# Sum of two Von Mises random variables

I know that the sum of two independent normally distributed random variables is also a normal random variable, but is this true of other distributions? For example, what probability distribution does the sum of two Von Mises distributions follow?

Yes, this property is true for some but not all other distributions. The distribution for the sum of two independent random variables is the convolution of the individual distributions. Some other distributions for which this holds includes the Poisson, Cauchy, and Gamma distributions (for the Gamma, they must have the same rate parameter). See here a list here.

My answer is incomplete because I don't know if this holds for the von Mises distribution.

Previous answer (retained for completeness) My answer is incomplete because I don't know what the distribution of the sum of independent Von Mises random variables is, but I do know it is not a Von Mises random variable: the Von Mises distribution has a compact support, and so the sum of two such random variables may therefore fall outside this interval.

• Could you explain what you understand the von Mises distribution to be? Usually this term is applied to a distribution of circular variables, which means "fall outside this interval" makes no sense.
– whuber
Commented Dec 24, 2019 at 16:53
• I think that @psboonstra means that if you sum two Von Mises deviates, $x$ and $y$, ignoring the fact that they are supposed to be circular, the range of $z=x+y$ is $[-2\pi,2\pi]$, not $[\pi,\pi]$. Commented Dec 24, 2019 at 17:50
• Perhaps--but that's irrelevant to demonstrating the sum does not have a von Mises distribution. For the von Mises distribution the sum must be computed modulo $2\pi.$
– whuber
Commented Dec 24, 2019 at 18:34
• Yes, you are correct @whuber. Thank you for correcting me. I was working from the Wikipedia article which states that the support can be chosen to be any length-2$\pi$ radiatns interval, but I see now that the typical assumption is that the support is equal to $[-\pi, \pi]$. Commented Dec 27, 2019 at 16:42
• You don't need to include the old text in your answer for completeness: all prior versions of an answer are preserved and can be viewed by clicking on the highlighted "edited ... ago" link under the answer.
– EdM
Commented Dec 27, 2019 at 16:42

Since the Von Mises distribution operates on a random angle we first need to be clear about what we mean when we refer to the "sum" of two Von Mises random variables. The most sensible thing to do is to view the sum as a new angle on the same interval, which means that the problem of different supports vanishes. Now, the Von Mises distribution is "circularly continuous" in the sense that it lets us work over any angular range. For simplicity in dealing with the modular arithmetic, let's consider two Von Mises random angles defined on the support $$X_1, X_2, \in [0, 2 \pi)$$ and define their sum on the same angular support as:

$$S \equiv (X_1+X_2) \text{ mod } 2\pi.$$

If you define the angular "sum" in this way then the sum is on the same support as the initial angles. This is the modular way to interpret the summation of angles, which makes sense in the context of this distribution. Since the sum of the angles is on the same support as the initial angles, it then makes sense to investigate whether it might have the same distribution as the initial angles.

Distribution of the angular sum: Taking $$X_1,X_2 \sim \text{IID VonMises}(\mu, \kappa)$$ (taken over the afforementioned support) we can use the law of total probability to obtain the density of the sum. For all $$0 \leqslant s < 2 \pi$$ we have:

\begin{align} f_S(s) &= \int \limits_0^{2\pi} p(S = s | X_2 = x) \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &= \int \limits_0^{2\pi} p((X_1+X_2) \text{ mod } 2\pi = s | X_2 = x) \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &= \int \limits_0^{2\pi} p((X_1+x) \text{ mod } 2\pi = s) \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &= \int \limits_0^{2\pi} [p(X_1 = s-x) + p(X_1 = 2\pi-x+s)] \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &= \int \limits_0^s \text{VonMises}(s-x | \mu, \kappa) \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &\quad \quad + \int \limits_s^{2 \pi} \text{VonMises}(2\pi-x+s | \mu, \kappa) \text{VonMises}(x | \mu, \kappa) \ dx \\[6pt] &= \frac{1}{4 \pi^2 I_0(\kappa)^2} \int \limits_0^s \exp \Big( \kappa \Big[ \cos(x-\mu) + \cos(s-x-\mu) \Big] \Big) \ dx \\[6pt] &\quad \quad + \frac{1}{4 \pi^2 I_0(\kappa)^2} \int \limits_s^{2 \pi} \exp \Big( \kappa \Big[ \cos(x-\mu) + \cos(x-s+\mu) \Big] \Big) \ dx \\[6pt] &= \frac{1}{4 \pi^2 I_0(\kappa)^2} \int \limits_0^s \exp \Big( 2 \kappa \cos(\tfrac{s}{2}-\mu) \cos(x-\tfrac{s}{2}) \Big) \ dx \\[6pt] &\quad \quad + \frac{1}{4 \pi^2 I_0(\kappa)^2} \int \limits_s^{2 \pi} \exp \Big( 2\kappa \cos(x-\tfrac{s}{2}) \cos(\tfrac{s}{2}-\mu) \Big) \ dx \\[6pt] &= \frac{1}{4 \pi^2 I_0(\kappa)^2} \int \limits_0^{2 \pi} \exp \Big( 2 \kappa \cos(\tfrac{s}{2}-\mu) \cos(x-\tfrac{s}{2}) \Big) \ dx \\[8pt] &= \frac{I_0(2 \kappa \cos(\tfrac{s}{2}-\mu))}{2 \pi I_0(\kappa)^2}. \\[6pt] \end{align}

(The third-last step of this derivation uses the sum-to-product trigonometric formula for cosines.) As you can see, this random variable does not follow a Von Mises distribution. Its distribution is and interesting composite function involving the modified Bessel function. The density of the distribution does not have a closed form, so its properties (moments, etc.) are likely to be quite complicated.

The operations of the two von Mises distributions can be found in these two papers.

[1] I. Marković and I. Petrović, "Bearing-only tracking with a mixture of von Mises distributions," 2012 IEEE/RSJ International Conference on Intelligent Robots and Systems, 2012, pp. 707-712, doi: 10.1109/IROS.2012.6385600.

[2] R. F. Murray and Y. Morgenstern, “Cue Combination on the Circle and the Sphere,” Journal of Vision, vol. 10, no. 11, pp. 1–11, 2010.