Derivation of Markov's Inequality











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I have some problems understanding the derivation of Markov's inequality.



$EX = int_{-infty}^{infty} xf_X(x)text{d}x$



$= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.



$geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.



$geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.



I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?






probability inequality







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    up vote
    2
    down vote

    favorite












    I have some problems understanding the derivation of Markov's inequality.



    $EX = int_{-infty}^{infty} xf_X(x)text{d}x$



    $= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.



    $geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.



    $geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.



    I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?






    probability inequality







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have some problems understanding the derivation of Markov's inequality.



      $EX = int_{-infty}^{infty} xf_X(x)text{d}x$



      $= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.



      $geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.



      $geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.



      I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?






      probability inequality







      share|cite|improve this question















      I have some problems understanding the derivation of Markov's inequality.



      $EX = int_{-infty}^{infty} xf_X(x)text{d}x$



      $= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.



      $geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.



      $geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.



      I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?






      probability inequality




      probability inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Falrach

      1,545223




      1,545223










      asked 2 hours ago









      Christian

      347




      347






















          5 Answers
          5






          active

          oldest

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          up vote
          3
          down vote



          accepted










          We could be more verbose in the third step as follows: Let $a > 0$. Then
          $$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
          Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is were integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
          $$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$



          Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
          $$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$






          share|cite|improve this answer























          • More verbose $ne$ More precise.
            – Did
            2 hours ago










          • Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
            – Falrach
            2 hours ago










          • You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
            – Did
            1 hour ago






          • 1




            Thank you Falrach. Your longer explanation was very helpful for me.
            – Christian
            56 mins ago


















          up vote
          3
          down vote













          $ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.






          share|cite|improve this answer




























            up vote
            3
            down vote













            The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.






            share|cite|improve this answer






























              up vote
              2
              down vote













              The last step uses the fact that the function
              begin{align} g : [a,infty) rightarrow [0, infty)\
              x mapsto x f_X(x)
              end{align}
              is always larger than
              begin{align} h : [a,infty) rightarrow [0, infty)\
              x mapsto a f_X(x)
              end{align}
              for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.






              share|cite|improve this answer




























                up vote
                2
                down vote













                No, there is a difference. Second to third step uses the fact that



                $int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$



                And



                $int_0^a xf_X(x) dx>0$



                Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration



                Hope it is helpful






                share|cite|improve this answer





















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                  5 Answers
                  5






                  active

                  oldest

                  votes








                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes








                  up vote
                  3
                  down vote



                  accepted










                  We could be more verbose in the third step as follows: Let $a > 0$. Then
                  $$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
                  Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is were integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
                  $$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$



                  Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
                  $$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$






                  share|cite|improve this answer























                  • More verbose $ne$ More precise.
                    – Did
                    2 hours ago










                  • Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                    – Falrach
                    2 hours ago










                  • You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                    – Did
                    1 hour ago






                  • 1




                    Thank you Falrach. Your longer explanation was very helpful for me.
                    – Christian
                    56 mins ago















                  up vote
                  3
                  down vote



                  accepted










                  We could be more verbose in the third step as follows: Let $a > 0$. Then
                  $$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
                  Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is were integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
                  $$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$



                  Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
                  $$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$






                  share|cite|improve this answer























                  • More verbose $ne$ More precise.
                    – Did
                    2 hours ago










                  • Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                    – Falrach
                    2 hours ago










                  • You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                    – Did
                    1 hour ago






                  • 1




                    Thank you Falrach. Your longer explanation was very helpful for me.
                    – Christian
                    56 mins ago













                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  We could be more verbose in the third step as follows: Let $a > 0$. Then
                  $$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
                  Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is were integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
                  $$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$



                  Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
                  $$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$






                  share|cite|improve this answer














                  We could be more verbose in the third step as follows: Let $a > 0$. Then
                  $$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
                  Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is were integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
                  $$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$



                  Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
                  $$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 hours ago

























                  answered 2 hours ago









                  Falrach

                  1,545223




                  1,545223












                  • More verbose $ne$ More precise.
                    – Did
                    2 hours ago










                  • Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                    – Falrach
                    2 hours ago










                  • You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                    – Did
                    1 hour ago






                  • 1




                    Thank you Falrach. Your longer explanation was very helpful for me.
                    – Christian
                    56 mins ago


















                  • More verbose $ne$ More precise.
                    – Did
                    2 hours ago










                  • Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                    – Falrach
                    2 hours ago










                  • You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                    – Did
                    1 hour ago






                  • 1




                    Thank you Falrach. Your longer explanation was very helpful for me.
                    – Christian
                    56 mins ago
















                  More verbose $ne$ More precise.
                  – Did
                  2 hours ago




                  More verbose $ne$ More precise.
                  – Did
                  2 hours ago












                  Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                  – Falrach
                  2 hours ago




                  Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
                  – Falrach
                  2 hours ago












                  You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                  – Did
                  1 hour ago




                  You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
                  – Did
                  1 hour ago




                  1




                  1




                  Thank you Falrach. Your longer explanation was very helpful for me.
                  – Christian
                  56 mins ago




                  Thank you Falrach. Your longer explanation was very helpful for me.
                  – Christian
                  56 mins ago










                  up vote
                  3
                  down vote













                  $ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote













                    $ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.






                    share|cite|improve this answer























                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      $ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.






                      share|cite|improve this answer












                      $ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      Kavi Rama Murthy

                      46.5k31854




                      46.5k31854






















                          up vote
                          3
                          down vote













                          The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.






                          share|cite|improve this answer



























                            up vote
                            3
                            down vote













                            The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.






                            share|cite|improve this answer

























                              up vote
                              3
                              down vote










                              up vote
                              3
                              down vote









                              The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.






                              share|cite|improve this answer














                              The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 hours ago

























                              answered 2 hours ago









                              drhab

                              95.7k543126




                              95.7k543126






















                                  up vote
                                  2
                                  down vote













                                  The last step uses the fact that the function
                                  begin{align} g : [a,infty) rightarrow [0, infty)\
                                  x mapsto x f_X(x)
                                  end{align}
                                  is always larger than
                                  begin{align} h : [a,infty) rightarrow [0, infty)\
                                  x mapsto a f_X(x)
                                  end{align}
                                  for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.






                                  share|cite|improve this answer

























                                    up vote
                                    2
                                    down vote













                                    The last step uses the fact that the function
                                    begin{align} g : [a,infty) rightarrow [0, infty)\
                                    x mapsto x f_X(x)
                                    end{align}
                                    is always larger than
                                    begin{align} h : [a,infty) rightarrow [0, infty)\
                                    x mapsto a f_X(x)
                                    end{align}
                                    for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.






                                    share|cite|improve this answer























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      The last step uses the fact that the function
                                      begin{align} g : [a,infty) rightarrow [0, infty)\
                                      x mapsto x f_X(x)
                                      end{align}
                                      is always larger than
                                      begin{align} h : [a,infty) rightarrow [0, infty)\
                                      x mapsto a f_X(x)
                                      end{align}
                                      for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.






                                      share|cite|improve this answer












                                      The last step uses the fact that the function
                                      begin{align} g : [a,infty) rightarrow [0, infty)\
                                      x mapsto x f_X(x)
                                      end{align}
                                      is always larger than
                                      begin{align} h : [a,infty) rightarrow [0, infty)\
                                      x mapsto a f_X(x)
                                      end{align}
                                      for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 2 hours ago









                                      Jonathan

                                      16012




                                      16012






















                                          up vote
                                          2
                                          down vote













                                          No, there is a difference. Second to third step uses the fact that



                                          $int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$



                                          And



                                          $int_0^a xf_X(x) dx>0$



                                          Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration



                                          Hope it is helpful






                                          share|cite|improve this answer

























                                            up vote
                                            2
                                            down vote













                                            No, there is a difference. Second to third step uses the fact that



                                            $int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$



                                            And



                                            $int_0^a xf_X(x) dx>0$



                                            Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration



                                            Hope it is helpful






                                            share|cite|improve this answer























                                              up vote
                                              2
                                              down vote










                                              up vote
                                              2
                                              down vote









                                              No, there is a difference. Second to third step uses the fact that



                                              $int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$



                                              And



                                              $int_0^a xf_X(x) dx>0$



                                              Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration



                                              Hope it is helpful






                                              share|cite|improve this answer












                                              No, there is a difference. Second to third step uses the fact that



                                              $int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$



                                              And



                                              $int_0^a xf_X(x) dx>0$



                                              Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration



                                              Hope it is helpful







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 2 hours ago









                                              Martund

                                              1,122211




                                              1,122211






























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