real sequence and convergence in probability












1












$begingroup$


$X_n$ is a sequence of random variables.$X_n equiv a_n$, $a_n $ is a real sequence.

Then prove that $X_n $ converges in probability iff $a_n$ converges and then $X_n to lim_{ntoinfty} a_n$ in probability.



I have a feeling that the above statement is trivially true but I am not so sure. If anyone can prove otherwise,please do so.






convergence







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  • 1




    $begingroup$
    write down the definition of convergence in probability and see what you can do
    $endgroup$
    – mm-aops
    May 30 '14 at 11:45










  • $begingroup$
    Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
    $endgroup$
    – DanZimm
    May 30 '14 at 11:55










  • $begingroup$
    no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
    $endgroup$
    – kris91
    May 30 '14 at 14:07
















1












$begingroup$


$X_n$ is a sequence of random variables.$X_n equiv a_n$, $a_n $ is a real sequence.

Then prove that $X_n $ converges in probability iff $a_n$ converges and then $X_n to lim_{ntoinfty} a_n$ in probability.



I have a feeling that the above statement is trivially true but I am not so sure. If anyone can prove otherwise,please do so.






convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    write down the definition of convergence in probability and see what you can do
    $endgroup$
    – mm-aops
    May 30 '14 at 11:45










  • $begingroup$
    Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
    $endgroup$
    – DanZimm
    May 30 '14 at 11:55










  • $begingroup$
    no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
    $endgroup$
    – kris91
    May 30 '14 at 14:07














1












1








1





$begingroup$


$X_n$ is a sequence of random variables.$X_n equiv a_n$, $a_n $ is a real sequence.

Then prove that $X_n $ converges in probability iff $a_n$ converges and then $X_n to lim_{ntoinfty} a_n$ in probability.



I have a feeling that the above statement is trivially true but I am not so sure. If anyone can prove otherwise,please do so.






convergence







share|cite|improve this question









$endgroup$




$X_n$ is a sequence of random variables.$X_n equiv a_n$, $a_n $ is a real sequence.

Then prove that $X_n $ converges in probability iff $a_n$ converges and then $X_n to lim_{ntoinfty} a_n$ in probability.



I have a feeling that the above statement is trivially true but I am not so sure. If anyone can prove otherwise,please do so.






convergence




convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 30 '14 at 11:43









kris91kris91

186112




186112








  • 1




    $begingroup$
    write down the definition of convergence in probability and see what you can do
    $endgroup$
    – mm-aops
    May 30 '14 at 11:45










  • $begingroup$
    Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
    $endgroup$
    – DanZimm
    May 30 '14 at 11:55










  • $begingroup$
    no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
    $endgroup$
    – kris91
    May 30 '14 at 14:07














  • 1




    $begingroup$
    write down the definition of convergence in probability and see what you can do
    $endgroup$
    – mm-aops
    May 30 '14 at 11:45










  • $begingroup$
    Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
    $endgroup$
    – DanZimm
    May 30 '14 at 11:55










  • $begingroup$
    no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
    $endgroup$
    – kris91
    May 30 '14 at 14:07








1




1




$begingroup$
write down the definition of convergence in probability and see what you can do
$endgroup$
– mm-aops
May 30 '14 at 11:45




$begingroup$
write down the definition of convergence in probability and see what you can do
$endgroup$
– mm-aops
May 30 '14 at 11:45












$begingroup$
Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
$endgroup$
– DanZimm
May 30 '14 at 11:55




$begingroup$
Hrmm if you're working with a general measure (which is a generalization of a probability measure) then convergence in measure is not equivalent to convergence almost everywhere. Is there something I'm missing about this specific question that allows for those two things to be equivalent?
$endgroup$
– DanZimm
May 30 '14 at 11:55












$begingroup$
no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
$endgroup$
– kris91
May 30 '14 at 14:07




$begingroup$
no.This was all there was to the question.I did write down the definition of convergence in probability and I think that this statement is trivially true but the thing is nothing is said about the nature of convergence of real sequence.
$endgroup$
– kris91
May 30 '14 at 14:07










1 Answer
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0












$begingroup$

For any real number $a$ and $varepsilon > 0$,
$$
Pr{leftlvert X_n-arightrvertgt varepsilon }=
begin{cases}
1&mbox{ if }leftlvert a_n-arightrvertgt varepsilon;\
0&mbox{ if }leftlvert a_n-arightrvertleq varepsilon.
end{cases}
$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    active

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    active

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    0












    $begingroup$

    For any real number $a$ and $varepsilon > 0$,
    $$
    Pr{leftlvert X_n-arightrvertgt varepsilon }=
    begin{cases}
    1&mbox{ if }leftlvert a_n-arightrvertgt varepsilon;\
    0&mbox{ if }leftlvert a_n-arightrvertleq varepsilon.
    end{cases}
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For any real number $a$ and $varepsilon > 0$,
      $$
      Pr{leftlvert X_n-arightrvertgt varepsilon }=
      begin{cases}
      1&mbox{ if }leftlvert a_n-arightrvertgt varepsilon;\
      0&mbox{ if }leftlvert a_n-arightrvertleq varepsilon.
      end{cases}
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For any real number $a$ and $varepsilon > 0$,
        $$
        Pr{leftlvert X_n-arightrvertgt varepsilon }=
        begin{cases}
        1&mbox{ if }leftlvert a_n-arightrvertgt varepsilon;\
        0&mbox{ if }leftlvert a_n-arightrvertleq varepsilon.
        end{cases}
        $$






        share|cite|improve this answer









        $endgroup$



        For any real number $a$ and $varepsilon > 0$,
        $$
        Pr{leftlvert X_n-arightrvertgt varepsilon }=
        begin{cases}
        1&mbox{ if }leftlvert a_n-arightrvertgt varepsilon;\
        0&mbox{ if }leftlvert a_n-arightrvertleq varepsilon.
        end{cases}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 14:57









        Davide GiraudoDavide Giraudo

        128k17156268




        128k17156268






























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