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Expected value

by webbster
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webbster
#1
Apr22-10, 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](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!
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EnumaElish
#2
Apr22-10, 05:23 PM
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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
webbster
#3
Apr23-10, 05:46 AM
P: 8
Quote Quote by EnumaElish View Post
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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