Probability and the Real Line

1. Jul 22, 2005

mruncleramos

Take the Interval [0,1] over the reals. Randomnly choosing a number, what is the probability that you will get an irrational number? A rational one?

2. Jul 22, 2005

Hurkyl

Staff Emeritus
Is this homework? Have you had any thoughts about how to do this?

Anyways, you haven't given enough information: you can't really talk about randomness until you specify a probability distribution.

3. Jul 23, 2005

EnumaElish

With a continuous distribution, the probability of picking any one number at all is zero, regardless of it being rational or irrational. So it has to be discrete. Now the tough question is, how do you specify a discrete distribution over a nonatomic domain (a domain without a "smallest unit")? May not be possible. E.g. the pseudorandom generators that are programmed into statistical software are defined on rationals only (for example, they will never return the number "pi," just a rational approximation to an arbitrary yet finite number of digits).

4. Jul 23, 2005

matt grime

it is possible to give a discrete distribution for this, but the probability of picking almost all numbers is zero.

whilst we do not have such a discrete dist. we do commonly abuse notation and appeal to measure theory for the answer

5. Jul 23, 2005

Hurkyl

Staff Emeritus
You don't need a discrete distribution to ask about the probability of picking a rational number vs a nonrational number. You don't even need it to be discrete for one (or both) of those probabilities to be nonzero.

6. Jul 23, 2005

EnumaElish

Can you give an example?

7. Jul 23, 2005

Hurkyl

Staff Emeritus
Ack, I can't believe I said nonrational.

Let's set A = the set of all rationals in [0, 1], and B = the set of all irrationals in [0, 1].

Now, for ANY probability measure P, we have that P(A) + P(B) = P([0, 1]).

For example, if P is the uniform distribution on [0, 1], then we have:

P(A) + P(B) = 1

In particular, the uniform distribution is simply given by $P(R) = \int_R \, dx$, so we know P(A)=0 and P(B)=1.

8. Jul 23, 2005

EnumaElish

Sure, you can always get a set of numbers, but you'll never get a number from a continuous distribution (with positive probability, that is). Remember, the original quote said "randomly choosing a number."

Last edited: Jul 23, 2005
9. Jul 23, 2005

Hurkyl

Staff Emeritus
Incorrect. I get a number with probability 1. :tongue:

The probability that I get any specific number, like 1/2, is zero1, but that's not what the question asked.

1: well, there are technicalities...

Last edited: Jul 23, 2005
10. Jul 23, 2005

EnumaElish

Okay, you may be right. It may take a little more to sink in.

Last edited: Jul 23, 2005
11. Jul 24, 2005

HallsofIvy

Staff Emeritus
EnumElish: there is a difference between "getting a specific number" and "getting a number". If I "pick a number at random" then I have to get a number!

For example, suppose we pick a number with the uniform probability distribution on the interval [0, 1]. That means that every number is equally likely to be picked, the probability of picking a number in a given interval is equal to the length of that interval, and the probability of picking a number in a given measurable set is equal to the measure of that set.
In this case, the probability of picking any specific number (like $$\pi$$ or 1/2) is 0. The probability of getting some number is, of course, one. In fact, since the set of all irrational numbers between 0 and 1 has measure 1,the probability that the number chosen will be irrational is 1, the probability that the number chosen will be rational is 0 (which does not mean that it can't be rational! Probabilities of 0 and 1 in infinite spaces do not mean "impossible" or "certain".)

12. Jul 24, 2005

EnumaElish

That's what I realized yesterday, after Hurkyl's last post. Thanks, all.

13. Jul 25, 2005

mruncleramos

Wow, so many responses. I was just wondering and all. I have another question now. This is not homework, I was just thinking. Lets say we take a definite integral from 0 to 1 of an arbitrary real valued function, and answer is x. Clearly if we take away any random point from the interval 0 to 1, the integral will still equal to x. Same thing with 2, 3, 4, 5 etc. What happens if we take away all of the rational numbers? Will the integral still be the same? It seems to me as if it should. But when I think about taking away all irrational numbers....

14. Jul 25, 2005

EnumaElish

My guess is, you'll still end up with the same integral value. I guess this will follow from Cantor's taxonomy of infinities: uncountable infinity (irrationals) + countable infinity (rationals) = uncountable infinity (reals). Therefore, U.I. = U.I. - C.I.

BTW, the integral you are describing is identical with the uniform probability distribution for 0 < x < 1. Under that measure, Hurkyl has suggested that P(rationals) = 0 and P(irrationals) = 1. Obviously P(reals) = 1. This imples P(irrationals) = 1 = P(reals). When you remove all rationals you are removing a set with measure zero.

15. Jul 26, 2005

matt grime

Removign a countable set doesn't affect the integral. Countable sets have measure zero. Let us prove it using some measure theory (as i know no measure theory this is hand wavy). First let us suppose that the function f is bounded and positive , ie 0<= f(x) <M for x in [0,1] and some M.

Now take any countable set of points

x_1, x_2, x_3,...

we want to show that this countable set of points conttributes "nothing" to the integral. Let us estimate the contribution.

Let d be some real positive number. About the point x_n put a little open set of width d2^{-n}. Then the x_i's can contribute no more than

$$\sum_n Md2^{-n}$$=

that is they contribute less than the rectangle of width d_2^{-n} and height M to the integral.

Now we can work out that geometric sumand it is Md. but d was arbitrary so the contribution must be less than Md for any d in the positive reals (and is not negative) so it must be zero.
Yes this is surprisig and no it isn't rigorous.

Your idea of throwing away ione point at a time shows that we can remove any finite number of points. it doesn't then follow that we can remove an infinite number of points. We see that we can always remove a countable number of points, but not an uncountable number (we could remove all the points and that would certainly affect the integral).

16. Jul 26, 2005

HallsofIvy

Staff Emeritus
To make sense of that question you have to, as Matt Grime did, change from the Riemann integral to the Lebesque integral.

17. Jul 26, 2005

mruncleramos

Measure Theory. That sounds pretty cool.

18. Jul 26, 2005

EnumaElish

I think what you mean is "we can remove a finite, and at most a countably infinite, number of points and still get the same sum."

19. Jul 26, 2005

matt grime

and what was the point of that post?

20. Jul 26, 2005

Hurkyl

Staff Emeritus
For the sake of precision, I would like to point out that there are some uncountable sets of points we can remove without affecting the integral. (e.g. a Cantor set -- uncountable, yet with measure zero)