On the limit of a continuous combination of sequences of random variables converging in distribution











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Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.



Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?



If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?










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    Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.



    Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?



    If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?










    share|cite|improve this question
























      up vote
      1
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      favorite









      up vote
      1
      down vote

      favorite











      Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.



      Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?



      If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?










      share|cite|improve this question













      Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.



      Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?



      If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?







      probability-theory measure-theory probability-distributions random-variables






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      user521337

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          Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.






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            For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).



            If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.



            Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              active

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              up vote
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              accepted










              Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted










                Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.






                share|cite|improve this answer























                  up vote
                  1
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                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.






                  share|cite|improve this answer












                  Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.







                  share|cite|improve this answer












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                  answered yesterday









                  Kavi Rama Murthy

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                  40.7k31751






















                      up vote
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                      down vote













                      For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).



                      If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.



                      Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).



                        If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.



                        Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).



                          If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.



                          Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.






                          share|cite|improve this answer












                          For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).



                          If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.



                          Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          angryavian

                          36.9k13178




                          36.9k13178






























                               

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