Is it ALWAYS the case that, for a unimodal probability distribution, the median is between the mode and mean?
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1$\begingroup$ "unimodal" can be an ambiguous term... en.wikipedia.org/wiki/… Anyway, this seems the thing: stat.purdue.edu/~dasgupta/publications/tr92-40.pdf $\endgroup$– leonbloySep 2, 2011 at 14:23
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$\begingroup$ First thing I think of is: what happens with log-normal distributions? $\endgroup$– Michael HardySep 2, 2011 at 18:36
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1 Answer
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The answer is "No," as the article linked to by leonbloy indicates. Here's an example of the kind of situation in which you would get mean < mode < median. The black rectangle below contains the mean, the blue rectangle the mode, and the red rectangle the median.
The black rectangle is $5 \times 1/5$, the blue $1/4 \times 4$, and the red $1 \times 3$. The median is in the red rectangle because the areas of the three rectangles are 1, 1, and 3. The blue rectangle clearly contains the mode. Placing the origin at the lower left corner of the blue rectangle, we see that the mean is at $(-2.5)(1) + (1/8)(1) + (3/4)(3) = -1/8$, and so the black rectangle contains the mean.
Of course, this can be scaled to produce an actual probability density function or smoothed to obtain a continuous pdf without changing mean < mode < median.
Added: One of our own users, Henry, has written a detailed article on the relationship between the mean, median, mode, and standard deviation in a unimodal distribution. It's definitely worth a look.
answered Sep 2, 2011 at 15:56
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1$\begingroup$ @Henry: I particularly like the graphs of the regions in which the various possibilities can occur. $\endgroup$ Sep 2, 2011 at 18:58
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$\begingroup$ This answer is super-cool, and I have enthusiastically up-voted and accepted it. Thank you! Last semester I taught an Intro Statistics course to high school students, and the textbook clearly implied that the median was always between the mean and the mode. This struck me as something worthy of proof or refutation, but I had forgotten about it until just now. In fairness to the textbook, I believe that they intended this remark to apply only to normal distributions. Should I submit a separate question about this? $\endgroup$ Sep 3, 2011 at 7:58
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$\begingroup$ @Mike: You're welcome. I'm glad the answer was helpful. As far as your question about normal distributions, they're all symmetric, and so for any normal distribution you have mean = median = mode. $\endgroup$ Sep 3, 2011 at 23:33
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1$\begingroup$ The link to the article of @Henry in the above answer is now broken. However, for anyone who is interested, I found what I think is a copy of the article here $\endgroup$ Jul 31, 2013 at 4:08
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$\begingroup$ @ColinTBowers: Thanks. I'll update the link to Henry's article in my answer. $\endgroup$ Jul 31, 2013 at 20:51