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Consider a joint angle $j$, such that $-180^{\circ} \lt j \lt 180^{\circ}$, with the true value unknown.

Given two estimates of $j$, I wish to combine them using a normalised weighted mean, such that the pair of weights sum to one.

The angle space wraps around ${\pm}180^{\circ}$, so the weightings are applied in a way that considers the smaller angular distance between the angle estimates. For example, if one estimate is $170^{\circ}$ and the other is $-160^{\circ}$, then, given an equal weighting ($0.5$ each), the weighted mean would be $-175^{\circ}$ and not $5^{\circ}$.

Based on this scenario, are the following statements correct?

1. The result of the weighted mean cannot be worse than *both*
   of the estimates.

2. The result of the weighted mean can only be better than the
   estimates if the true value lies between the estimates.
   (i.e. between the smaller angular region separating the estimates)

Now, if this was expanded to a set of joint angles, $J = \{j_1, \, j_2, \, \dots, \, j_n\}$, with two sets of estimates (i.e. two estimates per joint, as before) and still only two weightings (i.e. the elements within a set share the same weighting), then can the overall mean error of the weighted joint angle estimates be better than the mean errors of both sets? Can they be worse than that of both sets?

Can you please explain your answer with a simple anecdotal example.

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1 Answer 1

I think it might be smoother (and easier to get right) to create a vector sum of the various directions, with your weights providing the length of each of the vectors to add. Then look at the direction of the sum vector.

This ought to satisfy your axioms nicely. As a bonus, the ratio between the sum of the weights and the length of the sum vector will give you a reasonably well-behaved measure of "confidence", or how well your estimates agree.

For example, if you have two estimates of equal weight, your proposal (as far as I understand it) would map $(89^\circ,-89^\circ)$ to $0^\circ$, but would map $(91^\circ,-91^\circ)$ to $\pm180^\circ$, with no indication that your estimates are near a discontinuity and therefore probably shouldn't be trusted very much. On the other hand, the vector approach will give the same mean directions but additionally tell you that there is very little confidence in that direction.

(It is also not clear to me how your proposed averaging would deal with estimates that are all over the map, such as $0^\circ$, $120^\circ$, $-120^\circ$).

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