Expected value and nonnegative random variable

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SUMMARY

The discussion centers on the expected value (EV) of monotone functions in relation to nonnegative and nonpositive random variables. The EV of a nonnegative random variable \( t \) is defined as \( E(t) = \int (1 - F(t)) dt \), while for a nonpositive random variable \( v \), it is expressed as \( E(v) = -\int (1 - F(v)) dv \). The participant confirms that for a monotone decreasing function \( b(x) \), the expected value can be calculated as \( E(b(x)) = -E(z(x)) \), where \( z(x) = -b(x) \) is monotone increasing. The cumulative distribution function (CDF) \( F(.) \) is clarified to be the cumulative function corresponding to the probability density function (PDF) \( h(.) \).

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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 a lot!
 
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