Limits of a density function












2












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If the limit of a density function exists does it the follow that it is zero? To put is formally



$$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$






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    2












    $begingroup$


    If the limit of a density function exists does it the follow that it is zero? To put is formally



    $$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$






    pdf







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      If the limit of a density function exists does it the follow that it is zero? To put is formally



      $$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$






      pdf







      share|cite|improve this question









      $endgroup$




      If the limit of a density function exists does it the follow that it is zero? To put is formally



      $$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$






      pdf




      pdf






      share|cite|improve this question













      share|cite|improve this question











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      asked 4 hours ago









      Jesper HybelJesper Hybel

      921614




      921614






















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          $begingroup$

          Yes.



          Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.



          But then:



          $$
          int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
          $$



          So $f$ cannot be a density function.






          share|cite|improve this answer











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            $begingroup$

            Yes.



            Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.



            But then:



            $$
            int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
            $$



            So $f$ cannot be a density function.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              Yes.



              Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.



              But then:



              $$
              int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
              $$



              So $f$ cannot be a density function.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                Yes.



                Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.



                But then:



                $$
                int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
                $$



                So $f$ cannot be a density function.






                share|cite|improve this answer











                $endgroup$



                Yes.



                Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.



                But then:



                $$
                int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
                $$



                So $f$ cannot be a density function.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 hours ago

























                answered 3 hours ago









                Matthew DruryMatthew Drury

                25.8k262104




                25.8k262104






























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