Expected value and nonnegative random variable

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The discussion revolves around the expected value (EV) of monotone functions, specifically for nonnegative and nonpositive random variables. It establishes that the EV of a nonnegative random variable t is calculated using the formula E(t) = ∫(1-F(t))dt, while for a nonpositive random variable v, it is E(v) = -∫(1-F(v))dv. The conversation also explores the EV of a monotone increasing function g(x) and suggests that for a monotone decreasing function b(x), the EV can be expressed as E(b(x)) = -E(z(x)), where z(x) = -b(x). Clarification is sought regarding the relationship between the cumulative distribution function (CDF) F(.) and the probability density function (PDF) h(.), leading to a correction that F(.) should be H(.), representing the CDF corresponding to h(.). The discussion emphasizes the mathematical relationships in calculating expected values for different types of random variables.
webbster
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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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