Can the overall mean be higher than all subset means?

Question

I am investigating averages and have split my set into three sections and have found that the overall (global) mean is higher than the means of each of the three sections. Is this possible?

Answer 1

Not in the simple interpretation.

The global mean should be a weighted average of the section means and so should be in the range of the section means. If $m$ is the relevant mean then: $$m_g=\frac{w_n}{w_n+w_e+w_s}m_n +\frac{w_e}{w_n+w_e+w_s}m_e +\frac{w_s}{w_n+w_e+w_s}m_s \\ \le \max\{m_n,m_e,m_s\}$$

Answer 2

No, this is not possible, if the sections are disjunct - that is, there is no data point that exists in two regions at the same time.

This is quite simple to prove. If $A=A_1\cup A_2\cup\cdots \cup A_n$, and $A_i\cap A_j=\emptyset$ (that is, $A_i$ are pairwise disjunct), and we denote the mean of all values in a given set $X$ with $\mu(X)$, then some relatively basic algebraic manipulation can show you that $\mu(A)$ is a convex combination (also known as a weighted average) of the values $\mu(A_i)$.

In other words, there exists some collection of positive weights $w_1,\dots,w_n$ such that the sum of the weights is $1$ and

$$\mu(A) = w_1\mu(A_1)+\cdots + w_n \mu(A_n)$$

This is enough to see that $$\mu(A)\leq \max_{1\leq i\leq n}(\mu(A_i)).$$