Re: Probability of picking a positive rational number at random

On Thu, 13 Mar 2008 17:20:59 -0500, quasi <quasi@xxxxxxxx> wrote:

On Thu, 13 Mar 2008 15:03:57 -0700 (PDT), S_Paske@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
.

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