X is for continuous random variable and it's nonnegative. Then this is the formula.
$$E(X)=\int_0^\infty(1-F(x))dx$$
Does anyone know the proof? I appreciate any help.
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1Just integrate by part. (remember $\text{PDF}(x) = \frac{d}{dx}\text{CDF}(x)$ – achille hui Aug 02 '16 at 13:45
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1@achillehui A PDF might be lacking here. – drhab Aug 02 '16 at 13:46
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One of the most duplicated questions on the site? (And one needs no PDF, indeed.) – Did Aug 02 '16 at 14:22
2 Answers
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Hint:
Write RHS as $$\int_0^{\infty}\int_0^{\infty}1_{(x,\infty)}(y)dF(y)dx$$ and switch the order of integration.
drhab
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Provided there is a density function, you can do this:
$$\int_0^\infty (1-F(x)) \mathrm{d} x = \int_0^\infty \int_x^\infty f(y)\mathrm{d} y \mathrm{d} x = \int_0^\infty \int_0^y f(y) \mathrm{d}x\mathrm{d}y = \int_0^\infty y f(y)\mathrm{d}y = \mathrm{E}(X) $$
Otherwise look at drhab's answer.
Sanderr
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1$X$ has continuous distribution, but not necessarily absolutely continuous. It can be that no PDF exists. – drhab Aug 02 '16 at 13:49