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Question: Is there a way to represent the total percent change of multiple numbers (make it into one number that represents change) that isn't just the percent change between the first and last number of the set?

So I know percent change is new-old/old * 100. But that is just for 2 numbers. Is there a kind of percent change formula that represents the average change of a set of data. I can't seem to find something solid after some research. Someone recommended moving average to a similar question but I'm not sure that is the best way.

Possible Solution: I've tried the literal average of percent changes but there is a reason I don't think that fairly represents the improvement of the set of data. For example if you have the numbers 10, 7, 2, 5, 4 the percent changes between these numbers are-> -30, -71, 150, -20. The average of these 4 changes is 7.25. 3 out of the 4 changes are decreases and only 1 is an increase but the average makes it seem like there was a average increase just because the jump from 2 to 5 was larger than the decreases added together. So is there a better way to represent the average/total change?

Liger
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  • This question is hard to interpret and will probably have multiple answers as it stands depending on the specifics of what you want. Maybe you want to include what led you to ask this question- like what kind of problem were you trying to solve? – Bernard W Dec 22 '15 at 00:54
  • Perhaps in your example, instead of saying the changes are -30%, -71%, 150%, -20%, you could say your number is multiplied by 0.7, 0.29, 1.5, 0.8. Then find the arithmetic mean or geometric mean of those numbers, depending on what you are after. – turkeyhundt Dec 22 '15 at 02:01

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Given the sequence of five numbers $10, 7, 2, 5 \text{ and } 4$ consider the consecutive ratios $r_1=7/10$, $r_2 =2/7$, $r_3=5/2$ and $r_4=4/5$. The ratio $r=4/10$ (related to the overall change in your original sequence) can be written as the product of the consecutive ratios. That is $$r = r_1 \cdot r_2 \cdot r_3 \cdot r_4 \ .$$

Now if you think of each of the $r_i$'s as an estimate of some "typical" ratio $\hat{r}$ then $r=(\hat{r})^4$ so that $\hat{r} = r^{1/4}$, the geometric mean of the individual ratios.

Back to your original thinking, you defined percent change as $\frac{new-old}{old} \times 100 = (\frac{new}{old} - 1) \times 100$. Substitute $\hat{r}$ for the ratio $\frac{new}{old}$ and you get (for consecutive numbers in your sequence):

$$ \text{"Typical" Percent Change} = (\hat{r} - 1) \times 100 $$ $$ = (r^{1/4} - 1) \times 100 $$ $$ = \left(\left(\frac{4}{10}\right)^{1/4} - 1\right) \times 100 $$ $$ = -20.5\% $$

The individual percent changes are $-30\%, -71.4\%, 150\%, \text{ and } -20\%$ with the typical value as I have defined it falling somewhere in the middle. The typical value $-20.5\%$ is "that constant degree of change required between consecutive numbers in your sequence to get from the first number to the last." I think this is what you are looking for.

Generalizing to a sequence of positive numbers $x_1, x_2, ..., x_n$ the "typical" percent change is $$\left(\left(\frac{x_n}{x_1}\right)^{\frac{1}{(n-1)}} - 1\right) \times 100$$

This must seem a little unsatisfying because it appears as if only the first and last values in the sequence are used. But if you were given a sequence of ratios or percent changes (from which ratios could be derived) instead of the raw data, you would need all of those values in order to compute $\hat{r}$. Then your formula would be: $$ (\hat{r} - 1) \times 100 $$ where $\hat{r} = (r_1 \cdot r_2 \cdot ... \cdot r_{n-1})^{1/(n-1)}$.

SWilliams
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