Is it possible to calculate the arithmetic mean from the geometric mean?

Question

An estimation method would be acceptable, doesn't need to be exact (but obviously that would be preferable). I have a dataset of geometric means, need to calculate the arithmetic mean.

Answer 1

Unfortunately the AM-GM inequality is the best you can do. If your data is $\{x,\frac{1}{x}\}$ the geometric mean will be $1$, yet you can make your arithmetic mean any value in $[1,+\infty)$ by choosing $x$ large enough.

Answer 2

Since the geometric mean for both $(2,2)$ and $(1,4)$ is $2$, while the arithmetic means are $2$ and $2.5$, the answer is a clear no. The only thing you can say is that the geometric mean is smaller or equal to the arithmetic.

Answer 3

No it is not. Arithmetic mean gives you one equation. And there are two numbers to solve for. So there are infinite possibilities. For example, $$a=1,b=100$$ $$ a=0,b=101$$ They both have same A.M but widely different G.M

Answer 4

You can use the A.M. - G.M. inequality which is as follows- $$\frac{x_1+x_2+\cdots+x_n}{n}\ge\sqrt[\large n]{x_1\cdot x_2\cdots x_n}$$

Answer 5

Of course the answer is "no," not without more data. There's not a direct calculation.

In finance, what's interesting is that the arithmetic mean will always be a bit higher than the geometric for year on year stock returns.

From 1929 to 2013, the average was 11.41% (S&P return) yet the geometric mean was 9.43% nearly 2% lower per year. An understanding of the math behind this difference is helpful when someone mentions the market's return over a particular period.

Answer 6

If you could, hey, why do you think people would have defined two different notions of mean?

However you can compute the arithmetic mean from the geometric mean of other numbers. Namely the arithmetic mean of $x_1,\ldots,x_n$ is the natural logarithm of the geometric mean of $e^{x_1},\ldots,e^{x_n}$. Not sure why anybody would like to compute it this way though.

Answer 7

For an arithmetic mean $a$ and for a geometric mean $g$, we can see

1. $\sqrt{x\cdot y}=a \leftrightarrow x\cdot y=a^2$
2. $\frac{x+y}{2}=b \leftrightarrow x+y=2b \leftrightarrow y=2b-x$

Substituting (2) into (1) we get

$$x\cdot(2b-x)=a^2 \leftrightarrow -x^2+2bx-a^2=0 \leftrightarrow x=-\frac{4b^2\pm\sqrt{4b^2-4a^2}}{2}$$

Thus not even a calculation of gm from am is possible, but for every positive $(a;b)$ am, gm pair we can find the $(x,y)$ real number pair, whose am and gm are the given values!

Answer 8

Since you want a estmated relationship between AM and GM, and there's no further constraint on the data, I can present a widely used formula for approximation in finance. Hope this will light you up a bit.
Let $A=\frac{1}{n}\sum_{k=1}^n r_k$ denote AM of a series of return $r_1,r_2...r_k$, $G=[\prod_{k=1}^n (1+r_k)]^{1/n}-1$ denote GM of the returns, then $G\approx A-\frac{1}{2}V$, where $V$ is variance of these returns.
Proof:
We can take $1/n$ inside the square brackets, and $G=\prod_{k=1}^n (1+r_k)^{1/n}-1$ , use the Maclaurin series expansion for $(1+r_k)^{1/n}$ up to degree 2 and ignore the remainder: $$(1+r_k)^{1/n}\approx 1+\frac{1}{n}r_k+\frac{1-n}{2n^2}r_k^2$$ Substitute the expansion into $G$, we have $G\approx\prod_{k=1}^n(1+\frac{1}{n}r_k+\frac{1-n}{2n^2}r_k^2)-1$, expand the product and drop terms with degree 3 and above, that is, you can see the expansion as selecting one term in one bracket such that their product's degree is no more than 2, then we have another level of approximation of $G$: $$G\approx\prod_{k=1}^n(1+\frac{1}{n}r_k+\frac{1-n}{2n^2}r_k^2)-1\approx \frac{1}{n}\sum_{k=1}^nr_k+\frac{1}{n^2}\sum_{k\neq l}^nr_kr_l+\frac{1-n}{2n^2}\sum_{k=l}^nr_k^2$$
As for $V$, we have $$V=\frac{1}{n}\sum_{k=1}^n(r_k-A)^2=\frac{1}{n}\sum_{k=1}^nr_k^2-A^2=\frac{n-1}{n^2}\sum_{k=1}^nr_k^2-\frac{2}{n^2}\sum_{k\neq l}^nr_kr_l$$ Observe $V$ and the last two terms of $G$, here we are: $G\approx A-\frac{1}{2}V$.