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I have found these sentences in my book:

Two additional formulas are used in the computation of the expected value of a function of X when X ≥ 0. Let X have pmf or pdf f(x) with support S and cdf F(x), and let G(x) be monotone where G(x) ≥ 0. If X is continuous we assume G(x) is differentiable with $\frac{d}{dx}G(x)=g(x)$ and that E{G(X)} exists; then using integration by parts, the expectation takes the form $$E[G(x)]=G(0)+ \int_{S}^{}g(x)[1-F(x)]dx$$

I tried to use integration by parts to derive the formula but fail. Could someone explain how to get the formula from $E[G(x)]=\int_{S}^{}G(x)f(x)dx$?

NoahC.
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    This is due to integration by parts (a la Lebesgue) There is a discussion on integration by parts here – Mittens Mar 29 '22 at 14:59
  • The comment above is of course correct but a quick way to see this is $\int g(x) P(X \geq x)dx = \int E g(x)1(X \geq x) = E \int^X_0 g(x) dx = EG(X) - EG(0)$. You will have to justify the second equality with the Fubini theorem. – Galton Mar 29 '22 at 15:05
  • See also https://math.stackexchange.com/questions/172841/explain-why-ex-int-0-infty-1-f-x-t-dt-for-every-nonnegative-rando – Math1000 Mar 29 '22 at 22:29

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