Cantor Set Geometric Mean

Question

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. — , 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 $\alpha \to 0^+$. — , 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
Of possible related interest is center of mass of the Cantor set, a sci.math thread I started on 27 May 2010. — Dave L. Renfro, Aug 11 '16 at 20:12
http://www.artofproblemsolving.com/community/c7h1288021_am_gm_over_cantor_set_and_01 — , Aug 11 '16 at 20:15
Also Bungo, how would one evaluate the latter expression you have? — , Aug 11 '16 at 20:18
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? — hmakholm left over Monica, Aug 11 '16 at 20:20
@SanjoySmileyKundu Beats me, that's why it was only a comment instead of an answer :-) — , Aug 11 '16 at 20:21
Dang now I'm even wondering if its possible to actually numerically or analytically evaluate this? Also Henning, how did you come up with that sum? — , Aug 11 '16 at 20:23
@SanjoySmileyKundu: Numerically ought to be possible, at least. We can certainly approximate a geometric mean of the right half of the Cantor set (from $\frac23$ to $1$), and I have a sketchy argument that the geometric mean of the entire set ought to be $\frac19$ of that. — hmakholm left over Monica, Aug 11 '16 at 20:26
@SanjoySmileyKundu: The sum comes from the standard characterization of the Cantor set of those numbers between $0$ and $1$ that have a ternary fraction representation using only the digits 0 and 2. It seems reasonable to consider a "standard" probability measure on the Cantor set by saying that each ternary digit position is independent and uniformly distributed. — hmakholm left over Monica, Aug 11 '16 at 20:28
@HenningMakholm I was wondering if you could give a rough estimate as to what the geometric mean of the entire set might be? My ansatz for this problem originally was to construct the first 20 or so iterations of the Cantor Set and then find the geometric mean from there. It shouldn't be too different from the actual GM. — , Aug 11 '16 at 20:32
@SanjoySmileyKundu: Extremely roughly and without actually calculating anything, I get that it must be somewhere between $\frac2{27}$ and $\frac19$ ... — hmakholm left over Monica, Aug 11 '16 at 20:35

Robert Israel · Accepted Answer · 2016-08-11T21:11:24.103

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. — , Aug 11 '16 at 21:37

Cantor Set Geometric Mean

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