# Expectations of the geometric mean of a random sample from a uniform distribution

If I have a random sample of size n from a Uniform(0,1) and I define the geometric mean as G can anyone give me insight in to how I can find the expected value of G, E[G]? Once I can get my head around this I would like to extend it to the expected value of powers of G. Any help would be appreciated.

• Exponentiated both sides and ended up with E[exp(G)] = E[exp(mu)]. This was from an online presentation I found. Then I guessed that the mean of a uniform(0,1) is 0.5 and subbed that in for mu. I am far from convinced with this. – Jon Oct 27 '14 at 9:48

This is the same as finding $$E[(X_1 X_2 \cdots X_n)^{1/n}],$$ where $X_i$ are independent and uniformly distributed. Hence it is equal to $$E[X_1^{1/n}]\cdots E[X_n^{1/n}] = E[X_1^{1/n}]^n = \left(\frac{1}{1+1/n}\right)^n.$$
• That last step is known as the Law of the Unconscious Statistician, Jon. When $X$ has a Uniform$(0,1)$ distribution its distribution function is $dx$ (for $0\le x\le 1$). Therefore whenever $g$ is a measurable function of $X$ then $\mathbb{E}(g(x))=\int_0^1 g(x) dx$. Paulius has applied this to the function $g(X)=X^{1/n}$. (Welcome to our site, Paulius!) – whuber Oct 27 '14 at 14:50