Proving the Markov inequality for a non-negative random variable

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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 $$E[U]geq b P(Ugeq b),$$ where b is a constant?



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




probability-theory statistics random-variables




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    Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
    – Did
    Nov 3 '17 at 14:30















up vote
1
down vote

favorite












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



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




probability-theory statistics random-variables




share|cite|improve this question
















  • 1




    Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
    – Did
    Nov 3 '17 at 14:30













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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



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




probability-theory statistics random-variables




share|cite|improve this question












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



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





probability-theory statistics random-variables





share|cite|improve this question











share|cite|improve this question




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asked Nov 3 '17 at 13:09









polly9900

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212







  • 1




    Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
    – Did
    Nov 3 '17 at 14:30













  • 1




    Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
    – Did
    Nov 3 '17 at 14:30








1




1




Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
– Did
Nov 3 '17 at 14:30





Integrate the pointwise inequality $$Ugeqslant b,mathbf 1_Ugeqslant b$$ This is valid for every nonnegative random variable, even ones with no PDF.
– 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?



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



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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    Comment: Sketch of proof for a continuous random variable. Can you give a justification for each step?



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



    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.






    share|cite|improve this answer
























      up vote
      0
      down vote













      Comment: Sketch of proof for a continuous random variable. Can you give a justification for each step?



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



      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.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Comment: Sketch of proof for a continuous random variable. Can you give a justification for each step?



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



        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.






        share|cite|improve this answer












        Comment: Sketch of proof for a continuous random variable. Can you give a justification for each step?



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



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 4 '17 at 1:47









        BruceET

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