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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jon
    Oct 27, 2014 at 9:48

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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 $$E[X_1^{1/n}]\cdots E[X_n^{1/n}] = E[X_1^{1/n}]^n = \left(\frac{1}{1+1/n}\right)^n.$$

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  • $\begingroup$ Then I imagine I could replace the X with X^2 and so on to get the expected value for higher powers of G? $\endgroup$
    – Jon
    Oct 27, 2014 at 9:40
  • $\begingroup$ Yes, most certainly $\endgroup$
    – psarka
    Oct 27, 2014 at 9:55
  • $\begingroup$ I am afraid I don't understand the last step could you direct me to an appropriate reference? $\endgroup$
    – Jon
    Oct 27, 2014 at 11:17
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    $\begingroup$ 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!) $\endgroup$
    – whuber
    Oct 27, 2014 at 14:50

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