probability question

Mononoke · #1 by **Mononoke** » Jan 25, 2012 7:47 am

I haven't done this stuff in a while so can someone help me answer a simple question.

Consider two sets of rationalnumbers Q=[0,1] & R=[2,4]. Assuming a uniform probability distribution for each set, what can we say about the probability of randomly picking a number p(Q), p(R) from either of the sets.

Is p(Q)<p(R) or p(Q)=p(R)?

Please give some mathematical justifications for this, i don't want a proof but siting some relevant theorems would be helpful.

Also what would happen if Q is a proper subset of R for example Q=[0,1] & R=[-1,2]

Thanks in advance
mono

home_ · #2 by **home_** » Jan 25, 2012 10:38 am

What does Q=[0,1] mean? Interval of all rational numbers between 0 and 1? (that is Q={a/b | a,b natural & a<b}) In this case uniform distribution doesn't exist, because measure of whole probability space would be greater than 1 (in fact it would be greater then any real number). Or did you mean just that Q and R are sets with two elements? (which is somehow trivial, but I'm not sure if this was the question)

CarlPierce · #3 by **CarlPierce** » Jan 25, 2012 10:49 am

The probabilities are the same that is both are zero given both sets have an infinite number of members.
Ditto your second example.

Mononoke · #4 by **Mononoke** » Jan 26, 2012 3:11 am

CarlPierce wrote:The probabilities are the same that is both are zero given both sets have an infinite number of members.
Ditto your second example.

That's what i thought as well, but do you anything in the text books that suggest this

twistor59 · #5 by **twistor59** » Jan 26, 2012 8:28 am

My reading of the question is:

I take the real interval Q = [0,1], I pick a random number in this inverval. I want to know the probability that this number is rational. Call this probability P(Q).

Ditto for the real interval R = [2,4], I pick a random number in this interval. I want to know the probability that this number is rational. Call this probability P(R).

Then the previous answers are saying that both probabilities are zero, since we're talking about countable sets. But I came across this reference, which talks about some sort of non standard probability measures (taking values in hyperreals ?) for which the answer would appear to be different. Have I understood this correctly ?

http://www.math.upenn.edu/~pemantle/Hypreals%5B1%5D.rtf

. Bernstein and Wattenberg (1969). In an early paper, B & W construct a measure that assigns a “probability” in *[0,1) to every subset of real [0,1), even those that are not Lebesgue-measurable. (E.g., Vitali sets get infinitesimal probability.) But its standard part is Lebesgue measure for every Lebesgue-measurable set, and finitely additive in general. Moreover, the basic measure has other very natural features: e.g., every nonempty set has nonzero probability, and the probability of the set of rationals in [0,1/4) is exactly half the probability of the set of rationals in [0,1/2).

Can someone briefly say what this is about ? Has this approach ever had any application ?

susu.exp · #6 by **susu.exp** » Jan 27, 2012 4:38 pm

twistor59 wrote:Can someone briefly say what this is about ? Has this approach ever had any application ?

The two things I get from a first glance:
a) That "measure" is not a measure. Any measure is a pre-measure and any pre-measure is sigma-additive. If that thing is merely additive, it can´t be a measure.
b) The lack of sigma-additivity in the general case would cause issues with various types of convergence used in probability theory, in particular alsmost sure convergence and convergence in distribution don´t work in that case. This in turn means that strong laws of large numbers and the CLT in it´s various forms don´t work either.

b in particular is a point for a very limited practical use of this.

twistor59 · #7 by **twistor59** » Jan 27, 2012 5:02 pm

susu.exp wrote:
twistor59 wrote:Can someone briefly say what this is about ? Has this approach ever had any application ?

The two things I get from a first glance:
a) That "measure" is not a measure. Any measure is a pre-measure and any pre-measure is sigma-additive. If that thing is merely additive, it can´t be a measure.
b) The lack of sigma-additivity in the general case would cause issues with various types of convergence used in probability theory, in particular alsmost sure convergence and convergence in distribution don´t work in that case. This in turn means that strong laws of large numbers and the CLT in it´s various forms don´t work either.

b in particular is a point for a very limited practical use of this.

Thanks.

Yes it reminds me of one of those "non-standard analysis" things where they try to treat infinitesimals rigorously.
The proliferation of double quotes gives the game away !

andrewk · #8 by **andrewk** » Mar 05, 2012 12:14 am

Mononoke wrote:
CarlPierce wrote:The probabilities are the same that is both are zero given both sets have an infinite number of members.
Ditto your second example.

That's what i thought as well, but do you anything in the text books that suggest this

The proof of why they are zero is that you have assumed a uniform probability distribution on each of the intervals, and that is defined to mean that the probability of drawing a number in any set is equal to the Lebesgue measure of that set, divided by the Lebesgue measure of the sample space.
The Lebesgue measure of a single point is zero, and the Lebesgue measures of your two sample spaces are 1 and 2 respectively, so the probabilities of drawing any particular number are 0/1 and 0/2, ie both 0.

Edited to add the following:

Mononoke wrote:Also what would happen if Q is a proper subset of R for example Q=[0,1] & R=[-1,2]

This would make no difference: both are still zero, for the reason explained above.

andrewk · #9 by **andrewk** » Mar 05, 2012 12:28 am

I just noticed the bit about it being only rationals, which I had missed in making the above post. If we are talking rationals, we can't use Lebesgue measure, as the Lebesgue measure of the rationals in any interval is zero.
This then puts the original question in the undefined category. The OP needs to clarify what it means by "Assuming a uniform probability distribution for each set".

Defining the probability distribution is equivalent to defining a measure. Once we have a measure, we can just calculate the probability as the measure of the event in question (in this case a set consisting of a single point) divided by the measure of the sample space.