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I am trying to understand how to best describe left or negatively skewed data in terms of central tendency.

I have provided code below that simulates such a distribution, as well as the source code to calculate several different central tendency metrics (i.e. geometric, harmonic means) - note that the code for both means were obtained from here: https://github.com/cran/psych/tree/master/R

What I have been reading is that skewed distributions are best represented in terms of median or geometric mean, but when I plot the latter onto the histogram for the left skewed data, it does not really capture the "average" part of the data (see my code and said plot below).

I understand that part of the answer to this question is application dependent, and for my purposes I am trying to aggregated (or upscale) remote sensing raster data from a fine grid to a relatively coarser grid. In this context, it would appear the median will not work either, because the median does not include the information in the final answer.

So, back to my question, is there a "good" parameter that can be used to summarize negatively skewed data?

"geometric.mean" <- function(x,na.rm=TRUE){ 
if (is.null(nrow(x))) {    
    exp(mean(log(x),na.rm=TRUE))} 
else{
    exp(apply(log(x),2,mean,na.rm=na.rm))} 
}

"harmonic.mean" <- function(x,na.rm=TRUE,zero=TRUE){
 if(!zero) {
    x[x==0] <- NA
 }

if (is.null(nrow(x))) {
    1/mean(1/x,na.rm=na.rm)}
else {
    1/(apply(1/x,2,mean,na.rm=na.rm))} 
}

negskewdata<-rbeta(10000,5,2)
hist(negskewdata)
abline(v=mean(negskewdata),col="blue",lwd=2)
abline(v=median(negskewdata),col="red",lwd=2)
abline(v=geometric.mean(negskewdata),col="green",lwd=2)
abline(v=harmonic.mean(negskewdata),col="black",lwd=2)

legend("topleft",c("Arithmetic","Median","Geom. Mean","Harm. mean"),
       lty=1,col=c("blue","red","green","black"),bty="n")

enter image description here

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    $\begingroup$ It's difficult to see how this question would have any bearing on resampling a grid. Are you sure it's really the question you need to be asking? $\endgroup$
    – whuber
    Commented Feb 13, 2019 at 19:11
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    $\begingroup$ The mode is a serious candidate too stats.stackexchange.com/questions/176112/… (For those who struggle with red and green, the left-to-right sequence is harmonic, geometric, arithmetic, median.) $\endgroup$
    – Nick Cox
    Commented Feb 13, 2019 at 19:32

1 Answer 1

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The answer depends on your problem. There is no one number summery of the distribution that is universally the best.

I'll give you an example from mechanics. Suppose, you're modeling motion of celestial bodies of odd shapes:

enter image description hereenter image description here

You could argue that we really need to know the exact shape in order to model the motion of these objects. However, you're allowed to represent the whole thing with just two numbers. What would those numbers be?

Naturally, you'd suggest a mass and a coordinate of the central tendency. You'll model the motion of points with masses. The question is which point is the best representation? If you know physics, you know the answer already: the center of mass, indeed :)

For non-physicists, the center of a mass is the following point: $$\vec r_m=\int_V\rho(\vec r)dV,$$ where $\rho(.)$ is the mass density of a given point of the body.

You should immediately recognize the similarity to the mean: $$\bar x=\sum_i x_i p(x_i)$$ Yes, it's not a coincidence, this is the reason why PDF is called density. Note that for this problem regardless of the skewness of the density function $\rho(\vec r)$ the best central tendency is an analog of the mean in statistics, not the median or other metrics.

Hence, you need to consider what is exactly to be done with the central tendency in your problem in order to construct the best metric of such. The answer is problem specific, there is no general answer.

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