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Statistical Physics

  1. Aug 14, 2015 #1
    • New member warned about posting with no effort shown
    How to prove this "if p(a)=p(b)=p then p(ab) ≤ p^2


    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Aug 14, 2015 #2

    RUber

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    Do you know anything about p, a, or b?
     
  4. Aug 14, 2015 #3

    RUber

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    Please provide some information about what you have already tried or methods you are familiar with so we can point you in the right direction--i.e. fill in the template.
     
  5. Aug 14, 2015 #4
    Probability of a = probability b = p
     
  6. Aug 14, 2015 #5

    RUber

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    What does p(ab) mean? Both a and b happen? What if a is b? then p(a and b) = p(a and a) = p.
     
  7. Aug 14, 2015 #6
    No Ruber

    if p(a) = p(b) = p (let say some value) then prove or disprove p(a ∩ b) ≤ p^2
     
  8. Aug 14, 2015 #7

    RUber

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    So do you know anything about a and b? Assume a = b, then p(a ∩ b) = p(a) = p ≥ p^2 .
     
  9. Aug 14, 2015 #8
    nothing mention disjoint or not i need to prove this is right or not
     
  10. Aug 14, 2015 #9
  11. Aug 14, 2015 #10

    RUber

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    Clearly if a and b were disjoint, the probability of a and b happening together is zero which will surely be less than p.
    The key here would be if a and b were independent. If they are, then you might have something to prove...otherwise, you just have 0 ≤ p(a ∩ b) ≤ p.
     
  12. Aug 14, 2015 #11
    yes i got the point if and be disjoint then p(ab) =0

    let say they are not disjoint then how to prove that ?
     
  13. Aug 14, 2015 #12

    RUber

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    There is nothing to prove unless you know they are independent.
     
  14. Aug 14, 2015 #13

    RUber

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    If the events a and b are independent, then, by the definition of independence, p(a ∩ b) = p(a) p(b) = p^2.
    If they are not independent, then like I said before, they can be anywhere from disjoint to completely coincident, i.e. 0≤p(a ∩ b)≤p.
    Is p ≤ p^2?
     
  15. Aug 14, 2015 #14
    thing is no any hints (information) given in the question.. okay what if not ?
     
  16. Aug 14, 2015 #15
    hmm if p ≤ p^2

    then p= 1 kw i dont think in that way
     
  17. Aug 14, 2015 #16

    RUber

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    Think of a Venn diagram with two circles representing a and b, both the same size (p). What is the maximum size of the overlapping region?
    If no other information is given in the question, then you can assume that anything is possible other than what you know to be true.
    If you are to prove the statement, you need to show it holds true all the time. If you are to disprove it, you just need one counterexample.
     
  18. Aug 14, 2015 #17

    RUber

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    So, if p can be any value between 0 and 1, you have to prove that the statement is true for all values of p, not just p=1.
    I don't think you will be able to prove it to be true without more constraints or assumptions.
    Can you prove that it is not true?
     
  19. Aug 14, 2015 #18
  20. Aug 14, 2015 #19

    RUber

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    If p(a∩b) = 0, then p(a∩b) ≤ p^2.
    That is not a good counterexample.
    Similarly, if p = 0, then p(a∩b) ≤ p^2. So, that's no good.
     
  21. Aug 14, 2015 #20
    so you trying to explain that p(null set) = 0 so its not good example is it ?
     
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