Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ von Mises distribution ▷ Relationship to bivariate normal distribution

Theorem: Let $Y_1$ and $Y_2$ follow a bivariate normal distribution with means $\mu_1$ and $\mu_2$ and spherical covariance matrix:

\[\label{eq:bvn} Y = \left[ \begin{matrix} Y_1 \\ Y_2 \end{matrix} \right] \sim \mathcal{N}\left( \left[ \begin{matrix} \mu_1 \\ \mu_2 \end{matrix} \right], \sigma^2 \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right) \; .\]

Consider the polar-coordinate representation of the complex numbers $Y_1 + i Y_2$ and $\mu_1 + i \mu_2$ using the random variables $X$ and $R$ as well as the constants $x_0 \in [0, 2 \pi)$ and $r_0 \in [0, \infty)$:

\[\label{eq:Y12-mu12-rX-rX0} \begin{split} Y_1 &= R \cdot \cos X, \quad Y_2 = R \cdot \sin X \\ \mu_1 &= r_0 \cdot \cos x_0, \quad \mu_2 = r_0 \cdot \sin x_0 \; . \end{split}\]

Then, the circular random variable $X$, i.e. the angular direction of the random vector $Y$ follows a von Mises distribution with circular mean $\mu = x_0$ and reciprocal dispersion $\kappa = r_0 / \sigma^2$:

\[\label{eq:vm} X \sim \mathrm{vM}(\mu, \kappa) \; .\]

Proof: The probability density function of the bivariate normal distribution with the parameters specified in \eqref{eq:bvn} is equal to

\[\label{eq:bvn-pdf} \begin{split} p(y_1,y_2) &= \frac{1}{2 \pi \sqrt{\sigma^2 \sigma^2 - 0}} \cdot \exp \left[ -\frac{1}{2} \frac{\sigma^2 (y_1-\mu_1)^2 - 2 \cdot 0 \cdot (y_1-\mu_1)(y_2-\mu_2) + \sigma^2 (y_2-\mu_2)^2}{\sigma^2 \sigma^2 - 0} \right] \\ &= \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2} \frac{(y_1-\mu_1)^2 + (y_2-\mu_2)^2}{\sigma^2} \right] \; . \end{split}\]

Substituting the expressions from \eqref{eq:Y12-mu12-rX-rX0}, we obtain an unnormalized distribution over $X$ and $R$:

\[\label{eq:vm-pdf-s1} \begin{split} p(x,r) &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( (r \cos x - r_0 \cos x_0)^2 + (r \sin x - r_0 \sin x_0)^2 \right) \right] \\ &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( (r^2 \cos^2 x - 2 r r_0 \cos x \cos x_0 + r_0^2 \cos^2 x_0) + \right. \right. \\ &\hphantom{= \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \right]} \left. \left. (r^2 \sin^2 x - 2 r r_0 \sin x \sin x_0 + r_0^2 \sin^2 x_0) \right) \right] \; . \end{split}\]

Using the trigonometric identities

\[\label{eq:sin-cos} \begin{split} \sin^2 x + \cos^2 x &= 1 \\ \cos x \cos y + \sin x \sin y &= \cos(x-y) \; , \end{split}\]

this can be further developped into:

\[\label{eq:vm-pdf-s2} \begin{split} p(x,r) &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( r^2 + r_0^2 - 2 r r_0 (\cos x \cos x_0 + \sin x \sin x_0) \right) \right] \\ &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( r^2 + r_0^2 - 2 r r_0 \cos(x-x_0) \right) \right] \; . \end{split}\]

Conditioning on the unit circle, i.e. setting $r = 1$ and eliminanting this random variable, and collecting all terms not depending on $x$ into a constant, we get:

\[\label{eq:vm-pdf-s3} \begin{split} p(x) &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( 1 + r_0^2 - 2 r_0 \cos(x-x_0) \right) \right] \\ &\propto \frac{1}{2 \pi \sigma^2} \cdot \exp \left[ \frac{r_0}{\sigma^2} \cos(x-x_0) + \mathrm{const.} \right] \; . \end{split}\]

Finally, substituing $x_0 = \mu$ and $r_0 / \sigma^2 = \kappa$, and normalizing the right-hand side to get a probability density function, we obtain

\[\label{eq:vm-pdf-qed} \begin{split} p(x) &= \frac{1}{2 \pi I_0(\kappa)} \cdot \exp \left[ \kappa \cos(x-\mu) \right] \\ \mathrm{with} \quad I_0(\kappa) &= \frac{1}{2\pi} \int_0^{2\pi} \exp \left[ \kappa \cos(x) \right] \, \mathrm{d}x \end{split}\]

which is equivalent to the probability density function of the von Mises distribution.

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Metadata: ID: P536 | shortcut: vm-bvn | author: JoramSoch | date: 2026-04-22, 15:24.