I'm aware, that negative skew means long left tail and the mean < median < mode. Also aware, that negative skew is where most values are plotted on the right side of the graph.

But the following statement confuses me the most.

It is characterized by many small gains and a few extreme losses

Shouldn't it be many large positive gains (Most values are plotted on the right of the graph) and few small losses (Long left tail)

enter image description here

  • $\begingroup$ What exactly do you mean by the "right side of the graph," given you have named three distinct locations that can serve as the middle to split right from left? $\endgroup$
    – whuber
    Feb 12, 2022 at 21:15
  • $\begingroup$ Added a graph, indicating that most of the values occurs on the right of the distribution. $\endgroup$
    – nsivakr
    Feb 12, 2022 at 21:19
  • $\begingroup$ It's still completely unclear what you mean by "right" and "most." For instance, when we define (as is usual) the "right" to mean the "upper half," then by definition there are equal amounts on both sides. $\endgroup$
    – whuber
    Feb 12, 2022 at 21:25

1 Answer 1


I agree with whuber that your language is a bit unclear in the problem. However, I'm familiar with what you're talking about and I think I can answer your question.

The reason you're confused (I believe), along with potential readers, is because your plot doesn't have enough information. Let me replicate your plot with added info to make clear what "many small gains and few big losses means".

Obviously, please excuse the poor drawing, but take a look: enter image description here

The red line is the break-even point (not a gain, nor a loss). What you see is that the significant majority of points/returns are just to the right of break-even (small gains). You can see that there is basically no tail here. However, if you look to the left of the 0, you see a smaller number of points but it's a heavy tail - they are spread out much wider. So, there are fewer of them (fewer returns in the negative), but they are (on average) larger in magnitude.

I hope this explains the statement, with respect to a distribution of returns that looks like this.

  • $\begingroup$ Thanks @Vladimir. Now I understood very well. $\endgroup$
    – nsivakr
    Feb 12, 2022 at 21:48

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