Talk:Cantor distribution

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)

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)

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)

• 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)

• 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)

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)

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)

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