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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).

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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.

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