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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.

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$\{4,1}\$ and $\{2,2\}$ have the same geometric mean but have different arithmetic mean. The best you can do is the AM-GM inequality. – Jp McCarthy Jun 10 '14 at 14:14
AM - GM inequality?? $$ \frac{x_1+x_2+\cdots+x_n}{n}\ge\sqrt[\large n]{x_1\cdot x_2\cdots x_n}. $$ – Tunk-Fey Jun 10 '14 at 14:15
Some additional information might allow you to make a better estimate, for instance if you also knew the maximum value/an upper bound. – R.. Jun 10 '14 at 17:15
up vote 31 down vote accepted

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.

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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.

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Please correct: $AM(1,4)=2.5$ – gammatester Jun 10 '14 at 14:17
@gammatester: Done. Not sure why you didn't just fix it yourself, though, given that you have more than enough rep to edit. – Ilmari Karonen Jun 10 '14 at 15:49
No, the geometric mean is not always smaller than the arithmetic mean. $\frac{2+2}{2}=\sqrt{2\cdot 2}$. – user26486 Jun 10 '14 at 17:15
@mathh Thank you. fixed. – 5xum Jun 10 '14 at 18:17

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

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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

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While you are correct that there is an infinite amount of possibilities, your explanation of the cause is not exactly right. One equation with two numbers to solve for does not necessarily have an infinite solution set - $a^2+b^2=0$ is an example. – user26486 Jun 10 '14 at 17:19
$a^2+b^2=0$ have infinite solutions if you do not assume a & b to be real. Also in the above case it is a linear equation. – Edwin_R Jun 10 '14 at 17:40
The arithmetic mean is not the one given; the geometric mean is. Hence the arithmetic mean is not the one giving us an equation, as you claim. And yes, I assumed they are real (I agree I shouldn't have), I don't think the OP meant otherwise. – user26486 Jun 10 '14 at 18:37
Oh. I wrote it assuming given A.M. You are right then. I have not been exactly precise. – Edwin_R Jun 10 '14 at 18:43

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.

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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.

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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!

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