# 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?

## 1 Answer

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 Dec 24 '19 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 $[\pi,\pi]$. – sammosummo Dec 24 '19 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 Dec 24 '19 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 $[-\pi, \pi]$. – psboonstra Dec 27 '19 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 Dec 27 '19 at 16:42