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I A simple question about probability theory

  1. Aug 2, 2017 #1
    Hello everyone!
    I'm studying the physics of complex systems and I'm approaching probability theory.
    I understand that we need a ## \sigma-algebra ## and the Kolmogorov axioms in order to define a probability space but then I bumped into the following relation:
    $$ p(A_1 \cup A_2 ) = p( A_1 ) + p( A_2 ) - p( A_1 \cap A_2 ) $$
    where ## A_1, A_2 ## are two sets of events that satisfy Kolmogorov axioms. I used all the properties i know to understand it but I wasn't able to demonstrate it. The book says that this equation comes from Kolmogorov axioms...
    Can you help me?
     
  2. jcsd
  3. Aug 2, 2017 #2

    Krylov

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    Can you show that if ##A, B## are events with ##B \subseteq A##, then ##P(A \setminus B) = P(A) - P(B)##?
    Using this, now write ##A_1 \cup A_2 = A_1 \cup A_2 \setminus (A_1 \cap A_2)##. The right-hand side is a disjoint union and ##A_1 \cap A_2 \subseteq A_2##. Now you are almost there.

    (In these cases, it helps to draw Venn diagrams.)
     
  4. Aug 2, 2017 #3
    I understand that if ## A_2 \subseteq A_1 ##, then the equation it's easy to understand, more or less. Unfortunately the book doesn't give me this hypothesis. :frown:
    Is it possible that the authors forgot about it or is this hypothesis not necessary?
     
  5. Aug 2, 2017 #4

    Krylov

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    No, ##A_1## and ##A_2## are arbitrary events. It is not necessary to assume that ##A_2 \subseteq A_1##, this is not what I meant.

    Do you agree with
    $$
    A_1 \cup A_2 = A_1 \cup A_2 \setminus (A_1 \cap A_2),
    $$
    and that this is a disjoint union? If so, then by Kolmogorov's axioms,
    $$
    P(A_1 \cup A_2) = P(A_1) + P(A_2 \setminus (A_1 \cap A_2)),
    $$
    right? Now look at the set inside the second probability on the right. We have ##(A_1 \cap A_2) \subseteq A_2## so, by what I wrote on the first line of my initial reply,
    $$
    P(A_2 \setminus (A_1 \cap A_2)) = P(A_2) - P(A_1 \cap A_2),
    $$
    which is what you need.
     
  6. Aug 2, 2017 #5
    Thank you very much!
    I only need the hypothesis in writing the difference, and obviously ## A_1 \cap A_2 \subseteq A_2 ##. I also get that ## A_1 \cap [ A_2 \setminus ( A_1 \cap A_2 ) ] = \emptyset ##.
    I still don't get the first relation that you wrote. :(
     
  7. Aug 2, 2017 #6

    Krylov

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    As you have seen, in these cases it is often useful to write a set as a disjoint union of other sets. Here the same happens: For ##A,B## events with ##B \subseteq A##,
    $$
    A = (A \setminus B) \cup B,
    $$
    and this is a disjoint union, so by Kolmogorov,
    $$
    P(A) = P(A \setminus B) + P(B)
    $$
    and therefore,
    $$
    P(A \setminus B) = P(A) - P(B).
    $$
    Another exercise in this spirit is to show that any countable union ##A_1 \cup A_2 \cup \ldots## of events can be written as a countable disjoint union.

    Another often used technique is splitting out an event: You want to calculate ##P(A)## for some event ##A## and you have a partition ##B_1,B_2,\ldots## of the underlying sample space. Then write ##A = (A \cap B_1) \cup (A \cap B_2) \cup \ldots## as a disjoint union. This comes in handy when it is somehow easier to calculate ##P(A \cap B_n)## for all ##n \in \mathbb{N}##.
     
  8. Aug 2, 2017 #7
    Sorry, I was ambiguous! I meant that I don't understand
    $$ A_1 \cup A_2 = A_1 \cup [ A_2 \setminus ( A_1 \cap A_2 ) ] $$
    It seems like ## A_2 = A_2 \setminus ( A_1 \cap A_2 ) ##. Thank you again. :)
     
  9. Aug 2, 2017 #8

    Krylov

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    By itself, the second equality is not true. However, the first equality is true.

    To see that the first one is true, try to prove two inclusions: Prove that if ##x \in A_1 \cup A_2##, then ##x \in A_1 \cup [ A_2 \setminus ( A_1 \cap A_2 )]## and, conversely, if ##x \in A_1 \cup [ A_2 \setminus ( A_1 \cap A_2 )]## then ##x \in A_1 \cup A_2##.

    If you make a drawing (draw the sets as two intersecting circles, this is called a Venn diagram), then you can see it more clearly.
     
  10. Aug 2, 2017 #9
    Ok, I made a drawing and I realized my mistake. :sorry:
    I will also try to demonstrate the two inclusions, thank you for the big help! :biggrin:
     
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