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For simplicities sake (the actually problem is more complex)...Let say I have a set of n 3d points, whose position move over time. For all pairs, I have calculated the mean and standard deviation of the euclidean distance between them.

I would like an error metric which incorporates the following two properties and I can use to "score" each pair in an attempt to find the "best".

1) Pairs of points which on average over time are "close" to one another are preferred i.e small mean -> low error

2) Pairs of points whose distance between them over time varies little i.e small standard deviation -> low error

And I am not sure of the mathematically correct way of combining these two properties.

Any help much appreciated.

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1 Answer 1

One possibility is "Root Mean Squared Error", or RMSE. Defining $\mu_{ij}$ as the average difference between the means of two points $i$ and $j$ (as in your property 1) and $\sigma_{ij}$ as the standard deviation of the difference between two points (as in your property 2), $\text{RMSE}_{ij} = \sqrt{\mu_{ij}^2+\sigma_{ij}^2}$. It represents something that can be interpreted as the "standard deviation around $0$", rather than the standard deviation around the mean difference between $i$ and $j$. You will find it commonly used in statistics as a performance metric that incorporates both bias (property 1) and variance (property 2).

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Sorry if I am misunderstanding / poor wording of my initial post, but won't that just give me one metric for all the data. What I am looking for is the "error" between each pair, to find the "best" pair, not an overall measure of how good the data is. –  oracle3001 Jun 12 '12 at 20:44
    
You would get an RMSE for each pair of points. $\mu$ should really have been denoted $\mu_{ij}$ for the mean difference between $i$ and $j$, and similarly for $\sigma$. Sorry about that; I'll edit the answer. –  jbowman Jun 12 '12 at 21:50
    
Thanks...and does it make any sense if I want to weight one property more than another and if so how would I do that? Simply $\sqrt{ \alpha \mu^2 + (1-\alpha)\sigma^2}$ –  oracle3001 Jun 13 '12 at 8:13

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