Expectation of an infinite sum of indicator functions












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$begingroup$


Let $(X_n)$ be i.i.d. with $E|X_n| = infty$, $E$ being expected value. I'm trying to understand a proof that ${lim sup}_{n to infty} |X_n|/n = infty$.



In the proof the following inequality is used: for all $a in mathbb R$,



$$Eleft[sum_{n=1}^infty mathbb 1_{frac{|X_1|}{n} > a}right] geq Eleft[frac{|X_1|}{a} - 1right]$$



How should I understand this inequality? Where does it come from?






probability probability-theory expected-value







share|cite|improve this question









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    1












    $begingroup$


    Let $(X_n)$ be i.i.d. with $E|X_n| = infty$, $E$ being expected value. I'm trying to understand a proof that ${lim sup}_{n to infty} |X_n|/n = infty$.



    In the proof the following inequality is used: for all $a in mathbb R$,



    $$Eleft[sum_{n=1}^infty mathbb 1_{frac{|X_1|}{n} > a}right] geq Eleft[frac{|X_1|}{a} - 1right]$$



    How should I understand this inequality? Where does it come from?






    probability probability-theory expected-value







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $(X_n)$ be i.i.d. with $E|X_n| = infty$, $E$ being expected value. I'm trying to understand a proof that ${lim sup}_{n to infty} |X_n|/n = infty$.



      In the proof the following inequality is used: for all $a in mathbb R$,



      $$Eleft[sum_{n=1}^infty mathbb 1_{frac{|X_1|}{n} > a}right] geq Eleft[frac{|X_1|}{a} - 1right]$$



      How should I understand this inequality? Where does it come from?






      probability probability-theory expected-value







      share|cite|improve this question









      $endgroup$




      Let $(X_n)$ be i.i.d. with $E|X_n| = infty$, $E$ being expected value. I'm trying to understand a proof that ${lim sup}_{n to infty} |X_n|/n = infty$.



      In the proof the following inequality is used: for all $a in mathbb R$,



      $$Eleft[sum_{n=1}^infty mathbb 1_{frac{|X_1|}{n} > a}right] geq Eleft[frac{|X_1|}{a} - 1right]$$



      How should I understand this inequality? Where does it come from?






      probability probability-theory expected-value




      probability probability-theory expected-value






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 26 at 15:23









      D GD G

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          $begingroup$

          We observe that
          $$begin{eqnarray}
          sum_{n=1}^infty 1_{{|X_1|/n>a}}&=&sum_{n=1}^infty 1_{{n<|X_1|/a}}\
          &=&leftlceilfrac{|X_1|}{a}rightrceil-1\&ge& frac{|X_1|}{a}-1,
          end{eqnarray}$$           where $lceil xrceil$ denotes the least integer $geqslant x$. Therefore it follows
          $$
          Bbb Eleft[sum_{n=1}^infty 1_{{|X_1|/n>a}}right]ge Bbb Eleft[frac{|X_1|}{a}-1right].
          $$






          share|cite|improve this answer









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            $begingroup$

            We observe that
            $$begin{eqnarray}
            sum_{n=1}^infty 1_{{|X_1|/n>a}}&=&sum_{n=1}^infty 1_{{n<|X_1|/a}}\
            &=&leftlceilfrac{|X_1|}{a}rightrceil-1\&ge& frac{|X_1|}{a}-1,
            end{eqnarray}$$             where $lceil xrceil$ denotes the least integer $geqslant x$. Therefore it follows
            $$
            Bbb Eleft[sum_{n=1}^infty 1_{{|X_1|/n>a}}right]ge Bbb Eleft[frac{|X_1|}{a}-1right].
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We observe that
              $$begin{eqnarray}
              sum_{n=1}^infty 1_{{|X_1|/n>a}}&=&sum_{n=1}^infty 1_{{n<|X_1|/a}}\
              &=&leftlceilfrac{|X_1|}{a}rightrceil-1\&ge& frac{|X_1|}{a}-1,
              end{eqnarray}$$               where $lceil xrceil$ denotes the least integer $geqslant x$. Therefore it follows
              $$
              Bbb Eleft[sum_{n=1}^infty 1_{{|X_1|/n>a}}right]ge Bbb Eleft[frac{|X_1|}{a}-1right].
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We observe that
                $$begin{eqnarray}
                sum_{n=1}^infty 1_{{|X_1|/n>a}}&=&sum_{n=1}^infty 1_{{n<|X_1|/a}}\
                &=&leftlceilfrac{|X_1|}{a}rightrceil-1\&ge& frac{|X_1|}{a}-1,
                end{eqnarray}$$                 where $lceil xrceil$ denotes the least integer $geqslant x$. Therefore it follows
                $$
                Bbb Eleft[sum_{n=1}^infty 1_{{|X_1|/n>a}}right]ge Bbb Eleft[frac{|X_1|}{a}-1right].
                $$






                share|cite|improve this answer









                $endgroup$



                We observe that
                $$begin{eqnarray}
                sum_{n=1}^infty 1_{{|X_1|/n>a}}&=&sum_{n=1}^infty 1_{{n<|X_1|/a}}\
                &=&leftlceilfrac{|X_1|}{a}rightrceil-1\&ge& frac{|X_1|}{a}-1,
                end{eqnarray}$$                 where $lceil xrceil$ denotes the least integer $geqslant x$. Therefore it follows
                $$
                Bbb Eleft[sum_{n=1}^infty 1_{{|X_1|/n>a}}right]ge Bbb Eleft[frac{|X_1|}{a}-1right].
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 15:34









                SongSong

                18.5k21550




                18.5k21550






























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