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If U is a non-negative random variable and it has pdf $f_U(u)$, how can we prove the Markov inequality $$E[U]\geq b P(U\geq b),$$ where b is a constant?

Not sure how to prove this and haven't really gotten anywhere. Thanks for any help.

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    $\begingroup$ Integrate the pointwise inequality $$U\geqslant b\,\mathbf 1_{U\geqslant b}$$ This is valid for every nonnegative random variable, even ones with no PDF. $\endgroup$ – Did Nov 3 '17 at 14:30
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Comment: Sketch of proof for a continuous random variable. Can you give a justification for each step?

$$E(U) = \int_0^\infty\! xf(x)\,dx \ge \int_b^\infty\! xf(x)\,dx \ge \int_b^\infty\! bf(x)\,dx = b\int_b^\infty\! f(x)\,dx = bP(X \ge b). $$

The proof for a discrete random variable involves summations instead of integrals. Other types of integrals for a more general proof, if you know about them.

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