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I’m having an argument with another numerically literate person. We both saw a news site state that under no circumstances should you compare the median of one set with the mean of another.

It’s important to note what we mean by compare: no one is talking about constructing an hypothesis and test statistic to compare with a distribution. This is just a sort of layman, back-of-the-envelope rest of two distributions measured in dollars. One has a long trail sprinkled with upper outliers, the other does not and is presumably bell shaped. Also, I think we both agree that there are better ways to compare the two; the question is really about whether the news reporter should be chided for doing this mean-to-median comparison.

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  • $\begingroup$ you can compare median and mean when you have a perfect normal distribution $\endgroup$ – flobrr May 8 '19 at 15:50
  • $\begingroup$ Thank you @flobrr, but if you have two perfect normal distributions, the mean and median (and mode, for that matter) are all equivalent, so comparing any of them is equivalent to comparing the means. Note in the question that I’m specifically saying that one distribution is more non-normal than the others, what with the long, outlier-containing tail. $\endgroup$ – Sciolism Apparently May 8 '19 at 15:56
  • $\begingroup$ If you have the mean of the bell shaped distribution this should be similar to it's median. Hence you could argue that you are comparing medians of the two distributions. In the case you have the mean of the long tailed distribution then you can assume that the median of the bell shped one is similar to it's mean. Then you would be comparing their means. $\endgroup$ – Manuel May 8 '19 at 15:57
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    $\begingroup$ For the distribution that is bell-shaped, its mean and median should be pretty similar anyway. If it was perfectly bell-shaped, they would be the same. But there are certainly instances where it wouldn't make much sense to compare (in terms of descriptive statistics not hypothesis testing) the mean of one dataset with the median of another, so it's probably not best practice... $\endgroup$ – Izy May 8 '19 at 15:58
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    $\begingroup$ Let's not confuse distribution parameters with their estimators, @Manuel. Although the mean and median qua parameters of a Normal distribution will be equal, as estimators of the center they have different properties and will rarely agree. This isn't just a quibble, either, because the median resists the effects of outliers--which almost inevitably crop up in real dataset--and so the direct comparison of the mean of one dataset to the median of another is suspect in any event. $\endgroup$ – whuber May 8 '19 at 16:04
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"Under no circumstances" is too strong.

If (!) a distribution $F$ is symmetric, then the mean and the median coincide. This is the case, e.g., for the normal distribution.

Thus, you could compare a symmetric $F$'s mean to some other distribution $G$'s median (since you would in fact be comparing $F$'s median to $G$'s). Or you could compare $F$'s median to $G$'s mean (same thing).

Note that only one of the two distributions needs to be symmetric. And of course, symmetry is sufficient but not necessary for the mean and the median coincide; there are asymmetric distributions with equal means and medians.

However, I find it rather hard to imagine a situation where we would want to compare the central tendencies of a symmetric and an asymmetric distribution. In general, it's better to follow the advice not to compare apples and oranges (or means and medians).

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    $\begingroup$ I would like to draw your attention to a comment I posted to the question. It cautions against this confounding of parameters with estimators. $\endgroup$ – whuber May 8 '19 at 16:05
  • $\begingroup$ @whuber: as always, your comment is spot on. Do you want to expand it into an answer? $\endgroup$ – Stephan Kolassa May 8 '19 at 17:43

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