Re: Probability of picking a positive rational number at random

On Mar 13, 2:55 pm, quasi <qu...@xxxxxxxx> wrote:
On Thu, 13 Mar 2008 17:20:59 -0500, quasi <qu...@xxxxxxxx> wrote:
On Thu, 13 Mar 2008 15:03:57 -0700 (PDT), S_Pa...@xxxxxxxxxxx wrote:

I am thinking that if i picked a random integer n>=1 i could use this
distribution:
The probability of the exponent of its 2 prime factor=0 is 1/2
The probability of the exponent of its 2 prime factor=1 is 1/4
...
The probability of the exponent of its 3 prime factor=0 is 2/3
The probability of the exponent of its 3 prime factor=1 is 2/9
The probability of the exponent of its 3 prime factor=2 is 2/27
...
The probability of the exponent of its n'th prime factor=k is (Pn-1)/
Pn^k
Each prime factor would be independent of the others.

It's not a valid distribution since, for every natural number n, the
probability that the result of your proposed experiment yields the
value n is 0.

Any valid concept of "random" natural number amounts to choosing a
distribution on N. In other words, effectively it amounts to the
choice of a sequence

p_1, p_2, p_3, ...

of nonnegative real numbers such that

p_1 + p_2 + p_3 + ... = 1

Of course, the values of the sequence can't all be the same, hence
there is no such thing as a uniform distribution on N.

But clearly, there are uncountably many distinct non-uniform
distributions.

So suppose you choose one of them and use that distribution twice,
independently, once for the numerator, once for the denominator.

By symmetry, the probability of getting a rational greater than 1 is
the same as the probability of getting a rational less than 1,
although that common probability will be _less_ than 1/2, for the
simple reason that there is a positive probability of getting a
rational _equal_ to 1. Of course, the actual value of the common
probability will depend on the choice of p_1, p_2, p_3, ...

Similarly, by symmetry, the probability of getting a rational strictly
between 1/2 and 1 would equal the probability of getting a rational
strictly between 1 and 2, but once again, the actual value of that
common probability will depend on the choice of p_1, p_2, p_3, ...

Bottom line -- the question is a naive one. The answer depends
critically on the choice of distribution on N, and moreover, there is
no natural such choice which would qualify as a universal default.

quasi

A uniform probability distribution over the naturals would have that
the sum of a constant infinitely many times would equal one. Now,
that kind of notion is well reflected in that of arguments about the
differential, that the sum of infinitely many (constant) infinitesimal
width differential areas is a finite sum and exactly one. (The
inexistence of that constant leads to nonmeasurable sets, in for
example Vitali's argument.)

Arguments have been presented here as to why the existence of a
uniform distribution over R_[0,1] implies a uniform distribution over
N_[0, oo).

http://groups.google.com/group/sci.math/msg/2e346cb26bc92f7d

Asymptotically, a wide variety of distributions over the naturals go
flat to uniformity, just as many peak to Dirac's delta.

Imagine a distribution over the rationals by selecting an integer at
random (and from a population as large as to have the sample be
uniform with respect to features of the model) and then selecting a
rational, similarly uniformly, for example a censored uniformly random
real from [0,1], then the probability that said rational was in [0,1]
would be the same probability that the integer part was zero.

Now, those quantities would be equal, but infinitesimal, but positive,
and only commensurable to standard positive real quantities as being
less than them and standard zero as being greater.

ZF is inconsistent.

Whatever natural one there would be would fulfill all its properties.

Ross

--
Finlayson Consulting


.

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