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


*

*$s = (2(1 − R))^{-\frac{1}{2}}$


or


*$s_{0} = (−2 \ln{R})^{-\frac{1}{2}}$


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