Mean of Medians? Median of Means?

Question

This is a question I have wondered about for a long time and have never been able to find a full mathematical explanation behind this.

Suppose there are 100 countries. As an experiment:

• We give Person A the median income of each country
• We give Person B the mean income of each country

Now, suppose the following happens:

• Person A decides to take the mean of all median incomes
• Person B decides to take the median of all mean incomes
• Person C shows up out of nowhere and decides to take the median of all median incomes

My Question: Using mathematics, we can we demonstrate that perhaps some of these calculations are not very "meaningful"? For example, can we somehow show that some of these calculations lack important mathematical properties and are basically arbitrary?

Thanks!

Answer 1

The relationship between these calculations depends on the skewness of the underlying distributions.

If it is right skewed, then the average of the medians will be less than the median of the means, since mean > median.

The opposite is true of left skewed distributions.

The median of medians isn't really special, just another way to express the typical median income over all countries.

Answer 2

Well, from a very basic understanding of statistics, I'd say that the point to central tendency measurements (mean, median), is, find a few representative numbers for the data. The choice between mean and median to stand for a particular data is a matter of whether there are extreme values in the data or not (a basic course on statistics will tell you when mean should be preferred over the median and vice versa). The rationale being the median is less susceptible to extremes than the mean.

That out of the way, in the problem you pose, the means and medians of all the countries is just another data set. You apply the same rules, one being if there are outliers, either calculate the median or ignore them and calculate the mean.

My two cents ...