Re: Probability of picking a positive rational number at random

On Thu, 13 Mar 2008 19:12:43 -0700 (PDT), Ray Vickson
<RGVickson@xxxxxxx> wrote:

On Mar 13, 5:34 pm, S_Pa...@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.

No.

Yes -- he's not saying the distribution is uniform.

If you read my prior reply, I already pointed out that a uniform
distribution on N is not possible -- I think the OP is now aware of
that.

In reply, the OP chooses a particular distribution, and then re-asks
the original question.

So the correct objection is not that the OP's distribution fails to be
uniform (since it _can't_ be), but rather, that there is no particular
reason to accept the OP's distribution as a natural, default,
distribution on N. It's just an arbitrary choice of distribution.

If you have a countably infinite set of values V = {v} (whether
integer or not) it is impossible to have a UNIFORM probability
distribution over V. In other words, if all p(v) were equal to each
other, you would be asking for an infinite number of equal values that
sum to 1. There are no such values.

I said that in the reply before the one you replied to, and thus, the
OP already knows this.

Of course, there are lots of NON-UNIFORM distributions on V;
any infinite set of values {p(v)} that are
=0 and sum to 1 will do.

Again, that just repeats what I said -- let me quote from my previous
reply ...

quasi wrote:

"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 apparently, you missed part of the thread.

quasi
.

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