Counterexamples where Median is outside [Mode-Mean]

Question

This article is above my league but it talks about a topic which I am interested in, the relationship between mean, mode and median. It says :

It is widely believed that the median of a unimodal distribution is "usually" between the mean and the mode. However, this is not always true...

My question: can someone provide examples of continuous unimodal (ideally simple) distributions where the median is outside the [mode, mean] interval? For example a distribution such as mode < mean < median.

=== EDIT =======

There are already good answers by Glen_b and Francis, but I realized that what I am really interested in is an example where mode < mean < median or median < mean < mode (that is both median is outside [mode, mean] AND median is "on the same side" as mean of mode (i.e. both above or below mode)). I can accept the answers here are open a new question or maybe someone can suggest a solution here directly?

Answer 1

Sure, it's not hard to find examples - even continuous unimodal ones - where the median is not between the mean and the mode.

1. Consider $T_1,T_2$ i.i.d from a triangular distribution of the form $f_T(t) = 2(1-t) \mathbb{1}_{0<t<1}$

   Now let $X$ be a 60-40 mixture of $T_1$ and $-4T_2$.

   The density of $X$ looks like this:

   

   The mean is below 0, the mode is at 0, but the median is above 0. A minor modification of this would yield an example where even the density (rather than just the cdf) was continuous, but the relationship between location-measures was the same (edit: see 3. below).

2. Generalizing, let's put a proportion $p$ (with $0<p<1$) of the total probability into the right-hand-side triangle and a proportion $(1-p)$ into the left-hand side triangle (in place of the 0.6 and 0.4 we had before). Further, make the scaling factor on the left half $-\beta$ rather than $-4$ (with $\beta>0$):

   

   Now assuming $p>\frac12$, the median will always be in the interval covered by the right-triangle, so the median will exceed the mode (which will always remain at $0$). In particular, when $p>\frac12$, the median will be at $1-1/\sqrt{2p}$.

   The mean will be at $(p - \beta(1-p))/3$.

   If $\beta>p/(1-p)$ then the mean will be below the mode, and if $\beta< p/(1-p)$ the mean will be above the mode.

   On the other hand, we want $ (p - \beta(1-p))/3 < 1-1/\sqrt{2p}$ to keep the mean below the median.

   Consider $p=0.7$; this puts the median above the mode.

   Then $\beta=2$ would satisfy $\beta<p/(1-p)$ so the mean is above the mode.

   The median is actually at $1-1/\sqrt{1.4}\approx 0.1548$ while the mean is at $\frac{0.7-2(0.3)}{3}\approx 0.0333$. Hence for $p=0.7$ and $\beta=2$, we have mode < mean < median.

   (NB For consistency with my notation, the variable on the x-axis for both plots should be $x$ rather than $t$ but I am not going to go back and fix it.)

3. This is an example where the density itself is continuous. It is based on the approach in 1. and 2. above, but with the "jump" replaced with a steep slope (and then the entire density flipped about 0 because I want an example that looks right-skew).

   

   [Using the "mixture of triangular densities" approach, it can be generated as a mixture of 3 independent scaled variates of the triangular form described in section 1. We now have 15% $T_1$, 60% $-3T_2$ and 25% $5T_3$.]

   As we see in the diagram above, the mean is in the middle, as requested.

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4. Note that m_t_ mentions the Weibull in comments (for which the median is outside the $[\text{mode},\text{mean}]$ interval for a small range of the shape parameter $k$). This is potentially satisfying because it's a well-known unimodal continuous (and smooth) distribution with simple functional form.

   In particular, for small values of the Weibull shape parameter, the distribution is right-skew, and we have the usual situation of median between the mode and the mean, while for large values of the Weibull shape parameter, the distribution is left-skew, and we again have that "median in the middle" situation (but now with the mode at the right rather than the mean). In between those cases is a small region where the median is outside the mean-mode interval, and in the middle of that the mean and mode cross over:

         k                 order
    (0,3.2589)      mode < median < mean
     ≈ 3.2589       mode = median < mean
   (3.2589,3.3125)  median < mode < mean    (1)
     ≈ 3.3215       median < mode = mean
   (3.3215,3.4395)  median < mean < mode    (2)
     ≈ 3.4395       median = mean < mode
     3.4395+        mean < median < mode
     (≈3.60235      moment-skewness = 0)
   

   Choosing convenient values for the shape parameter in the intervals marked (1) and (2) above -- ones where the gaps between the location-statistics are about equal -- we get:

   

   While these fulfill the requirements, unfortunately the three location-parameters are so close together that we can't visually distinguish them (they all fall in the same pixel), which is a little disappointing -- the cases for my earlier examples are much more separated. (Nevertheless it does suggest situations to examine with other distributions, some of which might give outcomes which are more visually distinct.)

Answer 2

The following example is taken from Jordan Stoyanov's Counterexamples in Probability.

Given positive constant $c$ and $\lambda$, consider a random variable $X$ with density $$f(x)=\begin{cases}ce^{-\lambda(x-c)}, & x\in(c, \infty) \\ x, & x\in(0, c] \\ 0, & x\in(-\infty,0].\end{cases}$$ The mean $\mu$, median $m$ and mode $M$ of $X$ can be found as $$\mu=\frac{c^3}{3}+\frac{c^2}{\lambda}+\frac{c}{\lambda^2},\qquad m=1,\qquad M=c.$$ Note $f(x)$ is a density only if $$\frac{c^2}{2}+\frac{c}{\lambda}=1.$$ So if we let $c\to1$ then $\lambda\to2$. As a result, if we choose a $c>1$ that's close to $1$ (say $1.0001$), we can find that $\mu>c$ and $M=c$, so median $m$ does not fall between $\mu$ and $M$.

Answer 3

Take the exponential distribution with rate parameter a and density a exp(-ax) for 0<=x< infinity. The mode is at zero. Of course the mean and the median are greater than 0. The cdf is 1-exp(-ax). So for the median solve for exp(-ax)=0.5 for x. Then -ax=ln(0.5) or x = -ln(0.5)/a. For the mean integrate ax exp(-ax) from 0 to infinity. Take a=1 and we have a median = -ln(0.5)=ln(2) and mean = 1.

So mode < median < mean.