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How median Absolute deviation is immune to outliers?

R2 or coefficient of determination is affected by outliers. And how this MAD is going to remove that constraint?

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    $\begingroup$ The breakdown point of the MAD estimator is much larger than the breakdown point of say the MSE. The breakdown point of an estimator is explained in any decent statistics book dealing with robustness properties. $\endgroup$
    – mlofton
    Mar 7, 2019 at 2:32
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    $\begingroup$ "Immune" is the wrong word. The right description is "less sensitive". Think of it this way. If you have the lowest value in a data set become much lower it doesn't change the median. It is a little more complicated when you take medians of functions of the data. $\endgroup$ Apr 6, 2019 at 16:03

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MAD has nothing to do with $R^2$ or the coefficient of determination. It doesn't remove any constraints (whatever that means) from those measures. In fact, it isn't even comparable to those measures as it is an attempt to answer a completely different question. Rather, you could compare MAD to other measures of spread, the best known of which is the standard deviation.

The median absolute deviation is very resistant to outliers. We can show this by example. Let's start with a reasonable set of values, with the following R code (anything after a # is a comment)

set.seed(1234)

x < rnorm(100)   #Standard normal, n = 100
mad(x)  #1.08
sd(x)   #0.85

x2 <- c(x,20)
mad(x2) #1.11
sd(x2)  #2.16

x3 <- c(x,rnorm(10, 20, 2)) #add 20 outliers
mad(x3) #1.11
sd(x3)  #5.75

Intuitively, the reason is that the median is resistant to outliers, so the median absolute deviation will be resistant to outliers in the absolute deviation. In fact, you could make just under half the data outliers and MAD would not change much.

Someone else can probably prove this mathematically.

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Please remember that R^2 and MAD are measures for two completelly different things. R^2 measures correlation between two variables, while MAD deals with the dispersion of a single variable.

The reason why MAD is not affected by outliers is that it is calculated with a median instead of a mean like MSE (compare mean and median in a sample like 1,2,3,4,5,6,100000)

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