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Similar amusing math terms are "almost everywhere" and "almost surely"/"almost never". For example, if you pick a random real number between 0 and 1, it will al

Similar amusing math terms are "almost everywhere" and "almost surely"/"almost never". For example, if you pick a random real number between 0 and 1, it will almost never be exactly 0.5. It's possible that you'll get 0.5, but the probability of getting it is exactly 0. https://en.wikipedia.org/wiki/Almost_everywhere https://en.wikipedia.org/wiki/Almost_everywhere https://en.wikipedia.org/wiki/Almost_surely https://en.wikipedia.org/wiki/Almost_surely

perl4ever 8y ago
I think the proper interpretation of that is that you can't "pick a random real number between 0 and 1". After all, you would have to choose an infinite number of digits.
- mcphage 8y ago
  Well, you can do that if you have the Axiom of choice.
  - perl4ever 8y ago
    My brain is fuzzy tonight, and I'm not good at math, but something about the distinction between countable and noncountable infinities makes me question the axiom of choice. I can't precisely state the issue, but something seems off and inconsistent.
    - yantrams 8y ago
      A recent twitter thread that could be helpful - https://twitter.com/johncarlosbaez/status/1056200189130600449 https://twitter.com/johncarlosbaez/status/105620018913060044... And this from the same thread :) https://pbs.twimg.com/media/DqhjmPdU4AAb979.jpg https://pbs.twimg.com/media/DqhjmPdU4AAb979.jpg
- AaronFriel 8y ago
  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.
yantrams 8y ago
Remember coming across this concept while reading about Measure theory. There are also concepts like "almost well-ordered" and "almost complete" in graph theory if my memory serves right.
- anyfoo 8y ago
  Something applying to “almost all” elements in a (preferably infinite) set: Applies to all elements, with the exception of a finite number of elements. For example, almost all prime numbers are odd.
  - yantrams 8y ago
    Exactly! I can't remember where but I distinctly remember some theory quantifying this as almost-1, almost-2,... as a function of the number of finite (or countably infinite maybe) exceptions.
  - jhanschoo 8y ago
    A warning: in general mathematical usage, this is not the definition of "almost all"; the notion of "almost all" is with respect to some notion of measure on a set. In the set of real numbers, for example, with the usual Lebesgue measure on it, if all but a /countable/ number of elements has a property X, then almost all the reals has that property X. (Note that uncountable sets in R exists that have zero measure; i.e. uncountable sets that "almost no" points in R lie in that uncountable subset of R)
    - anyfoo 8y ago
      Ah. "With the exception of finitely many" [1] was the definition given to me in University. The generalization makes sense, however. [1] I don't think I'm doing a good job translating the original German definition I was given, which sounds much better: "Mit Ausnahme endlich vieler".
      - jhanschoo 8y ago
        Ah, in English "all but finitely many" is indeed very commonly used, and rarely is "almost all" defined to have that sense.

perl4ever 8y ago

I think the proper interpretation of that is that you can't "pick a random real number between 0 and 1". After all, you would have to choose an infinite number of digits.

mcphage 8y ago
Well, you can do that if you have the Axiom of choice.
- perl4ever 8y ago
  My brain is fuzzy tonight, and I'm not good at math, but something about the distinction between countable and noncountable infinities makes me question the axiom of choice. I can't precisely state the issue, but something seems off and inconsistent.
  - yantrams 8y ago
    A recent twitter thread that could be helpful - https://twitter.com/johncarlosbaez/status/1056200189130600449 https://twitter.com/johncarlosbaez/status/105620018913060044... And this from the same thread :) https://pbs.twimg.com/media/DqhjmPdU4AAb979.jpg https://pbs.twimg.com/media/DqhjmPdU4AAb979.jpg
AaronFriel 8y ago
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.

yantrams 8y ago

Remember coming across this concept while reading about Measure theory. There are also concepts like "almost well-ordered" and "almost complete" in graph theory if my memory serves right.

anyfoo 8y ago
Something applying to “almost all” elements in a (preferably infinite) set: Applies to all elements, with the exception of a finite number of elements. For example, almost all prime numbers are odd.
- yantrams 8y ago
  Exactly! I can't remember where but I distinctly remember some theory quantifying this as almost-1, almost-2,... as a function of the number of finite (or countably infinite maybe) exceptions.
- jhanschoo 8y ago
  A warning: in general mathematical usage, this is not the definition of "almost all"; the notion of "almost all" is with respect to some notion of measure on a set. In the set of real numbers, for example, with the usual Lebesgue measure on it, if all but a /countable/ number of elements has a property X, then almost all the reals has that property X. (Note that uncountable sets in R exists that have zero measure; i.e. uncountable sets that "almost no" points in R lie in that uncountable subset of R)
  - anyfoo 8y ago
    Ah. "With the exception of finitely many" [1] was the definition given to me in University. The generalization makes sense, however. [1] I don't think I'm doing a good job translating the original German definition I was given, which sounds much better: "Mit Ausnahme endlich vieler".
    - jhanschoo 8y ago
      Ah, in English "all but finitely many" is indeed very commonly used, and rarely is "almost all" defined to have that sense.