Proving an inequality of random variables












3












$begingroup$


I am currently reading a paper that claims the following fact:




Let $X$, $Y$ be $L^2$ random variables on some probability space. The $L^2$ norm is denoted by $| cdot |_{2}$. Then there exists $C>0$ such that
begin{eqnarray}
&& mathbb{E} bigg[ mathbf{1}_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| bigg] \
& leq & C |Y|_2 bigg( |Y|_2 + sqrt{|Y|_2} + sup_{A: mathbb{P}(A) leq |Y|_2} mathbb{E}[|X|^2 mathbf{1}_A]^{1/2} bigg).
end{eqnarray}




It seems to involve the Cauchy-Schwarz inequality, applying to the terms $ mathbf{1}_{{|Y| geq |Y|^{1/2}_2}}(1+|X|+|Y|)$ and $|Y|$. But I don't see how the $L^2$ norm of the first term can be bounded.






probability probability-theory lp-spaces







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$endgroup$

















    3












    $begingroup$


    I am currently reading a paper that claims the following fact:




    Let $X$, $Y$ be $L^2$ random variables on some probability space. The $L^2$ norm is denoted by $| cdot |_{2}$. Then there exists $C>0$ such that
    begin{eqnarray}
    && mathbb{E} bigg[ mathbf{1}_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| bigg] \
    & leq & C |Y|_2 bigg( |Y|_2 + sqrt{|Y|_2} + sup_{A: mathbb{P}(A) leq |Y|_2} mathbb{E}[|X|^2 mathbf{1}_A]^{1/2} bigg).
    end{eqnarray}




    It seems to involve the Cauchy-Schwarz inequality, applying to the terms $ mathbf{1}_{{|Y| geq |Y|^{1/2}_2}}(1+|X|+|Y|)$ and $|Y|$. But I don't see how the $L^2$ norm of the first term can be bounded.






    probability probability-theory lp-spaces







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I am currently reading a paper that claims the following fact:




      Let $X$, $Y$ be $L^2$ random variables on some probability space. The $L^2$ norm is denoted by $| cdot |_{2}$. Then there exists $C>0$ such that
      begin{eqnarray}
      && mathbb{E} bigg[ mathbf{1}_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| bigg] \
      & leq & C |Y|_2 bigg( |Y|_2 + sqrt{|Y|_2} + sup_{A: mathbb{P}(A) leq |Y|_2} mathbb{E}[|X|^2 mathbf{1}_A]^{1/2} bigg).
      end{eqnarray}




      It seems to involve the Cauchy-Schwarz inequality, applying to the terms $ mathbf{1}_{{|Y| geq |Y|^{1/2}_2}}(1+|X|+|Y|)$ and $|Y|$. But I don't see how the $L^2$ norm of the first term can be bounded.






      probability probability-theory lp-spaces







      share|cite|improve this question









      $endgroup$




      I am currently reading a paper that claims the following fact:




      Let $X$, $Y$ be $L^2$ random variables on some probability space. The $L^2$ norm is denoted by $| cdot |_{2}$. Then there exists $C>0$ such that
      begin{eqnarray}
      && mathbb{E} bigg[ mathbf{1}_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| bigg] \
      & leq & C |Y|_2 bigg( |Y|_2 + sqrt{|Y|_2} + sup_{A: mathbb{P}(A) leq |Y|_2} mathbb{E}[|X|^2 mathbf{1}_A]^{1/2} bigg).
      end{eqnarray}




      It seems to involve the Cauchy-Schwarz inequality, applying to the terms $ mathbf{1}_{{|Y| geq |Y|^{1/2}_2}}(1+|X|+|Y|)$ and $|Y|$. But I don't see how the $L^2$ norm of the first term can be bounded.






      probability probability-theory lp-spaces




      probability probability-theory lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 22 at 23:40









      RichardRichard

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      1,2211724






















          1 Answer
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          $begingroup$

          We can estimate each term separately. Observe the following:
          $$
          Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|right]le Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}frac{|Y|^2}{|Y|_2^{1/2}}right]le|Y|_2^{3/2},
          $$
          $$
          Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]le |Y|_2^2,
          $$



          $$
          Bbb P(|Y| geq |Y|^{1/2}_2)=Bbb P(|Y|^2 geq |Y|_2)le frac{Bbb E|Y|^2}{|Y|_2}=|Y|_2,
          $$           and
          $$begin{eqnarray}
          Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X||Y|right]^2&le& Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X|^2right]Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]\
          &le&sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]cdot |Y|^2_2
          end{eqnarray}$$           by Cauchy-Schwarz. This gives
          $$
          begin{eqnarray}
          mathbb{E} left[ 1_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| right] le |Y|_2left(|Y|_2+|Y|_2^{1/2}+left(sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]right)^{1/2}right)
          end{eqnarray}
          $$           as desired.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
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            active

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            2












            $begingroup$

            We can estimate each term separately. Observe the following:
            $$
            Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|right]le Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}frac{|Y|^2}{|Y|_2^{1/2}}right]le|Y|_2^{3/2},
            $$
            $$
            Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]le |Y|_2^2,
            $$



            $$
            Bbb P(|Y| geq |Y|^{1/2}_2)=Bbb P(|Y|^2 geq |Y|_2)le frac{Bbb E|Y|^2}{|Y|_2}=|Y|_2,
            $$             and
            $$begin{eqnarray}
            Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X||Y|right]^2&le& Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X|^2right]Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]\
            &le&sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]cdot |Y|^2_2
            end{eqnarray}$$             by Cauchy-Schwarz. This gives
            $$
            begin{eqnarray}
            mathbb{E} left[ 1_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| right] le |Y|_2left(|Y|_2+|Y|_2^{1/2}+left(sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]right)^{1/2}right)
            end{eqnarray}
            $$             as desired.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              We can estimate each term separately. Observe the following:
              $$
              Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|right]le Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}frac{|Y|^2}{|Y|_2^{1/2}}right]le|Y|_2^{3/2},
              $$
              $$
              Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]le |Y|_2^2,
              $$



              $$
              Bbb P(|Y| geq |Y|^{1/2}_2)=Bbb P(|Y|^2 geq |Y|_2)le frac{Bbb E|Y|^2}{|Y|_2}=|Y|_2,
              $$               and
              $$begin{eqnarray}
              Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X||Y|right]^2&le& Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X|^2right]Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]\
              &le&sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]cdot |Y|^2_2
              end{eqnarray}$$               by Cauchy-Schwarz. This gives
              $$
              begin{eqnarray}
              mathbb{E} left[ 1_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| right] le |Y|_2left(|Y|_2+|Y|_2^{1/2}+left(sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]right)^{1/2}right)
              end{eqnarray}
              $$               as desired.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                We can estimate each term separately. Observe the following:
                $$
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|right]le Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}frac{|Y|^2}{|Y|_2^{1/2}}right]le|Y|_2^{3/2},
                $$
                $$
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]le |Y|_2^2,
                $$



                $$
                Bbb P(|Y| geq |Y|^{1/2}_2)=Bbb P(|Y|^2 geq |Y|_2)le frac{Bbb E|Y|^2}{|Y|_2}=|Y|_2,
                $$                 and
                $$begin{eqnarray}
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X||Y|right]^2&le& Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X|^2right]Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]\
                &le&sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]cdot |Y|^2_2
                end{eqnarray}$$                 by Cauchy-Schwarz. This gives
                $$
                begin{eqnarray}
                mathbb{E} left[ 1_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| right] le |Y|_2left(|Y|_2+|Y|_2^{1/2}+left(sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]right)^{1/2}right)
                end{eqnarray}
                $$                 as desired.






                share|cite|improve this answer









                $endgroup$



                We can estimate each term separately. Observe the following:
                $$
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|right]le Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}frac{|Y|^2}{|Y|_2^{1/2}}right]le|Y|_2^{3/2},
                $$
                $$
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]le |Y|_2^2,
                $$



                $$
                Bbb P(|Y| geq |Y|^{1/2}_2)=Bbb P(|Y|^2 geq |Y|_2)le frac{Bbb E|Y|^2}{|Y|_2}=|Y|_2,
                $$                 and
                $$begin{eqnarray}
                Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X||Y|right]^2&le& Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|X|^2right]Bbb Eleft[1_{{ |Y| geq |Y|^{1/2}_2 }}|Y|^2right]\
                &le&sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]cdot |Y|^2_2
                end{eqnarray}$$                 by Cauchy-Schwarz. This gives
                $$
                begin{eqnarray}
                mathbb{E} left[ 1_{{ |Y| geq |Y|^{1/2}_2 }} (1+ |X| + |Y|)|Y| right] le |Y|_2left(|Y|_2+|Y|_2^{1/2}+left(sup_{A:Bbb P(A)le |Y|_2}Bbb E[X^2 1_A]right)^{1/2}right)
                end{eqnarray}
                $$                 as desired.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 0:21









                SongSong

                17.2k21145




                17.2k21145






























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