Average of abs value vs RMS

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TecHNooB

Diamond Member
Sep 10, 2005
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So I've always wondered about the obsession with RMS values outside the realm of power. I was told by a coworker that the RMS converges statistically and the average of the absolute values does not. Why is this true?

Clarification on absolute value: for a data sequence, the abs value relative to the mean of the data zero.
 
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schenley101

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Aug 10, 2009
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I'm an EE so I don't really know what people outside of electricity use RMS for, but the reason that you don't use the mean of the abs values is because it is not "statistically robust" ie it is "greatly" affected by outliers. I don't know anything about its convergence and what not.
 

Mark R

Diamond Member
Oct 9, 1999
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One of the problems with the mean of absolute values, however, is the difficulty in algebraic analysis. RMS (or at least squares the mean of the squares) is continuous and smooth, which means that it can be manipulated (e.g. differentiated) straightforwardly.

The absolute value operator is not smooth, in particular, the derivative is undefined at zero. This causes significant problems with any algebra that depends on calculus - any equations you derive algebraically may not be valid if the input data cross zero).

This problem of algebraic analysis is one of the reasons why simple algebraic solutions to least-squares fitting problems exist (e.g. linear regression) - but simple solutions are not available to least-sum-of-errors fitting.
 

CycloWizard

Lifer
Sep 10, 2001
12,348
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It's been a while since I thought about this, but I think the answer is related to convexity. The problem of optimizing parameters based on RMS is a quadratic problem, which has lots of favorable properties. The absolute value of the error, on the other hand, is not quadratic and not even strictly convex (though I believe it is still convex).
 

CP5670

Diamond Member
Jun 24, 2004
5,527
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Least-squares problems in general have nice mathematical properties and you can often find exact formulas for the solution. L2 minimization amounts to solving a linear equation, while L1 (absolute value) minimization involves linear programming.

The absolute value of the error, on the other hand, is not quadratic and not even strictly convex (though I believe it is still convex).

They're both convex (which means they both have basically good properties), but the strict convexity does give L2 problems a unique solution while L1 problems generally lack that.
 
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