# Right-skewed distribution with mean equals to mode?

Is it possible to have a right-skewed distribution with mean equal to mode? If so, could you give me some example?

• Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way. – whuber Apr 15 at 20:27
• @whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that. – beta1_equals_beta2 Apr 15 at 21:25

Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $$\times$$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.

The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)

: (0::10), binomialp(10, (0::10), 0.1)
1             2
+-----------------------------+
1 |            0   .3486784401  |
2 |            1    .387420489  |
3 |            2   .1937102445  |
4 |            3    .057395628  |
5 |            4    .011160261  |
6 |            5   .0014880348  |
7 |            6    .000137781  |
8 |            7   8.74800e-06  |
9 |            8   3.64500e-07  |
10 |            9   9.00000e-09  |
11 |           10   1.00000e-10  |
+-----------------------------+


Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.

• Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned. – Don Tawanpitak Apr 15 at 19:00
• By the way, do you know some other continuous distribution with finite support that can exhibit the same property? – Don Tawanpitak Apr 15 at 19:04
• @DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $\alpha = 3.3125, \beta = 1$ where $\alpha$ is the shape and $\beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$). – COOLSerdash Apr 15 at 19:12

If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function

• $$P(X=0) = 0.36$$
• $$P(X=1) = 0.40$$
• $$P(X=2) = 0.13$$
• $$P(X=3) = 0.10$$
• $$P(X=4) = 0.01$$

is right (i.e. positively) skewed and has both a mean and a mode of 1.

• Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that. – Thomas Cleberg Apr 15 at 18:47
• No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property. – beta1_equals_beta2 Apr 15 at 18:53