Proof that when $n to infty$, then $P(A_{n}cap B_{n})to p$












1












$begingroup$


Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.



I know there is a theorem that says:



If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.



I am almost sure that I have to use this Theorem but I am not able to do this.



Any help?






probability-theory elementary-set-theory convergence







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.



    I know there is a theorem that says:



    If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.



    I am almost sure that I have to use this Theorem but I am not able to do this.



    Any help?






    probability-theory elementary-set-theory convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.



      I know there is a theorem that says:



      If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.



      I am almost sure that I have to use this Theorem but I am not able to do this.



      Any help?






      probability-theory elementary-set-theory convergence







      share|cite|improve this question











      $endgroup$




      Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) to 1$ and $P (B_{n})to p$, when $n to infty$ then $P(A_{n}cap B_{n})to p$.



      I know there is a theorem that says:



      If the sequence $(A_{n})_{n>1}$, where $A_{n} in mathbb{A}$, decrease to the empty set, $A_{n} supset A_{n + 1}$ for all n, then $P (A_{n}) to 0$ when $n to infty$.



      I am almost sure that I have to use this Theorem but I am not able to do this.



      Any help?






      probability-theory elementary-set-theory convergence




      probability-theory elementary-set-theory convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 21:09









      Davide Giraudo

      126k16150261




      126k16150261










      asked Jan 8 at 19:46









      LauraLaura

      2288




      2288






















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          $begingroup$

          Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
          $$
          mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
          $$
          hence
          $$
          mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
          $$
          The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?






          share|cite|improve this answer









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            $begingroup$

            Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
            $$
            mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
            $$
            hence
            $$
            mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
            $$
            The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
              $$
              mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
              $$
              hence
              $$
              mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
              $$
              The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
                $$
                mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
                $$
                hence
                $$
                mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
                $$
                The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?






                share|cite|improve this answer









                $endgroup$



                Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that
                $$
                mathbb Pleft(A_ncap B_nright)+mathbb Pleft(A_n^ccap B_nright)=mathbb Pleft(B_nright)
                $$
                hence
                $$
                mathbb Pleft(A_ncap B_nright)=mathbb Pleft(B_nright)-mathbb Pleft(A_n^ccap B_nright).
                $$
                The only remaining question is: what is $lim_{nto +infty}mathbb Pleft(A_n^ccap B_nright)$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 19:57









                Davide GiraudoDavide Giraudo

                126k16150261




                126k16150261






























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