Hi All,(adsbygoogle = window.adsbygoogle || []).push({});

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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# Expected value

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