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Issue in understanding stochastic ordering

  1. May 3, 2012 #1
    This is not actually a problem but I need clarification for stochastic ordering

    From Wikipedia, there is stated: (http://en.wikipedia.org/wiki/Stochastic_ordering)

    Usual stochastic order:
    a real random variable A is less than a random variable B in the "usual stochastic order" if
    Pr(A>x) ≤ Pr(B>x) ,where Pr(°) denotes the probability of an event.

    Here comes the issue that I do not understand:
    Characterizations:
    The following rules describe cases when one random variable is stochastically less than or equal to another. Strict version of some of these rules also exist.

    1. A ≤ B if and only if for all non-decreasing functions u, E[u(A)] ≤ E[u(B)].

    Why it is not possible to attain Pr(A>x) ≥ Pr(B>x) for some x even if E[u(A)] ≤ E[u(B)] ?
     
  2. jcsd
  3. May 3, 2012 #2

    Ray Vickson

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    The inequality E[u(A)] ≤ E[u(B)] is assumed to hold for ALL non-decreasing functions. For the special case where u(t) = 0 if t ≤ x and u(t) = 1 if t > x (a non-decreasing function) we have E[u(A)] = Pr(A>x), etc.

    Note: we can assume the seemingly-weaker hypothesis that E[u(A)] ≤ E[u(B)] holds for all strictly increasing u; then, by a limiting argument we can prove it for all non-decreasing u. Therefore, the special u above is in no sense "artificial".

    RGV
     
  4. Nov 22, 2012 #3
    Late gratitudes! I believe I get it.
     
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