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From: Rob Johnson on 28 May 2010 11:12
In article <79454625-2297-4cc1-b844-ac18da67dff4(a)40g2000vbr.googlegroups.com>,
Pubkeybreaker <pubkeybreaker(a)aol.com> wrote:
>On May 28, 6:40 am, r...(a)trash.whim.org (Rob Johnson) wrote:
>> In article <2de3769a-6d5b-4c72-9e1f-43e877f41...(a)c11g2000vbe.googlegroups.com>,
>
>>
>> Looking back at the thread, I see that Robert Israel has come up
>> with the same answer using a probabilistic argument.
>
>Your posts always impress me.

Thank you. However, I hate it when posts leave an impression on my
bumper.

>Did you write: JUST THE ESSENTIALS OF ELEMENTARY STATISTICS?

It wasn't me. Nor did I sell my soul at the crossroads to play the
blues (enter "sell my soul at the crossroads to play the blues" into
Google).

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
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From: Robert Israel on 28 May 2010 15:39
Pubkeybreaker <pubkeybreaker(a)aol.com> writes:

> On May 28, 3:45=A0am, Robert Israel
> <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > "Dave L. Renfro" <renfr...(a)cmich.edu> writes:
> >
>
> >
> > It's most convenient to consider the circular Cantor set as the image
> > of the usual Cantor set under the map f: t -> exp(2 pi i t) from [0,1]
> > into the complex plane. =A0Now if X_j are independent Bernoulli random
> > variables with p=3D1/2 (i.e. flips of a fair coin),
> > Y =3D sum_{j=3D1}^infty (2/3^j) X_j is uniformly distributed on
> > the Cantor set.
>
> I do not see where Y comes from. Can you explain further?
> Excuse my ignorance.

The Cantor set is the set of all members of [0,1] that have a base-3
expansion consisting only of 0's and 2's. 2 X_j is the j'th base-3
digit of Y.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Pubkeybreaker on 28 May 2010 15:44
On May 28, 3:39 pm, Robert Israel
<isr...(a)math.MyUniversitysInitials.ca> wrote:
> Pubkeybreaker <pubkeybrea...(a)aol.com> writes:
> > On May 28, 3:45=A0am, Robert Israel
> > <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > > "Dave L. Renfro" <renfr...(a)cmich.edu> writes:
>
> > > It's most convenient to consider the circular Cantor set as the image
> > > of the usual Cantor set under the map f: t -> exp(2 pi i t) from [0,1]
> > > into the complex plane. =A0Now if X_j are independent Bernoulli random
> > > variables with p=3D1/2 (i.e. flips of a fair coin),
> > > Y =3D sum_{j=3D1}^infty (2/3^j) X_j is uniformly distributed on
> > > the Cantor set.
>
> > I do not see where Y comes from.  Can you explain further?
> > Excuse my ignorance.
>
> The Cantor set is the set of all members of [0,1] that have a base-3
> expansion consisting only of 0's and 2's.  2 X_j is the j'th base-3
> digit of Y.


Thanks.
From: Niels Diepeveen on 28 May 2010 22:46
Dave L. Renfro wrote:

> Here's an interesting problem I recently saw.
>
> Take the usual middle thirds Cantor set, constructed on the
> closed interval [0, 2*pi] instead of the closed interval
> [0,1], and bend it without stretching into a circle of
> radius 1 centered at the origin of the xy-coordinate plane
> so that the points 0 and 2*pi are glued together at (0,1).
> What are the xy-coordinates for the center of mass of the
> resulting circular Cantor set, assuming a uniform density
> for the Cantor set?
>
> Dave L. Renfro

I'm curious. When I first saw this post, I imagined the answer was
"Right next to the centre of mass of the rationals".
Usually, calculations of centres of mass are based on counting measure
for finite sets or Lebesue measure for infinite sets. Neither applies in
this case, yet everyone who replied seemed to make to make
(essentially) the same assumptions. Is there some general definition
that I'm not aware of, or is it really an ad hoc solution based on "it
stands to reason" given the symmetries.

--
Niels Diepeveen

From: Rob Johnson on 29 May 2010 03:55
In article <4c008003$0$22920$e4fe514c(a)news.xs4all.nl>,
Niels Diepeveen <n659474(a)dv1.demon.nl> wrote:
>Dave L. Renfro wrote:
>
>> Here's an interesting problem I recently saw.
>>
>> Take the usual middle thirds Cantor set, constructed on the
>> closed interval [0, 2*pi] instead of the closed interval
>> [0,1], and bend it without stretching into a circle of
>> radius 1 centered at the origin of the xy-coordinate plane
>> so that the points 0 and 2*pi are glued together at (0,1).
>> What are the xy-coordinates for the center of mass of the
>> resulting circular Cantor set, assuming a uniform density
>> for the Cantor set?
>>
>> Dave L. Renfro
>
>I'm curious. When I first saw this post, I imagined the answer was
>"Right next to the centre of mass of the rationals".
>Usually, calculations of centres of mass are based on counting measure
>for finite sets or Lebesue measure for infinite sets. Neither applies in
>this case, yet everyone who replied seemed to make to make
>(essentially) the same assumptions. Is there some general definition
>that I'm not aware of, or is it really an ad hoc solution based on "it
>stands to reason" given the symmetries.

There are several ways to characterize the uniform measure supported
on the Cantor set. One is defined in parallel to the way the Cantor
set is defined. For each stage of the middle thirds set, define the
measure to be the usual uniform measure on the remaining closed
intervals divided by their total measure. For example,

u_0 = X_[0,1]

u_1 = 3/2 X_{[0,1/3]U[2/3,1]}

u_2 = (3/2)^2 X_{[0,1/9]U[2/91/3]U[2/3,7/9]U[8/9,1]}

etc.

Each measure is absolutely continuous and as measures, they converge
to the Cantor measure.

Another way is using delta measures, as I did in my solution. The
following singular measures put delta masses at the left end of each
of the intervals of the previous measure

u_0 = d_0

d_0 + d_{2/3}
u_1 = -------------
2

d_0 + d_{2/3} d_0 + d_{2/9}
u_2 = ------------- * -------------
2 2

d_0 + d_{2/3} d_0 + d_{2/9} d_0 + d_{2/27}
u_3 = ------------- * ------------- * --------------
2 2 2

etc.

Using these measures and looking at the center of mass for each,
leads us to the center of mass of the limit measure, the Cantor
measure.

Thus the Cantor measure is the limit of absolutely continuous
measures and the limit of singular measures, as well.

Rob Johnson <rob(a)trash.whim.org>
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