# Can an average of an overall set be equal to the average of a subset?

I read in a publication that the average salary for lawyers in America is $$\bar x=\163,595$$. Of these salaries, the average for men is $$\bar x_m=\183,687$$, and for women, it is $$\bar x_w=\163,595$$. I'm thinking, how is it possible for the average of women's salary be equal the average of the entire set?

Note that $$\bar x_w = \bar x < \bar x_m$$, and $$\bar x_w,\, \bar x,\, \bar x_m > 0$$.

Can we prove that this is possible/impossible? Can we find 1 simple example where $$\bar x_w = \bar x$$ (given $$\bar x_w,\, \bar x,\, \bar x_m > 0$$, and $$\bar x_m > \bar x$$) is possible?

• No it's not possible unless there are $0$ men. The overall average will be a positive weighted sum of the two gender averages ( weighted by the portion of population of lawyers in each gender) which will be properly between the two subset averages. That is (overall average) = (men ave.)(men % of pop.) + (woman ave.)(woman % of pop.). I'm assuming these are actual means, not sample means or some other sort of estimate. – Ned May 25 '20 at 22:56
• @twosigma This would not work if the remaining elements have an average that is higher. – Gilead May 25 '20 at 22:56
• @ThomasWinckelman: the subsets would need to be mutually exclusive. In your example, the first subset is (1,3,5) with a mean of 3, but the second subset is (1,5) also with a mean of 3. In my example, the second (higher) subset needs to be strictly greater than the mean of the entire set, which it's not in this case. – Gilead May 25 '20 at 23:02
• @ThomasWinckelman The men and women subsets are disjoint and exhaustive (i.e. a partition) so that sort of example doesn't apply here. – Ned May 25 '20 at 23:03
• I was able to find those numbers at datausa.io/profile/soc/… but with the women's average at 132,637. This seems to be from some governmental source but I can't figure out which. I suspect your source comes from the same ultimate source as this one but with an error. – Michael Lugo May 26 '20 at 15:23

Let $$m$$ be the number of men, $$w$$ the number of women and $$a_m$$, $$a_w$$ respective averages. Then, the overall average is
$$a = \frac{ma_m + wa_w}{m+w}.$$ If $$a = a_w$$, from the above we get $$ma_m + wa_w = (m+w)a_w$$, i.e. $$m(a_m-a_w) = 0$$. Thus, either $$m = 0$$ or $$a_m = a_w$$.
Moreover, if $$a_m \geq a_w$$, then
$$a_ w = \frac{ma_w + wa_w}{m + w} \leq \frac{ma_m + wa_w}{m + w} \leq \frac{ma_m + wa_m}{m + w} = a_m.$$