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You absolutely can "pick a random real number between 0 and 1", at least for the sake of mathematical argument. But for an interval [a, b] a random real number

You absolutely can "pick a random real number between 0 and 1", at least for the sake of mathematical argument. But for an interval [a, b] a random real number will be x s.t. a <= x <= b is zero. That's because the set {x} has measure 0. An even more fun fact is that the probability that the randomly chosen number is rational is also zero, by the same argument. However, once one starts to try to describe a distribution over the reals, e.g.: "let X be a real number chosen from a uniform probability distribution over [0, 1]", then you run into serious problems. That example begs the question because there is no uniform probability distribution over that set.

thaumasiotes 8y ago
There definitely is a uniform probability distribution over the interval [0,1]. It is given by the probability density function f(x) = 1. There's no uniform probability distribution over the reals, or over the integers, but defining one over a closed real interval is trivially easy. That's what "pick a random real number between 0 and 1" means.
- AaronFriel 8y ago
  I don't believe there is a uniform probability distribution over any interval of the reals. Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero. The second axiom of probability states the probability of one of the numbers being chosen must be is one. The third axiom requires countable additivity. We can easily see that for our distribution, we cannot satisfy both at the same time. If we strive for the sum of probabilities to equal 1, then we end up with the sum(P(x)) = 1, but P(x) is everywhere zero. The countable additivity never enters into this, but for any uncountable set of events a uniform distribution is impossible.
  - thaumasiotes 8y ago
    > I don't believe there is a uniform probability distribution over any interval of the reals. Then I'll observe that you probably shouldn't be trying to comment on this topic. I didn't just assert that there is one, I pointed out to you what it was. The probability density function f(x) = 1, defined over the interval [0,1], is uniform (all values are equal, being 1) and covers an area of 1. That's all a uniform distribution is. It satisfies the axiom which I assume you're referring to as the "second axiom of probability" in that the definite integral of the pdf over the entire interval is 1. It satisfies the requirement you're confused about in that the probability of choosing a number that falls into either of two disjoint subintervals is equal to the sum of the probabilities of (1) choosing a number falling into the first subinterval; + (2) choosing a number falling into the second subinterval. > Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero. For fixed x, the probability of choosing x from a uniform distribution over the reals is 0. That is a special case of the probability that a value chosen from a uniform distribution over the reals will lie within an interval. The probability of drawing a value from within the interval [a,b], for any distribution, is the definite integral of the distribution's pdf from a to b. As you can see, when a = b, this value is 0, but when a ≠ b, it isn't. The probability of choosing an x such that 0 <= x <= 1 from a uniform distribution over [0,1] is 1, not 0.
    - AaronFriel 8y ago
      Oh gosh, you're correct. It's been a while since I've done this (well).
      - gustavpaul 8y ago
        Much respect for confirming to the rest of us reading this thread that your interlocutor's argument is the better one / yours is wrong. Having determined the correct answer, we all walk away correct, regardless of who was corrected.
  - maththrowawtays 8y ago
    You won't be able to find a countable union of sets of just one point which covers [0,1], so the third axiom you state is not violated. In fact, like the law says, if you take such a countable union, the measure of the set will be 0 (e.g. the rationals). If you're going to hold that the sum of measure for all points is 1, you're not going to be able to make any continuous distributions at all. The thing is, the assignment of measure does not work like this, as in you don't just add up the measure of all the points to get the measure of the whole set. Instead, measure is also assigned to subsets of [0,1] (in particular too all the sets in the sigma algebra of your choosing). There are some laws which prevent you from whatever probabilities to everything (such as the one you mentioned about countable unions), but there is nothing that says it has to be the sum of the probability of the points.

thaumasiotes 8y ago

There definitely is a uniform probability distribution over the interval [0,1]. It is given by the probability density function f(x) = 1. There's no uniform probability distribution over the reals, or over the integers, but defining one over a closed real interval is trivially easy. That's what "pick a random real number between 0 and 1" means.

AaronFriel 8y ago
I don't believe there is a uniform probability distribution over any interval of the reals. Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero. The second axiom of probability states the probability of one of the numbers being chosen must be is one. The third axiom requires countable additivity. We can easily see that for our distribution, we cannot satisfy both at the same time. If we strive for the sum of probabilities to equal 1, then we end up with the sum(P(x)) = 1, but P(x) is everywhere zero. The countable additivity never enters into this, but for any uncountable set of events a uniform distribution is impossible.
- thaumasiotes 8y ago
  > I don't believe there is a uniform probability distribution over any interval of the reals. Then I'll observe that you probably shouldn't be trying to comment on this topic. I didn't just assert that there is one, I pointed out to you what it was. The probability density function f(x) = 1, defined over the interval [0,1], is uniform (all values are equal, being 1) and covers an area of 1. That's all a uniform distribution is. It satisfies the axiom which I assume you're referring to as the "second axiom of probability" in that the definite integral of the pdf over the entire interval is 1. It satisfies the requirement you're confused about in that the probability of choosing a number that falls into either of two disjoint subintervals is equal to the sum of the probabilities of (1) choosing a number falling into the first subinterval; + (2) choosing a number falling into the second subinterval. > Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero. For fixed x, the probability of choosing x from a uniform distribution over the reals is 0. That is a special case of the probability that a value chosen from a uniform distribution over the reals will lie within an interval. The probability of drawing a value from within the interval [a,b], for any distribution, is the definite integral of the distribution's pdf from a to b. As you can see, when a = b, this value is 0, but when a ≠ b, it isn't. The probability of choosing an x such that 0 <= x <= 1 from a uniform distribution over [0,1] is 1, not 0.
  - AaronFriel 8y ago
    Oh gosh, you're correct. It's been a while since I've done this (well).
    - gustavpaul 8y ago
      Much respect for confirming to the rest of us reading this thread that your interlocutor's argument is the better one / yours is wrong. Having determined the correct answer, we all walk away correct, regardless of who was corrected.
- maththrowawtays 8y ago
  You won't be able to find a countable union of sets of just one point which covers [0,1], so the third axiom you state is not violated. In fact, like the law says, if you take such a countable union, the measure of the set will be 0 (e.g. the rationals). If you're going to hold that the sum of measure for all points is 1, you're not going to be able to make any continuous distributions at all. The thing is, the assignment of measure does not work like this, as in you don't just add up the measure of all the points to get the measure of the whole set. Instead, measure is also assigned to subsets of [0,1] (in particular too all the sets in the sigma algebra of your choosing). There are some laws which prevent you from whatever probabilities to everything (such as the one you mentioned about countable unions), but there is nothing that says it has to be the sum of the probability of the points.