Talk:Cantor distribution

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Untitled

Funny distribution this one, but nmot as funny as the word "eventuate" that was in the previous version of the article. --Lucas Gallindo 03:16, 5 October 2006 (UTC)[reply]

From the Oxford English Dictionary:

“

2. To be the issue; to result, come about.

1834 DE QUINCEY Coleridge Wks. (1863) II. 93 In the upshot, this conclusion eventuated (to speak Yankeeishly), that, etc. 1876 C. M. DAVIES Unorth. Lond. I. 25 If So-and-so were condemned, a schism in the National Church would eventuate. 1884 Law Times 14 June 121/1 When there was danger of a war eventuating with America.

”

Michael Hardy 16:11, 5 October 2006 (UTC)[reply]

Alternate definition?

Is this distribution also the limiting (as n goes to infinity) distribution of

{\frac {1}{2}}+\sum _{i=1}^{n}{\frac {R_{i}}{3^{i}}},\,\!

where the R_i are iid Rademacher distributions? Seems to me that it is, but my proof skills are quite rusty... Baccyak4H (Yak!) 20:15, 12 December 2006 (UTC)[reply]

Since any number in the Cantor set can be (uniquely) expressed, in base 3, as 0.abcd... where a,b,c,d,... are either 0 or 2, then a simple way to simulate this Cantor distribution would be to add iid Bernoullis, scaled to 2/3:

\sum _{i=1}^{n}{\frac {2}{3^{i}}}B_{i}({\frac {1}{2}})\,

Since Rademacher and Bernoulli are essentially the same distribution, just scaled ( $R_{i}=2B_{i}({\frac {1}{2}})-1\,$ ), then a simple substitution could prove your expression. Albmont (talk) 20:21, 17 July 2008 (UTC)[reply]

BTW, I think the characteristic function of the Cantor distribution is wrong. It seems like a series that diverges everywhere. Albmont (talk) 20:21, 17 July 2008 (UTC)[reply]

Thanks, I convinced myself in the interim time since asking, but it's good to know I haven't completely lost it.

About the series diverging, I don't see that... all the terms of the product are in [-1, 1], but the terms' limit is +/- 1, depending on the sign of t, so the product part cannot diverge, I would think (I cannot rule out it might be 0 in the limit, but that's OK). Could you elaborate? Baccyak4H (Yak!) 20:35, 17 July 2008 (UTC)[reply]

About the series diverging, I don't see that - neither do I now, but first I swear I saw a Σ instead of the Π! BTW, I tried to use the Bernoulli series to check the formula, but I only came as far as

\Pi (1/2+1/2\ \exp(2it/3^{n}))\,

. Albmont (talk) 20:49, 17 July 2008 (UTC)[reply]

Ouch! The derivation using Rademacher is obvious! Albmont (talk) 20:50, 17 July 2008 (UTC)[reply]

Request for clarification

This article currently states:

...it is not absolutely continuous with respect to Lebesgue measure,...

which is strictly true but awkwardly worded, misleading. The Cantor distribution, treated as a measure (shall we call it the "Cantor measure"?) is not absolutely continuous with respect to the Lebesgue measure. However, the Cantor distribution, treated as an "ordinary function", is absolutely continuous (with no reference to a measure), as currently defined in the article on absolute continuity. That is, for any ε>0 there exists a δ>0 such that any interval shorter than δ has max(f(x)) - min(f(x)) < ε on that interval. Indeed, for ε = 2^-n then δ = 3^-n-1 satisfies the definition given in the article on absolute continuity. I think these two contrasting results are sufficiently confusing that maybe the wording whould be chosen more carefully. I'd do it myself, except that I'm tired just right now, and fear making a mistake.linas (talk) 04:42, 11 February 2010 (UTC)[reply]

Mean

The article currently states "It is easy to see by symmetry that for a random variable X having this distribution, its expected value E(X) = 1/2, and that all odd central moments of X are 0." Is this adequate without a demonstration that the mean is finite (e.g., by symmetry, the standard Cauchy distribution has mean of 0, except it that doesn't since the integral doesn't converge). Rlendog (talk) 20:10, 7 March 2011 (UTC)[reply]