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I know this question has been asked with the case mean=median, but I did not find anything related to mean=mode.

If the mode equals the mean, can I always conclude this is a symmetric distribution? Will I be forced to know also the median for this manner?

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Mean = mode doesn't imply symmetry.

Even if mean = median = mode you still don't necessarily have symmetry.

And in anticipation of the potential followup -- even if mean=median=mode and the third central moment is zero (so moment-skewness is 0), you still don't necessarily have symmetry.

... but there was a followup to that one. NickT asked in comments if having all odd moments zero was enough to require symmetry. The answer to that is also no. [See the discussion at the end.$^\dagger$]

Those various things are all implied by symmetry (assuming the relevant moments are finite) but the implication doesn't go the other way - in spite of many an elementary text clearly saying otherwise about one or more of them.

Counterexamples are pretty trivial to construct.

Consider the following discrete distribution:

  x     -4    0    1    5
P(X=x)  0.2  0.4  0.3  0.1

It has mean, median, mode and third central moment (and hence moment-skewness) all 0 but it is asymmetric.

Asymmetric discrete pmf with mean=median=mode and moments skewness 0

This sort of example can be done with a purely continuous distribution as well. For example, here's a density with the same properties:

Asymmetric density with mean, median and mode 0 and zero moment-skewness

This is a mixture of symmetric triangular densities (each with range 2) with means at -6, -4, -3, -1, 0, 1, 2, 5 and mixture weights 0.08, 0.08, 0.12, 0.08, 0.28, 0.08, 0.08, 0.20 respectively. The fact that I just made this now -- having never seen it before -- suggests how simple these cases are to construct.

[I chose triangular mixture components in order that the mode would be visually unambiguous -- a smoother distribution could have been used.]


Here's an additional discrete example to address Hong Ooi's questions about how far from symmetry these conditions allow you to get. This is by no means a limiting case, it's just illustrating that it's simple to make a less symmetric looking example:

   x    -2    0    1    6
P(X=x) 0.175 0.5  0.32 0.005

plot of the above p.m.f.

The spike at 0 can be made relatively higher or lower without changing the conditions; similarly the point out to the right can be placed further away (with a reduction in probability) without changing the relative heights at 1 and -2 by much (i.e. their relative probability will stay close to the 2:1 ratio as you move the rightmost element about).


More detail on the response to NickT's question

$\dagger$ The all-odd-moments zero case is addressed in a number of questions on site. There's an example here (see the plot) based on the details here (see toward the end of the answer). That is a continuous unimodal asymmetric density with all odd moments 0 and mean=median=mode. The median is 0 by the 50-50 mixture construction, the mode is 0 by inspection -- all members of the family on the real half-line from which the example is constructed have a density that's monotonic decreasing from a finite value at the origin, and the mean is zero because all odd moments are 0.

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    $\begingroup$ I think the moral of the story is: symmetry is a strong property and can't be deduced from a few typical summary values of the distribution. $\endgroup$ – Kodiologist Sep 30 '16 at 1:35
  • $\begingroup$ An interesting question might be how "close" to symmetry can you get with these properties. Looking at your discrete example, it's kind-of sort-of symmetric with a hump in the middle. $\endgroup$ – Hong Ooi Sep 30 '16 at 2:06
  • $\begingroup$ @HongOoi I expect you mean to ask how far you can get rather than how close (since obviously you can make it perfectly symmetric any time you want). You can make it a good deal more asymmetric than my example -- it was just a convenient case. $\endgroup$ – Glen_b Sep 30 '16 at 2:21
  • $\begingroup$ @HongOoi I have added another example. $\endgroup$ – Glen_b Sep 30 '16 at 4:25
  • $\begingroup$ If all the (odd?) moments beyond the variance are 0, would that only happen iff there's a symmetric distribution? $\endgroup$ – Nick T Oct 1 '16 at 17:00
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Try this set of numbers:
\begin{align} X &= \{2,3,5,5,10\} \\[10pt] {\rm mean}(X) &= 5 \\ {\rm median}(X) &= 5 \\ {\rm mode}(X) &= 5 \end{align}

Histogram

I wouldn't call that distribution symmetrical.

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No.

Let $X$ be a discrete random variable with $p(X = -2) = \tfrac{1}{6}$, $p(X = 0) = \tfrac{1}{2}$, and $p(X = 1) = \tfrac{1}{3}$. Obviously, $X$ is not symmetric, but its mean and mode are both 0.

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To repeat an answer I gave elsewhere, but fits here too:

$$\mathbb{P}(X=n) = \left\{ \begin{array}{ll} 0.03 & n=-3 \\ 0.04 & n=-2 \\ 0.25 & n=-1 \\ 0.40 & n=0 \\ 0.15 & n=1 \\ 0.12 & n=2 \\ 0.01 & n=3 \end{array} \right.$$

enter image description here

which not only has mean, median and mode all equal, but also has zero skewness. Many other versions are possible.

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