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- Thread starter mruncleramos
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Hurkyl

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

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EnumaElish

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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).mruncleramos said:

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matt grime

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

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Hurkyl

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EnumaElish

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Can you give an example?Hurkyl said:

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Hurkyl

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

Now, for

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 [itex]P(R) = \int_R \, dx[/itex], so we know P(A)=0 and P(B)=1.

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EnumaElish

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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."

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Hurkyl

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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).

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

The probability that I get any specific number, like 1/2, is zero

1: well, there are technicalities...

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EnumaElish

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Okay, you may be right. It may take a little more to sink in.Hurkyl said:Incorrect. I get a number with probability 1. :tongue:

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HallsofIvy

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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

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EnumaElish

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That's what I realized yesterday, after Hurkyl's last post. Thanks, all.HallsofIvy said:EnumElish: there is a difference between "getting aspecificnumber" and "gettinganumber". If I "pick a number at random" then Ihaveto get a number!

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EnumaElish

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BTW, the integral you are describing is identical with the uniform probability distribution for 0

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matt grime

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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

[tex]\sum_n Md2^{-n}[/tex]=

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).

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HallsofIvy

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Measure Theory. That sounds pretty cool.

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EnumaElish

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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."matt grime said: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).

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matt grime

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and what was the point of that post?

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Hurkyl

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EnumaElish

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Do you mean "sets containing an uncountable number of points"?Hurkyl said:there are some uncountable sets of points

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EnumaElish

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You juggled finite vs. infinite, and then countable vs. uncountable. I was just paraphrasing you in a way that (I thought) is marginally more organized.matt grime said:and what was the point of that post?

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matt grime

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i "an uncountable set of poitnts" and "a set containing an uncountable number of poitns" are indeed synonymous though the former is perhaps better as it does not imply that we have defined what it means for a number to be uncountalby large.

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mruncleramos said:

As there are infinitely more irrational numbers than rational in this interval, the probabilily of getting a rational must be 0 while irrationals are obtained with prob = 1.

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matt grime

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"infintely more" isn't very rigorous is it?

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