# Question about the the median and the mode in skewed distributions

In intro statistics textbooks, the mode is typically described as least susceptible to skewness, followed by the median, which is in turn followed by the mean. The difference between the median and the mean is pretty straightforward to me, but I am a bit unclear about the mode. It seems that in discrete distributions, it is possible for the median and the mode to be the same.

For example, if I have a dataset as shown below:

#R code
median(rep(1:8, c(rep(1,3), rep(2, 2), 7, 1, 1)))
 1 2 3 4 4 5 5 6 6 6 6 6 6 6 7 8

hist(rep(1:8, c(rep(1,3), rep(2, 2), 7, 1, 1)),
breaks=seq(0.5, 8, length=8), freq=FALSE, main="", xlab="values") Here, the mean and the median are

mean(rep(1:8, c(rep(1,3), rep(2, 2), 7, 1, 1))); median(rep(1:8, c(rep(1,3), rep(2, 2), 7, 1, 1)))
 5.0625
 6


The mode is 6. So in this case, the median and the mode are identical. Can someone please elaborate on this? Thanks.

• If you have a discrete distribution the mode is essentially they single value that has the highest probability of happening. The median on the other hand is the value that you are just as like to be above or below. Dec 28, 2014 at 1:27

The values of the mode and median are equal in that sample ... but are they equally sensitive?

That depends on which end of the data values you play with.

If you take the smallest 8 values (1 2 3 4 4 5 5 and one of the 6's) and move them arbitrarily far down, the median can be moved as low as we like, but the mode doesn't change one whit.

On the other hand if you take the largest 6 values (8, 7 and four of the 6's) and move them up, the mode can be shifted up arbitrarily far, but the median doesn't change.

The statement can't be true in full generality, as your example data set shows. With a little care we can invent situations where the mode for discrete random variables is more sensitve or less sensitive that the median in both directions.

A simple example of a binomial distribution is pertinent. Here the code (from Stata's Mata) should be fairly transparent:

. mata : (0..5)', binomialp(5, (0..5), 0.2)'
1        2
+-------------------+
1 |       0   .32768  |
2 |       1    .4096  |
3 |       2    .2048  |
4 |       3    .0512  |
5 |       4    .0064  |
6 |       5   .00032  |
+-------------------+


Probability of success here is 0.2 and the probability of 0, 1, 2, 3, 4, 5 successes in 5 trials is tabulated. The mean is naturally 0.2 $$\times$$ 5 = 1. Note that the mode is also 1 and the median is also 1 (cumulative probability 0.32768 for values $$\le$$ 0, 0.72864 for values $$\le$$ 1).

It is often said that in a symmetric unimodal distribution mean, median and mode coincide. (Less careful authors omit "unimodal" and forget camel or other symmetrical shapes with two or an even number of humps.) But here's the nub: The converse isn't true: mean, median and mode being equal does not imply symmetry.

Psychologically as well as logically the example is cogent, as no-one statistically minded would (I hope) dismiss the binomial as a pathological or contrived example.