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Regarding continuous stochastic variables and probability

  1. Nov 27, 2012 #1
    One of my math teachers discussed stochastic ("random") variables today. In an example, he discussed the probability of picking a random number n, such that [itex]n\inℝ[/itex], in the interval [0,10]. He proceeded to say that the probability of picking the integer 4 ([itex]n = 4[/itex]) is 0, supporting his claim with the statement that [itex]\frac{1}{\infty} = 0[/itex].

    Now, I'm well aware that I am being extremely picky with this, especially since this is a high school class. However, if one wanted to be pedantic, one might say the following:

    1. [itex]P(4\in\textbf{Z}) = \frac{1}{|N|}[/itex], where Z denotes the set of integers and |N| denotes the cardinality of the set N, where N is defined as the set of all real numbers in the interval [0,10].
    2. The value of [itex]P(4\in\textbf{Z})[/itex] is infinitesimal, but not exactly zero.
    3. Though the cardinality of the set N (defined in #1) is infinite, it is not truly correct to use the lemniscate (∞) to denote the quantity.

    Are these all correct statements? If so, which one(s) are incorrect?

    Thank you in advance for your verifications/corrections. :tongue2:
  2. jcsd
  3. Nov 27, 2012 #2


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    If one wanted to be pedantic, I would say that ##P(4\in\textbf{Z})## reads "the probability that the number 4 is in the set of integers", which is 1. =)

    That said, saying that the probability that a continuous random variable x takes an exact value is zero because "##1/\infty = 0##" is most certainly a very hand-wavey argument and is not a rigorous way of showing it. Dealing with an infinite outcome space definitely requires some careful terminology and statements ("almost surely", "almost never", etc.).
  4. Nov 27, 2012 #3

    Stephen Tashi

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    You'd have to find an axiom system that defines division of a real number by an infinite cardinal number in order to say that. In the ordinary axioms for the real number system, there is no such definition. (You might find some support for such a statement in the axioms for "nonstandard analysis", where "infinitesimals" are defined, but I'm not an expert on that.)

    Correct. [itex] \infty [/itex] isn't the conventional notation for the cardinality of the real numbers (That cardinality is the same as the cardinality of the real numbers in [0,10]).

    However, the fundamental problem with rigor of the teacher's argument is that probability (when it is treated rigorously) isn't defined as a ratio. It isn't defined as "the number of favorable cases/ the number of possible cases". It is defined as a "measure on a sigma algebra of sets".

    In practical terms, the fact that any single value of a continuous random variable has probability zero doesn't contradict experience. There are no practical ways to pick an exact value from a continuum of values. If you try to do it by a physical measurement such as the voltage across a noisy resistor, you can only measure the voltage to a finite precision. So there is a "plus or minus" to the measurement and you only know the voltage is in some interval. I don't think you can write a computer program to generate pseudo-random numbers from a continuum of numbers.

    If we imagine tossing an infinite number of coins, we could use each coin to select one digit of a binary number such as .01001110.... . and thus pick a random number between 0 and 1. But this is only a theoretical method. (And it would assign zero probability to each individual infinite sequence of coin tosses.)
  5. Nov 27, 2012 #4
    If I may drop my 2 cents, perhaps a way of grasping it is the following:

    Continuous random variables are described using the "density function" of its distribution, and the probability of an event is given by a definite integral of this density function.

    For example, the probability of a continuous random variable falling in the interval [a,b] would be expressed as[tex]\int_a^b p(x) dx[/tex]where [itex]p(x)[/itex] was the density function of your random distribution.

    Then it's easier to see that this integral goes to zero when the two integration limits are equal (when a=b).
  6. Nov 27, 2012 #5


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    The probability of picking any number out of the interval [0 10] is zero because the measure of a point is zero. It takes up zero percent of the length of the interval.

    But on the integers there is no length or measure that assigns equal weight to each integer. If you do define a measure on the integers - and the weights will not be equal - then the probability of choosing 4 might be non-zero.

    You could say I guess that the probability of choosing 4 is the limit if 1/n as n goes to infinity where n is the number of integers in a set containing 4 and the probability distribution is uniform.

    You can say I guess that the probability of picking a number out of the reals is zero since it is zero for any finite interval.
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