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How can we calculate the standard error sigma and the chance of a sample being within N sigma (CDF) for values bound to [0, pi] (absolute angular errors)?

You can perhaps think of a blindfolded and/or drunken archer shooting at a target on the wall of a circular room. Good archers will get a tight distribution around the target, but that distribution will be not be strictly Gaussian when the values approach the limits [-pi, pi] creating a wrapped Gaussian distribution.

Yet a subset of that problem (with additional practical applications) is when is also not possible to measure the direction of the error, and therefore the space is additionally limited to [0, pi] (the absolute angular error) In this case, the regular average value is not the expected value of the distribution, which should be still, at least close to 0

Thanks already for any help

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    $\begingroup$ en.wikipedia.org/wiki/Directional_statistics $\endgroup$ – Jarle Tufto Oct 23 '20 at 12:12
  • $\begingroup$ What exactly do you mean with 'calculate the statistics'? Are you given data that was produced by a specific but known / unknown distribution or are you given the distribution or are you looking for a distribution that you can put on the interval [0,2pi]? $\endgroup$ – Fabian Werner Oct 23 '20 at 12:14
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    $\begingroup$ If this is about you asking 'what kind of distribution could I use on that interval to model the drunken archer' then what comes to my mind and 'kind of' looks like the normal (when choosing the correct parameters) is the Beta distribution: en.wikipedia.org/wiki/Beta_distribution $\endgroup$ – Fabian Werner Oct 23 '20 at 12:16
  • $\begingroup$ @FabianWerner It was a general question, but I do have data on the second subset I mentioned [0, pi] (The arc-cosine of cosine distances between n-dimensional vectors). In that case, it doesn't fit the Wrapped normal distribution cited by JarleTutto but thanks anyway! $\endgroup$ – Henrique Mendonça Oct 23 '20 at 12:29
  • $\begingroup$ @JarleTufto thanks for the link. That's pretty much what I was looking for, but calculating the pdf or the cumulative probability for a given angle doesn't seem trivial. Perhaps the wrapped Cauchy distribution gives a simpler solution... $\endgroup$ – Henrique Mendonça Oct 23 '20 at 13:14

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