Re: Probability of picking a positive rational number at random

On Fri, 14 Mar 2008 00:58:54 -0500, quasi <quasi@xxxxxxxx> wrote:

On Thu, 13 Mar 2008 21:22:32 -0600, Virgil <Virgil@xxxxxxx> wrote:

In article <u0kjt3p1jou8bmk31eommlhab2hf0davu7@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

On Thu, 13 Mar 2008 17:34:18 -0700 (PDT), S_Paske@xxxxxxxxxxx wrote:

On Mar 13, 5:20 pm, 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.

quasi

I was thinking it was a valid distribution because if i summed the
probabilities for each possible whole number the sum would be 1.

The probability of picking a 1 would be:
p1=(1/2 * 2/3 * 4/5 * 6/7 ...)
The probability of picking a 2 = p(1)/2
The probability of picking a 3 = p(1)/3
The probability of picking a 4 = p(1)/4
...
Thus the sum of all the probalities is (p(1) * harmonic series)
But p(1)= (1 / harmonic series) according to the Euler Product

Yes, sorry -- I misinterpreted.

It does seem like a valid distribution, but as I pointed out, it's
just one of uncountably many possible such distributions, thus
somewhat arbitrary.

Any answer you get for the probability of a rational being in a
specified interval would, in general, not be valid for another such
distribution.

With a given distribution on N for numerator and denominator, you can
always do a simulation to find the approximate probability of getting
a rational in a given interval, with accuracy depending on the number
of trials. Finding closed forms those probabilities seems problematic,
but for some distributions, you might be able to do it.

quasi

For a countable set, such as formed by any non-empty interval of
rationals, it is not possible to "choose one at random", i.e.,have a
probability distribution in which each member has the same probability
as every other member.

I already pointed that out, and I think the OP is aware of that.

A "random natural number" is random only relative to a specific,
chosen distribution -- I said exactly that.

In reply, the OP suggested a particular distribution, which of course,
is not uniform, but still a valid distribution.

Well, it's not valid after all -- all the probabilities are 0.

quasi
.

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