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Calculus and Probability

  1. Aug 28, 2012 #1
    In my probability class we were going over a method of visualizing probability as the area given two ranges of outcomes, for uncountable sample spaces.

    But if we measure the outcome of an exact value, we get a line segment, which of course has an area of 0. But that value is in no way impossible.

    Then I remembered that the sum of all events' probability is 1. So with an uncountably infinite amount of outcomes in a sample space, they must all sum to 1, yet when looking at one individually its probability is 0.

    It kind of paralleled integral calculus to me in that the probability of an exact value is analogous to a differential element of probability. Can anyone shed some light on this idea?
  2. jcsd
  3. Aug 28, 2012 #2
    So you reason that for all x in [0,1], we have [itex]P\{x\}=0[/itex]. So

    [tex]1=P([0,1])=P\left(\bigcup_{x\in [0,1]} \{x\}\right)=\sum_{x\in [0,1]} P \{x\} = 0[/tex]

    This is indeed a contradiction. But we made a mistake somewhere. Indeed, we did

    [tex]P\left(\bigcup_{x\in [0,1]} \{x\}\right)=\sum_{x\in [0,1]} P \{x\}[/tex]

    This is not valid since we are taking the union/sum over an uncountable set.

    The [itex]\sigma[/itex]-additivity states that if [itex](A_n)_n[/itex] is a countable collection, then we have

    [tex]P\left(\bigcup_n A_n\right)=\sum_n P(A_n)[/tex]

    But this does not hold anymore for uncountable collections, as your example shows.

    So while you can show that for every countable set [itex]A\subseteq [0,1][/itex], we have P(A)=0, we cannot do the same for uncountable sets.
  4. Aug 28, 2012 #3

    Stephen Tashi

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    Science Advisor

    Have you studied probability density functions yet? They are the functions that are integrated to get the probability of events, so I suppose they could be a "differential element", depending on what you mean by that. They are the function that appear inside the integrand.
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