# 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 Kundu 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

## 1 Answer

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 $.2749744970902444$.

• Thank you for your solution Robert! Interestingly, the maximum predicted value of the Geometric Mean of the Cantor Set divided by the Geometric Mean of the Cantor Set or all reals between (0,1) is roughly 74.7%. Accounting for error, since only 16 terms were used in place of an arbitrarily large amount, one can reasonably state that the Geometric Mean of the Cantor Set of all reals between (0,1] is ~75% of the GM of all reals between (0,1], which was 1/e. – Sanjoy Kundu Aug 11 '16 at 21:37