# how to properly calculate difference in percent

Suppose, you have

x=8
y=10

Then y is 25% more than x, if x is used as the "base" value. If y ist the base, then x is 20% less than y.

So if I want to say, that the difference between x and y is ... percent, what value could I use? I can think of using 22.5% being in the middle between 20% and 25%, but that does not feel well-founded.

Should I not use percentages at all?

• "the difference between x and y is ... percent" - of what? Jul 16, 2011 at 13:13
• @J.M., so you are asking for a "base" (do you call it that?) as well. I thought, I could do without. But @Sylverdrag 's answer suggests otherwise. Jul 16, 2011 at 13:37
• If you want to avoid a base, then you could take the logarithm of the quotient (i.e. the difference of the logarithms), since $\log_e \frac{10}{8} \approx 0.22314\ldots$ and $\log_e \frac{8}{10} \approx -0.22314\ldots$, but this is not a percentage. Jul 16, 2011 at 22:58
• This is an interesting idea, @Henry! Thank you. Jul 19, 2011 at 0:43

## 1 Answer

A percentage can not stand alone, it's always a percentage of something.

You have to specify which number you are using as a base. No way around it...unless of course you work in marketing.

• I think, I understand. So I could probably do the following: Take the mean between x and y and then calculate the ratio between y-x and the mean? What I mean is $\left| \frac{y-x}{(x+y)/2} \right|$ . Jul 16, 2011 at 13:40
• @Sebastian: You could do that. But then you would be using a convention that as far as I know no one else uses. Jul 16, 2011 at 13:48
• @André, ok, I hoped there was even a name for that formula. I should probably think twice, before using percentages if I want to described positive and negative "movements" (as in -20%, but +25%) at the same time. Jul 16, 2011 at 14:13
• @Sebastian: We can speak of percentage deviation from the mean. However, the context there is not a mere two observations. Using it on say two prices would be misleading. Jul 16, 2011 at 14:25