If $E[h(M_n)]to h(x)$ for all bounded and continuous functions $h$ then $M_nto x$ in probability












2












$begingroup$


Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
$$mathbb{E}[h(M_n)]to h(x)$$
as $ntoinfty$ for all bounded and continuous functions $h$.




How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?




What I thought:

Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.

Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?






probability-theory random-variables weak-convergence







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
    $$mathbb{E}[h(M_n)]to h(x)$$
    as $ntoinfty$ for all bounded and continuous functions $h$.




    How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?




    What I thought:

    Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.

    Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?






    probability-theory random-variables weak-convergence







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
      $$mathbb{E}[h(M_n)]to h(x)$$
      as $ntoinfty$ for all bounded and continuous functions $h$.




      How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?




      What I thought:

      Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.

      Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?






      probability-theory random-variables weak-convergence







      share|cite|improve this question











      $endgroup$




      Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
      $$mathbb{E}[h(M_n)]to h(x)$$
      as $ntoinfty$ for all bounded and continuous functions $h$.




      How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?




      What I thought:

      Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.

      Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?






      probability-theory random-variables weak-convergence




      probability-theory random-variables weak-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 18:34









      Did

      247k23222458




      247k23222458










      asked Dec 4 '18 at 9:51









      Yaroslav MlynárYaroslav Mlynár

      253




      253






















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.




          Direct proof. For any $delta>0$,
          begin{align}
          mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
          &le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
          end{align}
          where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
          $$
          1+h_{delta}(0)-h_{delta}(delta)=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
            $endgroup$
            – Yaroslav Mlynár
            Dec 4 '18 at 10:21













          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.




          Direct proof. For any $delta>0$,
          begin{align}
          mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
          &le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
          end{align}
          where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
          $$
          1+h_{delta}(0)-h_{delta}(delta)=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
            $endgroup$
            – Yaroslav Mlynár
            Dec 4 '18 at 10:21


















          3












          $begingroup$

          The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.




          Direct proof. For any $delta>0$,
          begin{align}
          mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
          &le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
          end{align}
          where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
          $$
          1+h_{delta}(0)-h_{delta}(delta)=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
            $endgroup$
            – Yaroslav Mlynár
            Dec 4 '18 at 10:21
















          3












          3








          3





          $begingroup$

          The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.




          Direct proof. For any $delta>0$,
          begin{align}
          mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
          &le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
          end{align}
          where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
          $$
          1+h_{delta}(0)-h_{delta}(delta)=0.
          $$






          share|cite|improve this answer











          $endgroup$



          The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.




          Direct proof. For any $delta>0$,
          begin{align}
          mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
          &le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
          end{align}
          where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
          $$
          1+h_{delta}(0)-h_{delta}(delta)=0.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 18:14

























          answered Dec 4 '18 at 10:06









          d.k.o.d.k.o.

          8,777628




          8,777628












          • $begingroup$
            But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
            $endgroup$
            – Yaroslav Mlynár
            Dec 4 '18 at 10:21




















          • $begingroup$
            But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
            $endgroup$
            – Yaroslav Mlynár
            Dec 4 '18 at 10:21


















          $begingroup$
          But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
          $endgroup$
          – Yaroslav Mlynár
          Dec 4 '18 at 10:21






          $begingroup$
          But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
          $endgroup$
          – Yaroslav Mlynár
          Dec 4 '18 at 10:21




















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