Can a Function Accurately Model a Random Array of Points on the x-y Plane?

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In summary, the conversation discusses the possibility of accurately modeling a set of points on the x-y plane with an equation. The use of an interpolating polynomial in the Lagrange form is suggested but may not work well with a random array of points. The question then shifts to whether any function, regardless of category, could be used to hit each point. The answer is yes, but a fourth-order polynomial may not be the most effective method. Different methods should be considered based on the expectations for the model.
  • #1
skyraider
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Hi,

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?

Thanks.
 
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  • #2
skyraider said:
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?

Thanks.

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.
 
  • #3
If you tell us what you expect from this "model", we can suggest various methods that are suited to the task.
 
  • #4
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?
 

1. What is "very robust regression"?

"Very robust regression" is a statistical method used to model the relationship between variables. It is designed to minimize the impact of outliers and other influential data points on the results of the analysis.

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"Very robust regression" should be used when the data contains outliers or influential data points that may significantly impact the results of the analysis. It is also useful when the underlying assumptions of traditional regression methods are not met.

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The main advantage of using "very robust regression" is that it provides more accurate and reliable results compared to traditional regression methods when the data contains outliers or influential data points. It can also help to identify and remove these influential points from the analysis.

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One limitation of "very robust regression" is that it can be computationally intensive and may require more time and resources to perform compared to traditional regression methods. Additionally, it may not be suitable for all types of data, and the choice of robust estimators and techniques may affect the results of the analysis.

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