Functions and sequence of random variables

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Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.



$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$




I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.




probability convergence




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    Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
    – copper.hat
    Jul 29 at 23:45














up vote
0
down vote

favorite













Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.



$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$




I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.




probability convergence




share|cite|improve this question















  • 1




    Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
    – copper.hat
    Jul 29 at 23:45












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.



$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$




I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.




probability convergence




share|cite|improve this question












Let $X_n_n geq 1$ be a sequence of random variables. Find a bounded function $g$ that doesn't preserve convergence in probability.



$X_n xrightarrowP X$ however $g(X_n) notxrightarrowP g(X)$




I know by the Continuous Mapping Theorem that functions continuous at all $ X(omega): omega in Omega$ aren't an option. However, I'm also failing when proposing discontinuous functions. I tried some piecewise functions that take constant values $a$ and $b$ accordingly to the value that $X_n$ takes. I don't have measure-theoretic background. Thanks for the help in advance.





probability convergence





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asked Jul 29 at 22:56









Sergio Andrade

143112




143112







  • 1




    Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
    – copper.hat
    Jul 29 at 23:45












  • 1




    Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
    – copper.hat
    Jul 29 at 23:45







1




1




Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45




Let $X_n = 1 over n$ and $X=0$. Define $g(x)$ to be one if $x neq 0$ and zero otherwise.
– copper.hat
Jul 29 at 23:45










1 Answer
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Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.






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    up vote
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    down vote



    accepted










    Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.






        share|cite|improve this answer













        Let $g(x)=I_[0,infty)$, $X_n=-frac 1 n$, $X=0$. Then $X_n to X$ in probability but $g(X_n)=0$ and $g(X)=1$.







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        answered Jul 29 at 23:45









        Kavi Rama Murthy

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