Re: Probability of picking a positive rational number at random

On Mar 17, 12:23 pm, S_Pa...@xxxxxxxxxxx wrote:
On Mar 17, 1:27 pm, quasi <qu...@xxxxxxxx> wrote:



On Mon, 17 Mar 2008 08:13:57 -0700 (PDT), S_Pa...@xxxxxxxxxxx wrote:
The question is if given a 'random rational r' in this way,

In _what_ way?

what is the probability that it is less than 1? How about < 2?

The rationals are a countable set, so there is no uniform distribution
on the rationals. There _are_ distributions on the rationals,
uncountably many of them. Specifying such a distribution is equivalent
to assigning a probability to _each_ rational (such that the
probabilities are nonnegative and the sum is 1).

Once you specify such a distribution, you can, in principle, answer
probability questions such as the ones you are asking, or, if a closed
form is too difficult, you can always do a simulation to approximate
the desired probabilities. But the choice of such a distribution is
_arbitrary_ (just so long as its a valid distribution), so none of
your questions has an answer which qualifies as universal. The answers
will _depend_ on the choice of distribution.

This has been spelled out to you many times. Why do you still not get
it?

quasi

I do get it. I fully understand that the question is meaningless
without a desnity function.
Here is the density function:

IntDensity=
2 3 5 7
0 1/2 2/3 4/5 6/7 ...
1 1/4 2/9 4/25 6/49 ...
2 1/8 2/27 4/125 6/343 ...
...


What do the row and column labels stand for? If your entries in row(i)
and column(j) are supposed to be probabilities, then this is not a
proper probability distribution. The entries sum to more than 1 (eg.,
1/2 + 2/3 > 1). Of course, discrete random variables do not possess
"densities" at all, but we can excuse your mi-use of the language
because it it not all that uncommon, even if incorrect. However,
having probabilities that sum to > 1 (seemingly, even to +infinity!)
is truly a blunder of major proportions.


So density of all prime factors in this random number <2 =
(1/2+1/4)*(2/3+2/9)*(4/5+4/25)...=(1-1/2^2)*(1-1/3^2)*(1-1/5^2)...=1/
zeta(2)=6/pi^2
Also, density of all prime factors <n = 1/zeta(n)
So obviously this density function yields a finite result when asked
this question. Is it arbitrary? Is it meaningless?

To speak about rationals one needs to say that there is an equay
probability of negative exponent as there is for positive exponent, so
i modify the density function for integers accordingly:
For each density given in integer density divide that density by 2 if
exponent>0
QDensity=
2 3 5 7
0 1/2 2/3 4/5 6/7 ...
-1 (1/4)/2 (2/9)/2 (4/25)/2 (6/49)/2 ...
+1 (1/4)/2 (2/9)/2 (4/25)/2 (6/49)/2 ...
-2
+2
...

Now, i can get the negative rationals by saying that the density of
negative rational picked in this way=density of positive rational.

Of course i have written simulators and i have posted the code.

I have determined that density(abs(random rational)<phi)=sqrt(phi)
experimentally, but i am not sure what the closed form for this is.

I am quite sure this density function converges for this question, but
i can not get the result. Is it meaningless? I guess that depends on
your philosophy.

It is not a question of Philosophy. There are some mathematical
properties that are just plain facts, not open to philosophical
disagreements. There are other properties that amount to agreed-upon
definitions, such as those related to probability. Then there are
other matters that arise logically out of definitions, such as
theorems about probabilities in various contexts. If your
"probabilities" have properties that violate the standards it means
that you must be using the word "probability" in a way that is
different from everybody else.

R.G. Vickson

.

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