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Very robust regression?

by skyraider
Tags: regression, robust
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Mar2-07, 09:37 AM
P: 3

I want to model a set of a few dozen points on the x-y plane where y can be anywhere from 0 to 100 and x increases by 1 for each point on the y-axis, ex:

(1, 26)
(2, 84)
(3, 2)
etc. . .

Is it possible to accurately model such a random array of points with an equation? Someone once suggested using an 'interpolating polynomial in the Lagrange form', but that does not appear to work well with such a random array of points.

If it can't be done with a known regression technique, here is my question:

Given the points (1, 26) (2, 84) (3, 2) (4, 100) (5, 50), could a function exist - any function of any category - which will hit each point?

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Mar2-07, 10:34 AM
P: 15,173
Quote Quote by skyraider View Post
Given the points (1, 26) (2, 84) (3, 2) (4, 100) (5, 50), could a function exist - any function of any category - which will hit each point?

This final question is an easy one: The answer is yes. A fourth-order polynomial will hit each point exactly:
[tex]-27x^4 + 323\frac1 3x^3-1335x^2+2204\frac2 3x-1140[/tex]

You generally don't want to do that, however. For example, this particular polynomial rapidly goes negative as x goes below 1 or above 5. In other words, it has very little extrapolative capability. You will quickly start to lose even interpolative capability with the exact-fit polynomial as the number of points increases. You want to develop a fit to a less expressive model.

There is no magic one-form-fits-all method. People can still get advanced degrees in statistics, after all.
Mar2-07, 08:07 PM
P: 1,295
If you tell us what you expect from this "model", we can suggest various methods that are suited to the task.

Mar2-07, 09:29 PM
Sci Advisor
P: 2,751
Very robust regression?

As Crosson says, obviuosly you must be expecting something from this model besides hitting all the points. You already have all the points so you must be expecting something additional, but what is it?

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