Cantor Set Geometric Mean

Find the Geometric Mean of all reals existing as part of the Cantor Set between (0,1]. I've been trying to solve this problem, but keep messing up the sets I construct for higher iterations. Any help would be appreciated.

https://en.wikipedia.org/wiki/Cantor_set

I'm not sure how to rigorously define Geometric Mean, I am just going by the definition posted on wikipedia, which is the n-th root of the product of n numbers.

https://en.wikipedia.org/wiki/Geometric_mean

Where the problem originated: http://www.artofproblemsolving.com/community/c7h1288021_am_gm_over_cantor_set_and_01

• The set is uncountable, so you'd have to rigorously define "geometric mean" – AJY Aug 11 '16 at 20:04
• I'm not sure how to rigorously define geometric mean, I'm just going by the standard definition of GM posted on wikipedia. – Sanjoy The Manjoy Aug 11 '16 at 20:07
• Analogously with this answer one might try writing $$\exp\left(\frac{1}{|C|}\int_C \log x\ dx\right)$$ where $C$ is the Cantor set and $|C|$ is its measure. But unfortunately, $|C| = 0$, so that idea doesn't work. I suppose you might compute $$\exp\left(\frac{1}{|C_{\alpha}|}\int_{C_{\alpha}} \log x\ dx\right)$$ where $C_{\alpha}$ is the fat Cantor set of measure $\alpha > 0$, and take the limit as $\alpha \to 0^+$. – Bungo Aug 11 '16 at 20:10
• In what context did this problem arise? And, how exactly was it phrased? – Noah Schweber Aug 11 '16 at 20:11
• A possible precise formulation would be: Let $X_1, X_2, \ldots, X_k, \ldots$ be a countable sequence of independent uniformly random bits. Then $\displaystyle \sum_{i=1}^\infty \frac{2}{3^i}X_i$ is a random variable whose range is the Cantor set. What is the expectation of its logarithm? – Henning Makholm Aug 11 '16 at 20:20

Using Henning Makholm's series, let $$Y = \sum_{n=1}^\infty \dfrac{2}{3^n} X_n = \dfrac{2}{3} X_1 + \dfrac{1}{3} Z$$ where $Z$ has the same distribution as $Y$ and is independent of $X_1$. Conditioning on $X_1$, \eqalign{ \mathbb E[\log Y] &= \dfrac{1}{2} \mathbb E\left[\log \left(\frac{Z}{3}\right)\right] + \dfrac{1}{2} \mathbb E\left[\log \left(\frac{2}{3} + \frac{Z}{3}\right)\right]\cr &= - \dfrac{\log(3)}{2} + \dfrac{1}{2} \mathbb E[\log Y] + + \dfrac{1}{2} \mathbb E\left[\log \left(\frac{2}{3} + \frac{Y}{3}\right)\right]\cr} so that $$\mathbb E[\log Y] = - \log(3) + \mathbb E\left[\log \left(\frac{2}{3} + \frac{Y}{3}\right)\right]$$ The expectation on the right can be nicely approximated using a few terms of the series. Using $16$ terms, I find that $$[-1.291076932952935 \le \mathbb E[\log Y] \le -1.291076923469643]$$ Your "geometric mean" is the exponential of this, thus between $.2749744944825810$ and $.2749744970902444$.