Averaging percentages

A stock grows by 20% in the first year and drops by 4% in the second year? What is the average performance of a stock per year?

We cannot just average +20% and -4% and get an increase of 16%. We can check that this is not correct. Assume the stock was worth 100 at start. Then, after the first year it is 120, and after the second year it would be 115.2. If we assume an annual increase of 16%, then this should have been

$\(100 * 1.16^2 = 134.56\).

Instead, we have to find what number to put in the formula above to replace 1.16 and get the value we were looking at:

\[100 \left(1 + r\right)^2 = 115.2\]

So, the actual average performance is just 7.33%, much much smaller than the arithmetic mean.

In general, if we have a starting value of \(S_0\) and \(n\) periods leading to a final value of \(S\), the average \(r\) we are looking at is the value that gives:

\[S_0 \left(1 + r\right)^n = S\]

That is:

\[r = \left(\frac{S}{S_0}\right)^\frac{1}{n} - 1\]

For two years, with performances of \(a\) and \(b\), we have

\[S = S_0(1+a)(1+b)\]

So, the average rate is

\[r = \sqrt{(1+a)(1+b)} - 1\]

This is different than the geometric mean (but will be the geometric mean of the actual multipliers – the values of form \(1 + \alpha\)). Using the geometric-arithmetic mean inequality, we see that \(r\) is greater than the geometric mean of \(a\) and \(b\).

Next article is a special number, so it should be longer.