Is the average of a sum equal to the average of the sum of the averages of parts of the original sum? Can anyone explain to me whether this is true or not?
$$\frac{\sum x_i,_k}{n_i,_k} = \frac{\frac{\sum x_i,_j}{n_i,_j} + \frac{\sum x_j,_k}{n_j,_k}}{2}$$
Logically, it makes sense that they should equal, but when I plug in numbers I get differing results. Am I thinking about this completely wrong?
(btw not necessarily 2 in the denominator, could split up the original sum even more)
Thanks
 A: No, in general the average of sub-averages is not the average of the original set of numbers. For example, take the simple set if numbers 3, 98, 100. One way you can split things up is by having (3), (98,100), which means your sub averages are 3, 99 which when averaged gives you 51, not even close to the true average of 67. The bias is even more heavily demonstrated by more asymetric splits in other series, for example (1), (998, 1000, 1005, 1010, 1025). 
The fact that they're not identical can be shown by "simplifying" the sums. 
$$\frac{\sum_{m=i}^k x_m}{n_i,_k} = \frac{x_i}{n_i,_k} + \frac{x_{i+1}}{n_i,_k} ... + \frac{x_k}{n_i,_k}$$
$$\frac{\frac{\sum_{m=i}^{j-1} x_m}{n_i,_j} + \frac{\sum_{m=j}^k x_m}{n_j,_k}}{2} = \frac{x_i}{2n_i,_j} + \frac{x_{i+1}}{2n_i,_j} ... + \frac{x_{j-1}}{2n_i,_j} + \frac{x_{j}}{2n_j,_k} ... + \frac{x_k}{2n_j,_k} $$
Note that each of the $x_m$s only occur one in each sum, so for arbitrary $x_m$s, the two will only be equal if $2n_i,_j = 2n_j,_k = n_i,_k$. That is, if the two groups each have half the original set of terms. The problems arise when you have an unequal number of terms in each split.
The more general solution is to do a weighted average of the sub-averages, weighting by the relative fraction of the number of subterms. That is:
$$\frac{\sum_{m=i}^k x_m}{n_i,_k} = \frac{n_i,_j}{n_i,_k}\left(\frac{\sum_{m=i}^{j-1} x_m}{n_i,_j}\right) + \frac{n_j,_k}{n_i,_k}\left(\frac{\sum_{m=j}^k x_m}{n_j,_k}\right)$$
If you "simplify" out this sum by terms, you should see that the two are equal, regardless of the values of the $x_m$s.
