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All Integers are Interesting (with Proof)
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Reef Fish
science forum Guru Wannabe


Joined: 28 Apr 2005
Posts: 200

PostPosted: Thu Apr 28, 2005 7:54 pm    Post subject: Re: Measuring Spread of Combinations Reply with quote

Yaroslav Bulatov wrote:
Quote:

In your case there are no regular runs, but there is a run with step
3
(43,46,49). So if you devised a runs test which used runs of this
kind
before seeing the data, and then saw the sequence, you'd have
evidence
against the process being random.

No! Not in a SINGLE occurrence of a random combination of 5 numbers
from 1 to 52.

Each combination has the same probability of being observed.

Quote:

An important issue is to devise the test before seeing the data.

That's a non-issue in the winning number observed by the OP.

Read my original follow-up, the 2nd post in this thread, on April 24,
2005.
This is NEIHER a probability or statistics problem based on the
occurrence
of a SINGLE combination.

-- Bob.
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Reef Fish
science forum Guru Wannabe


Joined: 28 Apr 2005
Posts: 200

PostPosted: Thu Apr 28, 2005 9:29 pm    Post subject: OT: Why some integers are interesting Reply with quote

Carl G. wrote:

Quote:
I actually find some integers interesting (for example: 7101001000).

Since I had already proved that every integer is interesting, there
must be many reasons why 7101001000 is interesting.

It is interesting because it is 1000 times the prime number 7101001.

What's your reason?

-- Bob.
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mensanator@aol.compost
science forum Guru


Joined: 24 Mar 2005
Posts: 826

PostPosted: Thu Apr 28, 2005 9:55 pm    Post subject: Re: OT: Why some integers are interesting Reply with quote

Reef Fish wrote:
Quote:
Carl G. wrote:

I actually find some integers interesting (for example:
7101001000).

Since I had already proved that every integer is interesting, there
must be many reasons why 7101001000 is interesting.

It is interesting because it is 1000 times the prime number 7101001.

What's your reason?

It's rich with harmonic resonance:

: you have one 7 (lucky) and six (perfect) 0's

: perfect (6) + lucky (7) = unlucky (13)

: you also have three (sacred) 1's, which, when reversed (profane)
becomes 13!

A very scary number, eh?

Quote:

-- Bob.
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Carl G.
science forum beginner


Joined: 18 May 2005
Posts: 18

PostPosted: Thu Apr 28, 2005 9:58 pm    Post subject: Re: Why some integers are interesting Reply with quote

"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114730956.418320.26000@g14g2000cwa.googlegroups.com...
Quote:

Carl G. wrote:

I actually find some integers interesting (for example: 7101001000).

Since I had already proved that every integer is interesting, there
must be many reasons why 7101001000 is interesting.

It is interesting because it is 1000 times the prime number 7101001.

What's your reason?

-- Bob.

From one of my posts (whether it is interesting or not depends on the
reader):

Define the ten digit "checksum" of a ten digit sequence as follows:

If all ten digits in the sequence are the same, the checksum is undefined,
otherwise the first digit in the checksum is the number of zeros in the
sequence, the second digit in the checksum is the number of ones in the
sequence, ..., and the tenth digit in the checksum is the number of nines in
the sequence.

For example, if the ten digit sequence was "2718281828" the checksum would
be "0230000140", because there are 0 zero's , 2 ones, 3 twos, 0 threes, 0
fours, 0 fives, 0 sixes, 1 seven, 4 eights, and 0 nines.

Problem 1: Which ten digit sequence is same as its checksum?

Problem 2: I selected a particular ten digit sequence. I then replaced
this sequence with its checksum. I kept replacing the sequence with its
checksum until the resulting checksum matched one of the previously
encountered checksums. Following this procedure, I was able to obtain over
seven different checksums before stopping. What was the final (different
from previous) checksum?

Problem 3: What is most number of times that one can replace a sequence by
its checksum before the result is either undefined or matches a previously
encountered sequence?

Carl G.
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William Elliot
science forum Guru


Joined: 24 Mar 2005
Posts: 1906

PostPosted: Thu Apr 28, 2005 10:58 pm    Post subject: Re: OT: Why some integers are interesting Reply with quote

On Thu, 28 Apr 2005, Reef Fish wrote:

In a package of integers, how many crums are there?
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Reef Fish
science forum Guru Wannabe


Joined: 28 Apr 2005
Posts: 200

PostPosted: Thu Apr 28, 2005 11:04 pm    Post subject: Re: OT: Why some integers are interesting Reply with quote

mensanator@aol.com wrote:
Quote:
Reef Fish wrote:
Carl G. wrote:

I actually find some integers interesting (for example:
7101001000).

Since I had already proved that every integer is interesting, there
must be many reasons why 7101001000 is interesting.

It is interesting because it is 1000 times the prime number
7101001.


The notion of interesting numbers originated from those well versed
in Number Theory in Mathematics, in which PRIME numbers play a
dominant roll. These include Gauss, Fermat, Goldbach, and more recent
mathematicians as Ramanujan, Erdos, and others.

Not many people would know that 7101001 is a prime number! :-)

The number 7101001000 is interesting because it has this <interesting>
prime-number factorization:

7101001000 = 2^3 x 5^3 x 7101001.

Quote:

What's your reason?

It's rich with harmonic resonance:

: you have one 7 (lucky) and six (perfect) 0's

: perfect (6) + lucky (7) = unlucky (13)

: you also have three (sacred) 1's, which, when reversed (profane)
becomes 13!

A very scary number, eh?

Only to the Numerologist. :-)

But I think that's a better reason than that advanced by Carl G, who
obviously accidentally got lost from alt.geek.comp.lang.java into one
of the mathematical ngs. Number Theory is already OT here. His
"check-sum" reason is interesting, but at least (OT)^2.

-- Bob.
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Reef Fish
science forum Guru Wannabe


Joined: 28 Apr 2005
Posts: 200

PostPosted: Thu Apr 28, 2005 11:19 pm    Post subject: OT: Re: OT: Why some integers are interesting Reply with quote

William Elliot wrote:
Quote:

In a package of integers, how many crums are there?

You wrote elsewhere, for an Abelian group G with cyclic
subgroup H of G, and cosets X and Y,

WE> Do you mean 'for x,y in G, let X = x+H, Y = y+H' ?
WE> Then -Y = -y+H, X-Y = x-y + H, c(X-Y) = c(x-y) + cH ?


You're definitely FULL of crums, William. :-)

-- Bob.
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Odysseus
science forum addict


Joined: 05 Jun 2005
Posts: 60

PostPosted: Fri Apr 29, 2005 2:33 am    Post subject: Re: All Integers are Interesting (with Proof) Reply with quote

Reef Fish wrote:
Quote:

snip

"The Dyslexic Agnostic stayed up nights wondering if there is a Dog."
^

Insomniac
--
Odysseus
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Bob Pease
science forum beginner


Joined: 29 Apr 2005
Posts: 47

PostPosted: Fri Apr 29, 2005 4:44 am    Post subject: Re: All Integers are Interesting (with Proof) Reply with quote

"Odysseus" <odysseus1479-at@yahoo-dot.ca> wrote in message
news:4271B941.A239FBE7@yahoo-dot.ca...
Quote:
Reef Fish wrote:

snip

"The Dyslexic Agnostic stayed up nights wondering if there is a Dog."
^
Insomniac
--
Odysseus

I read this several times and didn't understand what was so fnnuy

RJP
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Pavel314
science forum addict


Joined: 29 Apr 2005
Posts: 78

PostPosted: Fri Apr 29, 2005 9:03 am    Post subject: Re: Measuring Spread of Combinations Reply with quote

"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114725261.938824.303720@l41g2000cwc.googlegroups.com...
Quote:

This is NEIHER a probability or statistics problem based on the
occurrence
of a SINGLE combination.

This has been an interesting and informative thread; thanks to everyone for
their contributions.

However, what I was originally trying to get at is not a test of the
randomness of the draw or the odds of getting a specific combination but
just a way of measuring the spread within a combination. Apparently I didn't
do a good job of stating my goal.

Starting with the original premise of drawing five numbers without
replacement from the set of integers 1-52. If you got the combination
1-2-3-4-5, you'd probably think it a bit unusual that the numbers are packed
so closely together. Not that it indicates an un-random process or that
these combinations are less likely to occur than any others, just that they
look strangely compact.

If you drew the numbers 5-13-27-39-48, you'd probably think that they were
fairly evenly spread out over the set from which you drew, even though you
know statistically that this combination is no more or less likely to turn
up than is 1-2-3-4-5.

So what I was looking for was just a measure of relative spread between two
combinations. If I take the ratios of the standard deviation to the mean of
each of the two groups above, I get .527 for the "compact" group and .673
for the "dispersed" group. The "dispersed" ratio is only 28% greater than
the "compact" ratio. It just feels that this is not enough of a difference
in spread measurement between the two combinations so I wondered if there is
some generally accepted method of computing such a concept.

Paul
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Duncan Smith
science forum beginner


Joined: 29 Apr 2005
Posts: 21

PostPosted: Fri Apr 29, 2005 5:04 pm    Post subject: Re: Measuring Spread of Combinations Reply with quote

Pavel314 wrote:
Quote:
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114725261.938824.303720@l41g2000cwc.googlegroups.com...

This is NEIHER a probability or statistics problem based on the
occurrence
of a SINGLE combination.


This has been an interesting and informative thread; thanks to everyone for
their contributions.

However, what I was originally trying to get at is not a test of the
randomness of the draw or the odds of getting a specific combination but
just a way of measuring the spread within a combination. Apparently I didn't
do a good job of stating my goal.

Starting with the original premise of drawing five numbers without
replacement from the set of integers 1-52. If you got the combination
1-2-3-4-5, you'd probably think it a bit unusual that the numbers are packed
so closely together. Not that it indicates an un-random process or that
these combinations are less likely to occur than any others, just that they
look strangely compact.

If you drew the numbers 5-13-27-39-48, you'd probably think that they were
fairly evenly spread out over the set from which you drew, even though you
know statistically that this combination is no more or less likely to turn
up than is 1-2-3-4-5.

So what I was looking for was just a measure of relative spread between two
combinations. If I take the ratios of the standard deviation to the mean of
each of the two groups above, I get .527 for the "compact" group and .673
for the "dispersed" group. The "dispersed" ratio is only 28% greater than
the "compact" ratio. It just feels that this is not enough of a difference
in spread measurement between the two combinations so I wondered if there is
some generally accepted method of computing such a concept.

Paul


Why divide by the mean?

Duncan
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Pavel314
science forum addict


Joined: 29 Apr 2005
Posts: 78

PostPosted: Fri Apr 29, 2005 6:48 pm    Post subject: Re: Measuring Spread of Combinations Reply with quote

"Duncan Smith" <buzzard@urubu.freeserve.co.uk> wrote in message
news:d4u0d8$1df$1@newsg4.svr.pol.co.uk...
Quote:
Pavel314 wrote:
So what I was looking for was just a measure of relative spread between
two combinations. If I take the ratios of the standard deviation to the
mean of each of the two groups above, I get .527 for the "compact" group
and .673 for the "dispersed" group. The "dispersed" ratio is only 28%
greater than the "compact" ratio. It just feels that this is not enough
of a difference in spread measurement between the two combinations so I
wondered if there is some generally accepted method of computing such a
concept.

Paul


Why divide by the mean?

Duncan

I don't know why I did that. It doesn't make much sense, now that you
mention it. When I take the standard deviations of 1-2-3-4-5 and
5-13-27-39-48, I get 1.58 and 17.77, respectively. The second one is 11.24
times the first, which sure seems like a better description of the relative
compactness. Thanks.

Paul
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Stig Holmquist
science forum beginner


Joined: 30 Apr 2005
Posts: 48

PostPosted: Fri Apr 29, 2005 10:36 pm    Post subject: Re: Measuring Spread of Combinations Reply with quote

On Fri, 29 Apr 2005 07:03:47 -0400, "Pavel314" <Pavel314@comcast.net>
wrote:

Quote:
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114725261.938824.303720@l41g2000cwc.googlegroups.com...

This is NEIHER a probability or statistics problem based on the
occurrence
of a SINGLE combination.

This has been an interesting and informative thread; thanks to everyone for
their contributions.

However, what I was originally trying to get at is not a test of the
randomness of the draw or the odds of getting a specific combination but
just a way of measuring the spread within a combination. Apparently I didn't
do a good job of stating my goal.

Starting with the original premise of drawing five numbers without
replacement from the set of integers 1-52. If you got the combination
1-2-3-4-5, you'd probably think it a bit unusual that the numbers are packed
so closely together. Not that it indicates an un-random process or that
these combinations are less likely to occur than any others, just that they
look strangely compact.

If you drew the numbers 5-13-27-39-48, you'd probably think that they were
fairly evenly spread out over the set from which you drew, even though you
know statistically that this combination is no more or less likely to turn
up than is 1-2-3-4-5.

So what I was looking for was just a measure of relative spread between two
combinations. If I take the ratios of the standard deviation to the mean of
each of the two groups above, I get .527 for the "compact" group and .673
for the "dispersed" group. The "dispersed" ratio is only 28% greater than
the "compact" ratio. It just feels that this is not enough of a difference
in spread measurement between the two combinations so I wondered if there is
some generally accepted method of computing such a concept.

Paul

Please refer to my earlier post. You need to establish a tentative
mean std.dev. for all 310 draws and also its variance, from which you
will get secand std.dev. from this mean. You can then compare any

specific set of five to the tentative mean and determine how many
units it differs from the mean. Treat it as a regular normal curve.
If I had the means to do this I would do it myself. If you do the test
of all 310 draws, please let us know the mean and variance.

Good luck,

Bertil
Quote:



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Cuvelier Etienne
science forum beginner


Joined: 30 Apr 2005
Posts: 4

PostPosted: Sat Apr 30, 2005 6:07 am    Post subject: Re: Bivariate/Multivariate Joint PDF Reply with quote

"Sebastian Stern" <sebastianstern@wanadoo.nl> a écrit dans le message de
news:4273389f$1$10991$dbd45001@news.wanadoo.nl...
Quote:
I have several random variables, each of which has a univariate normal
marginal PDF. Now I would like to compute the bivariate (or multivariate)
_joint_ PDF. At first I thought it would always be normal too, but the
book
"Introduction to Bivariate and Multivariate Analysis" (Lindeman, Merenda
and
Gold) states (p. 45):

| If X and Y have a bivariate normal distribution, then the
| marginal distributions of of both X and Y are normal.
| However, if both X and Y have univariate normal _marginal_
| distributions, it is _not_ necessarily true that X and Y
| have a _bivariate_ normal distribution.

(The book also states that the same holds, mutatis mutandis, for
multivariate normal distributions.)

1. What other bivariate distributions are possible, given that their
univariate distributions are normal? (I cannot imagine what they would
look
like, but the quote says they are possible.)
The response is : Copula


A Copula is a Uniform multivariate distribution over [0,1]^n, and Abe Sklar
shows the following theorem :

Let H be an n-dimensional distribution function with margins
F1,...,Fn. Then there exists an n-copula C such that for
all x in R^n

H(x1,...,xn) = C(F1(x1),...,Fn(xn)).

If F1,...,Fn are all continuous, then C is unique; otherwise,
C is uniquely determined on Range of F1 X...X Range of Fn.

In fact the copula captures the dependence structure of the
distribution,and coupling
the margins for rebuild the mutivariate distribution.

In practice, it is in the bivariate case that it's work better now.
An interesting class of copulas is the Archimedean copulas : they have a
parameter
and you could find the best parameter for a given sample.

Cuvelier Etienne
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Stig Holmquist
science forum beginner


Joined: 30 Apr 2005
Posts: 48

PostPosted: Sat Apr 30, 2005 7:48 pm    Post subject: Re: Measuring Spread of Combinations Reply with quote

On Fri, 29 Apr 2005 20:36:50 -0400, Stig Holmquist
<stigfjorden@hotmail.com> wrote:

Quote:
On Fri, 29 Apr 2005 07:03:47 -0400, "Pavel314" <Pavel314@comcast.net
wrote:

"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114725261.938824.303720@l41g2000cwc.googlegroups.com...

This is NEIHER a probability or statistics problem based on the
occurrence
of a SINGLE combination.

This has been an interesting and informative thread; thanks to everyone for
their contributions.

However, what I was originally trying to get at is not a test of the
randomness of the draw or the odds of getting a specific combination but
just a way of measuring the spread within a combination. Apparently I didn't
do a good job of stating my goal.

Starting with the original premise of drawing five numbers without
replacement from the set of integers 1-52. If you got the combination
1-2-3-4-5, you'd probably think it a bit unusual that the numbers are packed
so closely together. Not that it indicates an un-random process or that
these combinations are less likely to occur than any others, just that they
look strangely compact.

If you drew the numbers 5-13-27-39-48, you'd probably think that they were
fairly evenly spread out over the set from which you drew, even though you
know statistically that this combination is no more or less likely to turn
up than is 1-2-3-4-5.

So what I was looking for was just a measure of relative spread between two
combinations. If I take the ratios of the standard deviation to the mean of
each of the two groups above, I get .527 for the "compact" group and .673
for the "dispersed" group. The "dispersed" ratio is only 28% greater than
the "compact" ratio. It just feels that this is not enough of a difference
in spread measurement between the two combinations so I wondered if there is
some generally accepted method of computing such a concept.

Paul

Please refer to my earlier post. You need to establish a tentative
mean std.dev. for all 310 draws and also its variance, from which you
will get secand std.dev. from this mean. You can then compare any
specific set of five to the tentative mean and determine how many
units it differs from the mean. Treat it as a regular normal curve.
If I had the means to do this I would do it myself. If you do the test
of all 310 draws, please let us know the mean and variance.

Good luck,

Bertil

My son just did a partial statistical study of the last 104 draws.

He found the s.d ranged from 5.57 to 22.66 with a mean of
15.53+/-3.55.
Also, the mean sum was 134.4 +/-29.5. There was no correlation
between sums and std.dev. because the highest possible sum 250
will have the lowest s.d. of 1.58.>>

Ideally, you should test at least 260 draws but better yet 2600 or
26000 draws, but they are not available for a long time if ever.
So you would need to simulate the draws by a computer program.

Note that na one of the very low s.d values below 4 came up nor
did any of the very high above 24.

Hope this gives you an idea of what is possible to do.

Stig Holmquist
Quote:

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