# Standard error of circular standard deviation

How to calculate the standard error of circular standard deviation estimated with equation

1. $s = (2(1 − R))^{-\frac{1}{2}}$

or

1. $s_{0} = (−2 \ln{R})^{-\frac{1}{2}}$

where R is the length of the mean resultant vector (from here).

It should first be noted that the problem is not fully defined currently, because one would need to know the (assumed) distribution of the data before getting to a result that we can actually calculate. I will assume you are using the von Mises distribution.

Most of the theory devoted to this problem is in terms of the distribution of $R,$ $\bar{R} = R/n,$ or in the specific case of the von Mises distribution $\kappa$.

From Mardia & Jupp (2000), 5.3.17, we can get the standard error for $\hat\kappa$ following

$$n \text{var}(\hat\kappa) \approx \frac{1}{1 - A(\kappa) - A(\kappa)/\kappa},$$

so the standard error is

$$\text{sd}(\hat\kappa) \approx \frac{1}{\sqrt{n}} \frac{1}{\sqrt{1 - A(\kappa) - A(\kappa)/\kappa}},$$

where $A(\kappa) = \frac{I_1(\kappa)}{I_0(\kappa)},$ and $I_j(\kappa)$ the modified Bessel function of the first kind and order $j$.

From there, note that $A(\hat\kappa) = \bar{R},$ so $s = (2(1 - R))^{-1/2} = g(\hat\kappa) = (2(1 - nA(\hat\kappa)))^{-1/2}.$

You can see that I use a function $g(\hat\kappa)$ to show clearly that $s$ can be calculated directly from $\hat\kappa.$

We can then use the delta method to get

$\text{var}(s) \approx \left(\frac{d g(\hat\kappa)}{d\kappa}\right)^2 \text{var}(\hat\kappa) = (-2n A'(\kappa))^2 \text{var}(\hat\kappa)$

and finally the approximation to the standard error

$\text{sd}(s) \approx (-2n A'(\kappa)) \text{sd}(\hat\kappa) = (-2n A'(\kappa)) \frac{1}{\sqrt{n}} \frac{1}{\sqrt{1 - A(\kappa) - A(\kappa)/\kappa}}$

Whether this approximation is any good remains to be seen, but at least it's a start! A problem is that estimates of $\hat\kappa$ are usually biased, which will likely make this approximation a bit worse.