Expected Geometric Mean (Extended Version)

For n randomly chosen numbers $a_1,a_2,...,a_n$ , the expected value can be found by integrating the geometric mean function over every 0 to 1 interval:

$\int_{0}^{1}\int_{0}^{1}...\int_{0}^{1}\sqrt[n]{a_1a_2...a_n}\, da_1da_2...da_n$

Each integral provides a factor of $\frac{n}{n+1}$ , giving the total integral of $(\frac{n}{n+1})^{n}$ .

Now we can evaluate the limit as n approaches infinity; manipulating the expression, we get:

$\lim_{n \to \infty }(\frac{n+1}{n})^{-n}$ = $\lim_{n \to \infty }((1+\frac{1}{n})^{n})^{-1}$ = $e^{-1} \approx 0.367879$

I had no idea how to solve this. I gave a shot at 1/e. Sorry folks... :)

Let $a_k$ be random numbers uniformly distributed between 0 and 1. Then:

$\lim_{n\to\infty} \sqrt[n]{a_1a_2...a_n}=\exp(\frac{1}{n}\sum_{n=1}^{\infty} \log{(a_n)})$

Then from Monte Carlo integration it follows that:

$\frac{1}{n}\sum_{n=1}^{\infty} \log{(a_n)}=\int_{0}^{1} \log{(x)}dx= \log{(1)}-1 = -1$

And so finally

$\sqrt[n]{a_1a_2...a_n}=\exp(-1)$

Using logs and rules of indices you get
$\prod _{ n=0 }^{ \infty }{ G_n } = e^{\sum_{n = 0}^{\infty} \log(G_n)}$

by finding the (arithmetic) average of $\log(G)$ via rules of indices, it should follow that you can exponentiate it and the geometric average of $G$

the arithmetic mean of a function over a range can be attained by integrating it over the range and dividing it by the length of the range (which happens to be 1) $\bar{\log(G)} = \int_0^1 \log(x).dx = \lim_{a \to 0} \left[ x\ \left(log(x) - 1\right) \right]^1_a = \lim_{a \to 0}\left(-1 - a \left(\log(a) - 1\right)\right) = -1$

if we exponentiate $-1$ we get $\frac{1}{e}$

If numbers are picked randomly between 0 and 1, then those numbers are in the form 1/n, 2/n, ... n/n. If we do the GM of this sequence we obtain (n!/n^n)^(1/n) = (n!)^(1/n)/n. By Stirling then n! ~ k sqrt(n) (n/e) so taking the limit as n-> + infty we get 1/n..

My approach, as engineer by the trade, is a kind of heterodox .

You can consider the natural logarithm of geometrical mean as`equivalent to the edge of of an hyper-cube which volume equal to the product of the N values to be considered.

So by taking logarithm at both sides of the geometrical mean formula we have Ln ( Gm) = [ Ln (x1)+Ln(x2)+...Ln(xN) ] / N. Being infinite and equally probable all values x(j) between 0 and the numerator of quantity under bracket is precisely the area of function Ln(x) between 0 and 1 which values -1. N = 1 since is the interval between [0,1].

Therefore Ln(Gm) = -1, then Gm = 1/e

It is possible to write $G_{n}$ as

$G_{n} = e^{\frac{1}{n} \sum_{i=1}^n \log(a_i)} = e^{\mathrm{E}(\log(X_{n}))}$

where $X_{n}$ is the uniform discrete random variable in ]0;1[ and E denote the mean(expected value) of the random variable $X_{n}$ .

To calculate $\lim_{n \to +\infty}G_n$ it is sufficient to calculate the mean of the continuos random variable log(X), because for $n \to +\infty$ the discrete uniform random variable $X_{n}$ tends to continuos uniform random variable $X$

Let $Y = \log(X)$ , then it is possible to calculate the probability density function(pdf) $f_{Y}$ of $Y$ know the pdf of $X$ variable $f_{X}$ with the following formula ( random variables )

$f_{Y}(Y) = f_{X}(Y) \frac{dX}{dY}$

the calculation give

$f_{Y}(Y) = e^Y$ with $Y \in ]-\infty;0[$

Then the Y is a random variable with exponetial distribution and $E[Y]=E[\log(X)] = -1$ . substituting we obtain the desidered result

$\lim_{n \to +\infty}G_n = \lim_{n \to +\infty} e^{\mathrm{E}(\log(X_{n}))} = e^{\mathrm{E}(\log(X))} = e^{\mathrm{E}(Y)} = e^{-1} \approx$ 0.367879........