Is there any formula for computing the expected value E(f(X)) using the cumulated distribution function (CDF), rather than the probability density function (PDF)?
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1Check this http://math.stackexchange.com/questions/172841/integral-of-cdf-equals-expected-value – m_gnacik Nov 16 '16 at 15:41
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1If $P(X\geqslant0)=1$, then $$E(f(X))=f(0)+\int_0^\infty f'(x)(1-F_X(x))dx$$ – Did Nov 16 '16 at 16:38
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thank you for the answer. what about P(X⩾0)<1, this is too restricting – stat Nov 17 '16 at 11:08
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2$$E(f(X))=f(0)+\int_0^{+\infty}f'(x)(1-F_X(x))dx-\int_{-\infty}^0f'(x)F_X(x)dx$$ – Did Nov 18 '16 at 10:41