# Expected value

by webbster
Tags: expected
 P: 8 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!
 HW Helper Sci Advisor P: 2,482 What is F(.)? How is it related to h(.)? Why not use E[v] = [int] v h(v) dv? http://en.wikipedia.org/wiki/Expecte...cal_definition
P: 8
 Quote by EnumaElish What is F(.)? How is it related to h(.)? Why not use E[v] = [int] v h(v) dv? http://en.wikipedia.org/wiki/Expecte...cal_definition
oh i am sorry... just a typo.

F(.) should be H(.) and represent the cdf to the pdf h(.)...

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