Relationship between Euler's number, Uniform distribution and geometric mean

Question

Recently I was trying to combine p-values (e.g. by Fisher's method), but during that investigation I got led astray. I started looking at the distribution of the geometric mean of sets of independent Uniform(0,1) variables. I noticed that as the set size increased, the mean of the geometric mean’s distribution appeared to converge to a number. It appeared to converge to the inverse of Euler's number, 1/exp(1) = 0.3678794.

For set size n, I created 5000 n-vectors containing independent Uniform(0,1) draws. I calculated the geometric mean of each of the 5000 n-vectors. Then I calculated the sample mean of those 5000 geometric means. I did this for a range of set sizes from 1 to 1,000,000. Here are the results for the smallest and largest set sizes.

> head(dfRes)  
iGroupSize      mean  
1          1 0.4856412  
2          2 0.4599790  
3          3 0.4188242  
4          4 0.4110794  
5          6 0.3960357  
6         10 0.3839790  
> tail(dfRes)  
iGroupSize      mean  
26     100000 0.3678654  
27     158489 0.3678691  
28     251189 0.3679015  
29     398107 0.3679053  
30     630957 0.3678569  
31    1000000 0.3678630   

As you can see, for n=1 the mean geometric mean is close to 0.5 as expected, and when n is large (in fact for n>100) it was very close to 1/exp(1).

I think this connection between Uniform(0,1) and exp(1) if true is amazing. I can’t see the connection. Is this result well known? Can anyone prove this result?

Thanking you in advance.

Desmond

Answer 1

Neat find! I think what's happening is this: You're looking at the average of the geometric mean, which is really just $\mathbb{E}[(X_1X_2...X_n)^{1/n}]$ for $X_1,...,X_n$ independent $U(0,1)$ random variables. This can be directly calculated, since the probability density is just 1: $$\mathbb{E}[(X_1X_2...X_n)^{1/n}]=\int_0^1dx_1...\int_0^1dx_n x_1^{1/n}...x_n^{1/n}$$ $$=\left(\int_0^1x_1^{1/n}dx_1\right)...\left(\int_0^1x_n^{1/n}dx_n\right)$$ $$=\left(\frac{1}{1+\frac{1}{n}}x_1^{1+\frac{1}{n}}\vert_0^1\right)...\left(\frac{1}{1+\frac{1}{n}}x_n^{1+\frac{1}{n}}\vert_0^1\right)$$ $$=\frac{1}{(1+\frac{1}{n})^n}$$ One way to define $e$ is $e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$, so that's a proof that this converges to $1/e$. Whether that answers your question of why this converges to $1/e$, I don't know.