Expected value

  1. Hi All,

    i got a short question concerning the ev of a monotone decreasing function.


    when i got a nonnegative random variable t, then its ev (with a continuous density h(.)) is given by
    E(t)=[int](1-F(t))dt
    Then if v is a nonpositive random variable, is its ev given by
    E(v)=-[int](1-F(v))dv
    ???
    Hence,
    i got that the ev of a monotone increasing function g(x) is:
    E(g(x))=[int]g'(x)(1-F(x))dx

    Now, let b(x) denote a monotone decreasing function. Therefore: z(x)=-b(x) is a monotone increasing function.
    Am I correct, that it got the ev of b(x) by
    E(b(x))=-E(z(x))
    and thus
    E(b(x))= - [int]z'(x)(1-F(x))dx
    ???

    any thoughts are highly appreciated!

    thanks alot!
     
  2. jcsd
  3. EnumaElish

    EnumaElish 2,483
    Science Advisor
    Homework Helper

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