution we have the log-density:

\begin{align} \log \text{VonMises}(x|\mu, \kappa) &= \kappa \cos(x-\mu) - \log(2 \pi) - \log I_0(\kappa) \\[6pt] &= \kappa [\cos(x)\cos(\mu) + \sin(x)\sin(\mu)] - \log(2 \pi) - \log I_0(\kappa) \\[6pt] &= \begin{bmatrix} \kappa \cos(\mu) \\ \kappa \sin(\mu) \end{bmatrix} \cdot \begin{bmatrix} \cos(x) \\ \sin(x) \end{bmatrix} - \log(2 \pi) - \log I_0(\kappa), \\[6pt] \end{align}

where the support can be taken over any interval of length $$2 \pi$$. This is the form required for an exponential family, so it is an exponential family distribution.

Update: Hat-tip to CechMS for correcting a major error in an earlier version of this answer.

Here are two alternative paths to arrive at the solutions in the other answers. They are interesting to mention as they dive a bit deeper into the background.

### von Mises-Fisher distribution

The von Mises distribution is a special case of the von Mises-Fisher distribution. Using the expression for the latter shows quickly that it is an distribution in the exponential family.

$$f({\bf x}; \boldsymbol{\mu}, \kappa) = h(\kappa) e^{\kappa \boldsymbol{\mu}^T {\bf x} }$$

For the von Mises distribution you have a one-dimensional variable $$\theta$$ instead of the vector $${\bf x}$$ and $$\boldsymbol{\mu} = \begin{bmatrix} \cos(\mu_\theta) \\ \sin(\mu_\theta) \end{bmatrix} \qquad \boldsymbol{x} = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix}$$

### Maximum entropy distribution

Note that the von Mises distribution is a maximum entropy distribution. A maximum entropy distribution can be expressed as an exponential function $$p(x) \propto e^{\lambda_1f_1(x)+\lambda_2f_2(x)+ \dots + \lambda_n f_n(x)}$$ Here the functions $$f_i(x)$$ relate to functions whose expectation values are constrained (e.g. particular values for the moments).

The von Mises distribution is the case where the real and imaginary parts of the first circular moments are constrained. That is we have constraints for:

$$E[\mathcal{Re}(e^{ix})] \qquad \text{and} \qquad E[\mathcal{Im}(e^{ix})]$$

So the von Mises distribution can be described as

$$f(x) \propto e^{a \mathcal{Re}( e^{ix}) + b \mathcal{Im}( e^{ix} ) } = e^{a \cos(x) + b \sin(x) } = e^{A \cos(x+\phi) }$$

where $$A = \sqrt{a^2+b^2}$$ and $$\phi = \text{atan}(b/a)$$.