# Average percent increase not equal to total percent increase?

I tried searching around for this but it was difficult to boil down the search terms. Plus nothing seemed to be showing up anyway.

What's an easy way to show that the average percentage increase of n numbers will not equal the total percentage increase of the before/after summed n numbers? Yes, one could work out an example but it still doesn't make it apparently clear.

I worked out a small example to illustrate my intuition.

Initial    %       Final
10         1       10.1
12         2.5     12.3
11         2       11.22

Inital Sum  = 33
Final Sum   = 33.62

Average %   = (% Sum)/ # of %'s
= (1+2.5+2)/3
= 1.833

Total %
Increase    = (Final Sum - Inital Sum) / (Initial Sum) * 100
= (33.62 - 33)/33 * 100
= 1.87879


These percentages are close but not the same. My intuition tells me this is correct but I would like to develope a proof to show this is actually true.

Any help/guidance would be greatly appreciated.

-- Edit 1 --

My question contains an incorrect premise. I shouldn't be using the average of the percentages. Rather, I should be using the weighted average.

Using Excel and this example I was able to generate the total percent increase using the all the percentages.

A proof would still be helpful in understanding the problem.

-- Edit 2 --

Given initial values $a_k$ and percentages $p_k$, the weighted average for the percentages would be: $$\frac{\sum_{k=1}^{n} a_k *p_k}{\sum_{k=1}^{n}p_k}$$

Hopefully that's the correct notation. Like I stated in Edit 1, that was pulled from How to calculate weighted averages in Excel.

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What do you exactly mean by the weighted average of the percentages? –  Aryabhata Jan 26 '12 at 2:00

## 1 Answer

Perhaps the easiest way to see what’s going on is to look at an extreme case. Suppose that you start with just two initial values, $10$ and $1000$, and increase one of them by $10$% while leaving the other unchanged. The average of the two percentage increases is $\frac12(10+0)=5$% no matter which of the two numbers you increased. The percentage increase in the total, however, very clearly depends on which of the two was increased: increasing the $10$ by $10$% increases the total by $1$, from $1010$ to $1011$, but increasing the $1000$ by $10$% increases the total by $100$, from $1010$ to $1110$. Clearly the latter is a much larger percentage increase than the former; the actual figures are about $0.099$% in the first case and about $9.9$% in the second.

The point is that when you increase one of the items by a certain percentage, the effect on the total depends not only on the percentage, but also on the size of the item.

Edit: And that’s exactly why your weighted averages work: the individual percentage increases need to be weighted by the sizes of the original items. Let the original amounts be $a_1,\dots,a_n$, and let the percentage increases, expressed as fractions, be $p_1,\dots,p_n$. The amounts after the increase are $a_k(1+p_k)=a_k+a_kp_k$ for $k=1,\dots,n$, so the the total increase is $$a_1p_1+a_2p_2+\cdots+a_np_n\;.$$ Let $A=a_1+a_2+\cdots+a_n$ be the original total; then the fractional increase is

$$\frac{a_1p_1+a_2p_2+\cdots+a_np_n}A=\frac{a_1}Ap_1+\cdots+\frac{a_n}Ap_n\;,$$

which is indeed the weighted mean of the percentages when they are weighted by the relative sizes of the original amounts.

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The expanded total increase you list does not include the original values. Wouldn't there need to be all the original values as well: $... + a_np_n + a_1 + a_2 + ... + a_n$? Also, it looks like you were going to multiply something at the end or the a mistake with notation. –  JSuar Jan 29 '12 at 1:43
@Gumbo: No, I was giving the increase, not the new total. It’s correct as it stands. –  Brian M. Scott Feb 1 '12 at 23:04