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A basic probability inequality

  1. Oct 22, 2009 #1
    I am confused on a basic probability inequality, could anyone help me on this:
    If X>Y>t, then is P(X>t) larger or smaller than P(Y>t)?

    Thanks
     
    Last edited: Oct 22, 2009
  2. jcsd
  3. Oct 22, 2009 #2

    HallsofIvy

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    Staff Emeritus
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    I am confused about this. You start by asserting that "X> Y> t". Given that, P(x> t) and P(Y> t) are both 1.0!

    If you just mean that X and Y are random variables, requiring that X> Y, and t is some number, then, yex, P(X> t) is greater than or equal to P(Y> t) (not necessarily larger- they might be equal). If Y> t, then X> t follows from X> Y. But X> t may be true even if Y> t is not.
     
  4. Oct 22, 2009 #3

    statdad

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    Homework Helper

    This may not be the point of the question about the r.vs, but: we say that the random variable [tex] Y [/tex] is stochastically larger than the random variable [tex] X [/tex] provided that

    [tex]
    P(Y > t) \ge P(X > t), \quad \forall t
    [/tex]

    This idea is one way of discussing two-sample location problems.
     
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