
#1
Apr2210, 08:12 AM

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](1F(t))dt Then if v is a nonpositive random variable, is its ev given by E(v)=[int](1F(v))dv ??? Hence, i got that the ev of a monotone increasing function g(x) is: E(g(x))=[int]g'(x)(1F(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)(1F(x))dx ??? any thoughts are highly appreciated! thanks alot! 



#2
Apr2210, 05:23 PM

Sci Advisor
HW Helper
P: 2,483

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 



#3
Apr2310, 05:46 AM

P: 8

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


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