# Problem on interest rates -- Math proof interesting

1. Jul 1, 2014

### lesdavies123

Hi, I have a problem that I need to solve, it goes like this: Using the fact that the geometric mean of a collection of positive numbers is less than or equal to the arithmetic mean, show that if annual compound interest rates over an n-year period are i1 in the first year, i2 in the second year,...,in in the nth year, then the average annual compound rate of interest for the n-year period is less than or equal to 1/n x (the summation of all the interest rates.

So that's the question, so I figure out that 1/n and the summation is the arithmetic mean of the interest rates. But to me, the geometric mean would be the nth root of all products of all the i. But the average annual compound rate would be the nth root of all the products of all the (1+i) - 1 in the end to get the interest rate. So that's it now I can't figure out how to do the proof as the average annual compound rate is not exactly the geometric mean but the nth root of the products of the (i+1) - 1 as I said earlier. So how can I put these together and do the proof if anybody figures it out! Thank you!

2. Jul 2, 2014

### pasmith

You aren't applying the GM-AM inequality to the correct set of numbers.

If you invest $X$, then after $n$ years you have$$A_{\mathrm{var}} = (1 + i_1)(1 + i_2) \dots (1 + i_n)X = X\prod_{i=1}^n (1 + i_n).$$ If instead you had invested $X$ at a constant rate of $r$ you would have $$A_{\mathrm{fixed}} = X(1 + r)^n.$$ By definition, the average rate $\bar \imath$ is such that $A_{\mathrm{fixed}} = A_{\mathrm{var}}$, so $\bar \imath$ must satisfy$$\prod_{i=1}^n (1 + i_n) = (1 + \bar \imath)^n.$$ Thus $1 + \bar \imath$ is the geometric mean of the numbers $\{1 + i_1, 1+ i_2, \dots, 1 + i_n\}.$

3. Jul 2, 2014

### pwsnafu

Just to add to this, in actuarial science we define the geometric mean rate of return calculated over n periods of equal length to be
$(1+g)^n = \prod_{t=1}^{n}(1+i_t).$
As pasmith notes, its the mean of the accumulations not the rates themselves.

Last edited: Jul 3, 2014