After recently having delved into the world of robust measures (for location, mean being the classical case), I have had difficulty understanding robust measures' core dynamic. Basically, what are you giving away in order to become more resilient to extreme outliers (when estimating means or other moments)?

I am trying to phrase myself in such terms because that is how I understand dynamics like the bias/variance trade-off, of giving away one type of estimation error for another. After reading Wilcox's Introduction to Robust Estimation and Hypothesis Testing, as well as another cross-validated post which uses similar language, I am still confused.

Are there, ideally, good simulation studies of various models with different types of noise put on them, testing the different estimators to see how they perform under noisy conditions? Or a resource that summarizes such findings? The only ones I have seen have seemed limited for me, like the image referenced at the bottom of this medium post

Thanks ahead of time to anyone that replies.


Following BruceET, I'll be clearer on specifically what I am worried about concerning robust measures. Basically, what if A) there are types of contaminations on a normal that won't be picked up by these measures, and I support this by the fact that noise can come from any combination of distributions (of the 100+ that exist), and B), what is empirically meant by saying "they strike a balance between efficiency and robustness", as many resources I have read do, since this sounds a bit hand-wavey?


1 Answer 1


Here is a simple example to start. Suppose I have data on a process that is normally distributed with $\mu = 100$ and $\sigma = 10$ when everything is running properly. But for unknown reasons the process goes briefly 'out of control' about 5% of the time, and when that happens values have $\mu_b = 300$ and $\sigma_b = 20.$ Here is a histogram of such data:

enter image description here

Mean and Median. If I find the descriptive statistics of all the data (in and out of control), then I get:

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  57.71   93.72  100.50  110.01  108.00  366.95 

In particular, the sample mean $\bar X = 110.01$ has been pulled up above the sample median $H = 100.50.$ So, If I want an average of values from the entire process (including the bad values), it seems best to look at $\bar X = 110.01.$ But the median 100.50 is a more robust measure of location, less influenced by the bad values.

Trimmed mean. The 5% trimmed mean is somewhat of a compromise between the mean and the median. Roughly speaking, the lowest 5% and highest 5% of the data are deleted first and then the mean of the remaining 90% is computed. The result is 101.1286, mainly representative of the process when it is in control.

Such a trimmed mean is not inherently biased towards higher or lower values because values have been deleted from both tails of the distribution. If the percentage of high bad values is greater, beyond a 'breakdown point', then trimming 5% from each tail may not be enough. [Loosely speaking, you might say that the median is a 50% trimmed mean, because it is is just the middle value of the sorted data (or average of the middle two).]

mean(x, trim=.05)
[1] 101.1286

Outliers. Here is a boxplot of the data. Notice the large number of 'outliers' towards the right. We could consider removing the far outliers. But they are all in the upper tail, so taking a mean after deleting just these high values might give a biased view of the process when it is in control. Generally speaking, it is not a good idea to remove 'outliers' from a dataset without investigating what might have caused them.

enter image description here

Just how many high outliers could be removed without giving a false impression depends on the data, so this not a good procedure to use in general. In this particular case, it does little damage to remove just the values above 200; the mean of what's left is 99.98. But if I take the average of the values below 125, I get a perhaps sightly deceptive downward-biased value of 99.81--not extremely misleading is this dataset, but perhaps a similar deletion would be more misleading in other datasets.

mean(x[x < 200])
[1] 99.98124
mean(x[x < 125])
[1] 99.8055

Tradeoffs. If you have uncontaminated normal data, then using either the median or the 5% trimmed mean gives a less precise estimate of the population mean than does the sample mean. In the simulation below, I find means a, 5% trimmed means t, and medians h of a million normal samples of size $n = 50$ with $\mu = 100,\, \sigma=10.$

All three are unbiased estimators of the population mean $\mu$. It is well-known that the standard error of the sample mean is $\sigma/\sqrt{n} = 10/\sqrt{50} = 1.414.$ The standard error 1.428 of the trimmed mean is a little larger and the SE 1.749 of the median is even larger.

set.seed(1022); m = 10^6
a = t = h = numeric(m)
for(i in 1:m) {
  x = rnorm(50, 100, 10)
  a[i] = mean(x);  t[i] = mean(x, trim=.05);  h[i] = median(x) }
mean(a); mean(t); mean(h)
[1] 100.0018  # all unbiased: consistent with 100
[1] 100.002
[1] 100.0026
sd(a); sd(t); sd(h)
[1] 1.414471  # aprx 1.414 as shown above
[1] 1.427883  # slightly larger for sample trimmed mean
[1] 1.749267  # largest for sample median

Note: Those are a few of the basic issues. And your links mention a few more. If you would like a discussion of particular additional robust measures of location, please give names and details so someone can move on from this initial answer.

  • $\begingroup$ Thanks for this elaboration. Even though I would have preferred some studies that have exhaustively performed many simulations like this, I appreciate the effort and clarity. $\endgroup$
    – Coolio2654
    Feb 27, 2019 at 1:32
  • 1
    $\begingroup$ Perhaps you can find relevant references by googling 'M-estimator','Robust estimation', 'Huber estimator', 'Trimmed mean', and 'Truncated mean'. Some of the key references pre-date the routine use of extensive simulation to assess the properties and uses if robust estimates in applications. $\endgroup$
    – BruceET
    Feb 27, 2019 at 17:24

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