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I have a set of points that lie on a circle, and I know their angles. Is there a way that I can test whether they are closer to another point x on the circle than expected by chance (assuming a uniform distribution on the circle)?

Can I calculate mean arc length between xand the data, and then calculate the same for each of many simulated data sets generated by the uniform distribution?

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  • $\begingroup$ Yes, there are specific tests for this. Which one you choose depends on how $x$ was determined. Is it (a) a location established a priori by a theory or (b) an estimate based on data independent of this set of points or even (c) an estimate based on data that are not independent of these points? $\endgroup$
    – whuber
    Jul 5 '14 at 20:13
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    $\begingroup$ Is this a homework question? If so, you'd be better off posting the actual problem, especially since it may be a trick question along the same lines as Bertrand's Paradox. Actually, you should first take a look at this post and then proceed with the post-editing... $\endgroup$
    – Steve S
    Jul 6 '14 at 1:52
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The question as posed sounds like it's setting up a hypothesis test.

Distance in the case of circular data would (presumably) simply be angular distance which is the same as your suggested arc length when measured in radians on a unit circle. That is, for example based on the absolute difference in angle (which I think is your suggestion), or maybe something based on the squared difference in angle or some similar measure.

The null is that the distribution of such distances should be that for a uniform distribution of points, while the alternative is that the distances to $x$ will be smaller (since you said you're checking for clustering around $x$). If you wanted to check for being further from $x$ you'd look for the alternative to be larger, not smaller.

Without loss of generality, you might as well just take $x$ at the centre of your uniform*; it won't change the null distribution.

So you'd be trying to work out the distribution of the mean absolute angle for a uniform distribution (or perhaps the mean of the squared angle for a uniform distribution).

And yes, you can certainly do that by simulation, but I suspect the algebra isn't very hard at all. (Edit: actually, I just figured it out now for your suggested statistic. It's pretty straightforward - it will be amenable to somewhat mechanical direct solution, but it's already one which is quite well-known. Practically speaking, in anything but quite small samples you'd probably just use a normal approximation unless your significance level was very low.)

[Actually, this would be a case on which one could use the trick in Fisher's method to base a statistic off the sum of the log-absolute-angle-distance. That has the (modest) advantage of having a standard table as the distribution of the test statistic.]

* if angles are on $(-\pi,\pi]$, you could take $x=0$, while if on $[0,2\pi)$, you could take $x=\pi$.


If you measure the absolute angle from $x$, you get a uniform on $[0,\pi]$ (measured in radians). Any other angle-origin is easy to convert to this. The average of the absolute angles will therefore be a convolution of $n$ uniforms. For small $n$ this is easy to write explicitly (e.g. for $n=2$ the average is triangular, for $n=3$ it's a smooth-looking hill-shape composed of three quadratic segments). As $n$ increases it rapidly approaches normality except in the extreme tails.

You can base a test off that quite easily; the alternative (growing away) would correspond to larger-than-average absolute angle. Growing toward would be smaller-than-average absolute angle.

If you're happy using simulation, I see no problem with that, but if $n$ is bigger than about 15 or so you can probably manage quite well with the normal approximation.

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  • $\begingroup$ My question is not from a book or homework. I'm trying to test whether tree roots grow away from their nearest tree that they compete with. That competitor is the xpoint I mentioned. Does it make sense to do what I proposed? I'm in a bit of a time crunch, so I'm willing to sacrifice analytic elegance and use a simulation. I'm new to stats/probability so it would take me a while to figure out what the null distribution would look like. Alternatively, do you have some keywords that could lead me to the tests I might make use of? I don't even know where to begin. $\endgroup$
    – Biomath
    Jul 6 '14 at 18:56
  • $\begingroup$ Yes, using average absolute angle as a test statistic makes sense (I thought that was clear in my answer before; perhaps I should have been more specific). I don't think it would be the most powerful test possible under your assumptions, but I really don't think that matters; it will have power to detect your alternative. I'll make some additions to my answer. $\endgroup$
    – Glen_b
    Jul 6 '14 at 21:41
  • $\begingroup$ Edits complete. I hope that helps. $\endgroup$
    – Glen_b
    Jul 6 '14 at 23:13
  • $\begingroup$ A distribution on a circle has no mean. The distribution on an interval in $\mathbb{R}^1$ has, of course. So the mapping between both is statistically invalid. See my answer, s.v.p. $\endgroup$ Jul 7 '14 at 0:33
  • $\begingroup$ @Horst a circle doesn't inherently have an origin, which is why in general we can't really talk meaningfully about means for circular data. But as soon as it has a meaningful origin/center, one where we can talk about distances around the circle from that position such that this carries meaning, then mean distances from that origin do make sense. In this case, we do have a meaningful origin. Which is in turn why we can get a statistic that has the right significance level and reasonable power properties in spite of your insistence on its invalidity. $\endgroup$
    – Glen_b
    Jul 7 '14 at 2:39
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If your data scatter only in a small region of the circle, you can take the tools of euclidean statistics. You would then use that a manifold --here the circle $S_1$-- is locally euclidean. This is your and Glen_b'a approach.

However, you exclude this case in your model: Your null hypothesis says that the data scatter around the whole circle. So you have to use directional statistics. The mathematical reason why you can not use euclidean statistics by mapping each point on the circle to it's distance from a particular $x\in S_1$, let's call it north pole, is that there is no homeomorphism $S_k \rightarrow \mathbb{R}^k$. That's why a distribution on a circle has no mean in the classical euclidean sense: What's the mean of 270° and 90°? Is it the south pole or the north pole?

Fortunately, there are already tests for your hypothesis. See e.g. Ajne, "A Simple Test for Uniformity of a Circular Distribution", Biometrika, 1968, 55 (2), p. 343.

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    $\begingroup$ D'Agostino and Stevens' Goodness-of-Fit-Techniques mentions four statistics for uniformity on the circle (sec 8.10.3.2, p350): $U^2$, $V$, Watson's (1976) statistic, and Ajne's $\endgroup$
    – Glen_b
    Jul 7 '14 at 3:10

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