1

It can be derived from here by the triangle inequality that for unimodal distributions, the distance between the mean and the mode satisfies the bound

$$\left|\overline{X} - \text{mode}(X) \right| \leq (1 + \sqrt{5})\sqrt{\frac{3}{5}}\sigma.$$

Is it possible to improve upon this bound if we also know the skewness of the distribution? What about if we know even higher moments?

Thoth
  • 1,213
  • 9
  • 24
  • 1
    @user603, can you not use the triangle inequality like I did? – Thoth Oct 14 '16 at 12:03
  • [This](http://stats.stackexchange.com/a/3790/603) answer points to $\sqrt{3}$ factor between the lhs and the $\sigma$ (slightly smaller than the bound from the triangle inequality). I seem to recall that the J&R 1951 paper cited in that answer has some results pertaining to skew, might be worth checking. In general though you would expect skewness (in the VanZwet sense) to be unrelated to the relative disposition of the median and mode. – user603 Oct 14 '16 at 12:10

0 Answers0