Proof of a.s. uniqueness of probability convergence limit [closed]












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Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.



Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.






probability limits proof-verification







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closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.


















    1












    $begingroup$


    Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.



    Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.






    probability limits proof-verification







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      1












      1








      1


      0



      $begingroup$


      Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.



      Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.






      probability limits proof-verification







      share|cite|improve this question











      $endgroup$




      Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.



      Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n overset{P}{rightarrow} X, X_noverset{P}{rightarrow} Y$). Prove that $mathbb{P}(X=Y)=1$.






      probability limits proof-verification




      probability limits proof-verification






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 11:02







      J. Doe

















      asked Jan 4 at 19:07









      J. DoeJ. Doe

      1737




      1737




      closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser Jan 5 at 0:23


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, amWhy, Lee David Chung Lin, jgon, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






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          1












          $begingroup$

          Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
          $$
          \ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
          \ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
            $endgroup$
            – Lost1
            Jan 4 at 23:58




















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
          $$
          \ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
          \ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
            $endgroup$
            – Lost1
            Jan 4 at 23:58


















          1












          $begingroup$

          Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
          $$
          \ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
          \ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
            $endgroup$
            – Lost1
            Jan 4 at 23:58
















          1












          1








          1





          $begingroup$

          Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
          $$
          \ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
          \ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
          $$






          share|cite|improve this answer









          $endgroup$



          Let $varepsilon>0$. Let $omegainOmega$ (if there exists such $omega$) such that $|X^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus, for all $ninmathbb{N}$, definitely $|X^{-1}(omega)-X_n^{-1}(omega)|>varepsilon$ or $|X_n^{-1}(omega)-Y^{-1}(omega)|>varepsilon$. Thus,
          $$
          \ {|X-Y|>varepsilon}subseteq({|X_n-X|> {varepsilonover 3}}cup{|X_n-Y|>{varepsilonover 3}})
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>{1over n})leqmathbb{P}(|X_n-X|>{varepsilonover 3}cup|X_n-Y|leq{varepsilonover 3})
          \ leqmathbb{P}(|X_n-X|>{varepsilonover 3})+mathbb{P}(|X_n-Y|>{varepsilonover 3})underset{ntoinfty}{rightarrow} 0
          $$
          Thus,
          $$
          \ mathbb{P}(|X-Y|>0)=0Rightarrow mathbb{P}(X=Y)=1
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 19:07









          J. DoeJ. Doe

          1737




          1737








          • 1




            $begingroup$
            yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
            $endgroup$
            – Lost1
            Jan 4 at 23:58
















          • 1




            $begingroup$
            yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
            $endgroup$
            – Lost1
            Jan 4 at 23:58










          1




          1




          $begingroup$
          yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
          $endgroup$
          – Lost1
          Jan 4 at 23:58






          $begingroup$
          yes, as with many standard uniqueness proofs, it is the triangle inequality... this proof is correct
          $endgroup$
          – Lost1
          Jan 4 at 23:58





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