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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

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  • $\begingroup$ As is I don't think this question is answerable. It is not clear what the summations are over. Also you keep saying average of a sum, when I believe you mean average of a sequence of numbers. $\endgroup$ – jlimahaverford Sep 13 '15 at 0:09
  • $\begingroup$ I definitely meant average of sum. But thanks for your input. The summations are over n: the number of elements being summed. n is either in its entirety (i,k) or as a subgroup of (i,k); in this case (i, j) and (j, k) $\endgroup$ – Jona Sep 13 '15 at 0:43
  • $\begingroup$ Jona, I'm not trying to be rude, but terminology and notation are important in math, especially when conveying ideas to others. A sum is one number, therefore discussing the average of a sum does not make sense. Dividing a sum by the number of elements being summed gives you the average of the terms in the sum. When I asked what the sum was over, I meant out of all of these indices, which ones are actually changing in the sum. Is the first sum $x_{1,k} + x_{2,k} + \ dots$? If so, why does I appear in $n_{i,k}$. To me this equation does not carry clear meaning. $\endgroup$ – jlimahaverford Sep 13 '15 at 7:48
  • $\begingroup$ I can see what you're trying to do but haven't you double counted the $j$? Was that intentional? $\endgroup$ – Silverfish Sep 13 '15 at 9:40
  • $\begingroup$ Admittedly, I did write this late at night and I see where I could have been more clear. Apologies for my curtness. Thankfully, R.M was able to wade through my lack of clarity and provide a good answer $\endgroup$ – Jona Sep 14 '15 at 7:40
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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.

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    $\begingroup$ your intro was great, thanks. I have to study your simplification a little more, but thank you for the overall explanation! $\endgroup$ – Jona Sep 13 '15 at 0:45

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