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Almost certain = 1?

  1. Jun 14, 2009 #1
    Why is "almost certain" := EXACTLY 1? Shouldn't it be 1 minus an infinitesimal (reciprocal of an infinite cardinal) for a dense space? (Why aren't surreal numbers acceptable?)
     
  2. jcsd
  3. Jun 15, 2009 #2

    Office_Shredder

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    Because the reciprocal of a cardinal has no meaning here...

    The probability of an event A occuring is defined by a measure function, which essentially is a way of measuring the area that A occupies with respect to W. The measure function is defined to take a real value between [0,1], you could pick other measure functions but then you're not following the definition of a probability.

    One standard example is if you have the interval [0,1] and you pick a number at random, what is the probability that you pick a rational number?

    We define the probability of picking a number inside of an interval (a,b) as b-a (and as [0,1] has length 1 we see this defines a probability). Then since the rationals are countable, we can list them [itex]q_1, q_2,...[/itex] and around [itex]q_i[/itex] we can construct an interval of length [itex] \frac{e}{2^n}[/itex] for any e>0. Then adding up all the lengths of the intervals gives us at least the probability that we pick a number inside one of those intervals (if there's overlap, then we overestimated, but that's OK). We have geometric series, so this comes out to a total probability of e. But e was arbitrary, so the probability of picking a rational number < e for all e>0, so it must be zero. Hence the probability of picking an irrational number is 1.

    So while it is possible that you pick a rational number, the probability is 0. A possible interpretation in practical terms is that even if you keep picking numbers at random, your chance of picking a rational number never increases, whereas if the probability was anything other than 0, your chance of eventually picking one would increase to 1 as you make more picks

    As an aside, notice that Q is dense in R but the probability was 0, not 1. Knowing a set is dense doesn't tell you anything about its probability
     
  4. Jun 15, 2009 #3

    CompuChip

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    Colloquially, "almost" can mean "with great (but non unity) probability", as in: "the probability that he will be late for the meeting is almost 100%".

    So I will assume that you mean "almost" in the mathematical sense here, which involves measure theory. Let X be the set of possible outcomes of an experiment, and we define a measure μ on X such that μ(X) = 1. Now we can define events as subsets A of X (note that there is implicitly a sigma algebra floating around here which I am not mentioning) and the probability of event A is then μ(A). Then by definition, we can say that A is almost certain to happen, when the probability that A is not going to happen is zero. That is, if the complement is a null set: μ(AC) = 0.

    Example. What is the probability that if we pick a random number from [0, 1], it will be irrational? Well, in most applications you would tacitly assume that [0, 1] is equipped with Lebesgue measure λ. Well, the set of rational numbers is a null set for this measure: λ(Q) = 0. So the probability we're interested in is λ(QC) = 1 - λ(Q) = 1.

    This definition makes sense, although it may be a little counter-intuitive at first. As you suggest, the most logical way to overcome this would be to work with infinitesimal numbers, but then we have to give up the idea that we can express probabilities as a percentage between 0 and 100 (i.e. a probability measure is a map from the sigma algebra to the interval [0, 1]).
     
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