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How should I intepret P(X<Y<Z)?

  1. Dec 8, 2013 #1
    For X, Y and Z are iid,
    I have actually seen many versions of solutions, and personally I can write out the multi-variable integration one:
    ∫∫∫f(x,y,z)dxdydz, for x from -inf to y, then y from -inf to z, and z for all values, generally.

    But from a book there is another method:
    P(X<Y<Z)=P(X<min(Y,Z))P(Y,Z), which is neat and easier to compute.

    May I know how would it be possible?
    Can I say P(X<Y<Z)=P(X<Y) and P(Y<Z)?
    And what I could come out with was P(X<Y<Z)=P(X<Y)P(Y<Z), which looks incorrect, may I know what is wrong in this?


    Many thanks.
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    haruspex

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    That makes no sense. Do you mean P(X<Y<Z)=P(X<min(Y,Z))P(Y<Z)?
    No, those would not be independent.
    The easiest is to realise that all six possible orderings are equally likely.
     
  4. Dec 8, 2013 #3
    Oh yes, I am very sorry for the typo.
    For the 6 possible orderings, did you mean X<Y, Y<Z, X<Z, Z<X, Z<Y, Y<X?
    Out of them, what can I do with that?
     
  5. Dec 8, 2013 #4

    Ray Vickson

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    P(X < min(Y,Z)) = P(X < Y < Z) + P(X < Z < Y). However, if the random variables are iid and continuous you can use symmetry to conclude that P(X < Y < Z) = (1/2)P(X < min(Y,Z)) and P(Y<Z) = 1/2. Thus, P(X<Y<Z)=P(X<min(Y,Z))P(Y,Z) for iid continuous random variables (but not if they have nonzero point probabilities at some places).
     
    Last edited: Dec 8, 2013
  6. Dec 8, 2013 #5

    D H

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    You are implicitly assuming P(X=Y)=P(X=Z)=P(Y=Z)=0 here. There are more than six possible orderings if that that isn't the case. That would mean a pathological probability distribution, which is probably beyond the scope of this question.

    No. He meant X<Y<Z is one of the possible orderings. X<Z<Y is another. There are six altogether (assuming a non-pathological distribution).
     
  7. Dec 8, 2013 #6
    Thanks to all for your inputs,
    for the P(X<min(Y,Z)),
    can I say that it is:
    P(X<min(Y,Z))=P(X<Y|Y<Z)+P(X<Z|Z<Y)?
    and also would P(X<Y<Z) be equal to P(X<Y,Y<Z)?
     
  8. Dec 8, 2013 #7

    D H

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    No!

    Yes.
     
  9. Dec 8, 2013 #8
    Thanks so much for your clarification, I have been spending so much time on all these seemingly very basic concepts, but still I have trouble getting them.

    Would there be a more efficient way in acquiring the sense in probabilistic calculations?
    Somehow it's hard for me to observe P(X<Y<Z) and then come up with all these..........
     
  10. Dec 8, 2013 #9

    D H

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    You can add two probabilities only if they are mutually exclusive events from the same sample space.

    In your previous post, you wanted to use P(X<Y|Y<Z)+P(X<Z|Z<Y). That's not valid because P(X<Y|Y<Z) and P(X<Z|Z<Y) represent rather different sample spaces. Adding those probabilities doesn't make sense.

    You can multiply two probabilities only if they are independent events. In your first post you asked about P(X<Y<Z)=P(X<Y)P(Y<Z). That's not valid because P(X<Y) and P(Y<Z) are not independent events.

    If the CDF is continuous there are six mutually exclusive events with non-zero probabilities. X<Y<Z is one of them. What are the others? Is there *any* reason to think that these six events have different probabilities?
     
  11. Dec 8, 2013 #10
    Thanks a lot for your statements, they are very clear and much better than any books I have read.
    And for the second concern, I understand that, after you all have stated the range of possible outcomes, how the equation of P(X<min(Y,Z) was computed; I was just sad about not being able to observe that at the beginning, as it seems to me very anti-intuitive..........just wondering how I should learn probability in order to acquire a sense like you all have.
     
  12. Dec 8, 2013 #11

    D H

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    The reasons you can use P(X<Y<Z) = P(X<min(Y,Z))*P(Y<Z) are that
    • X<Y<Z is equivalent to X<min(Y,Z) and Y<Z, and
    • Assuming a non-pathological distribution, X<min(Y,Z) and Y<Z are independent events, so you can multiply the probabilities P(X<min(Y,Z)) and P(Y<Z) to yield P(X<Y<Z).

    Note that another way to express X<Y<Z is X<Y and Y<Z. The problem here is that X<Y and Y<Z are not independent events. Given iid X,Y,Z, P(X<Y)=P(Y<Z)=1/2, so P(X<Y)*P(Y<Z)=1/4. This is not P(X<Y<Z).
     
    Last edited: Dec 8, 2013
  13. Dec 9, 2013 #12

    Ray Vickson

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    Even if the random variables are not continuous there are still six possible (strict) orderings, all having equal probability in the iid case. What will change is that each strict ordering no longer has probability 1/6, because there will be nonzero probabilities of equalities. For this to happen the random variables don't need to be "pathological"; there is nothing particularly pathological about binomial or Poisson or hypergeometric distributions, or even about mixed discrete-continuous distributions. Such distributions occur all the time in applications.
     
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