Characterization of convergence in probability












0












$begingroup$


I have the following proposition:




Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
$$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$




In the proof I have this equality



$$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$



Why is it true?






probability convergence







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have the following proposition:




    Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
    $$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$




    In the proof I have this equality



    $$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$



    Why is it true?






    probability convergence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following proposition:




      Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
      $$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$




      In the proof I have this equality



      $$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$



      Why is it true?






      probability convergence







      share|cite|improve this question









      $endgroup$




      I have the following proposition:




      Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
      $$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$




      In the proof I have this equality



      $$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$



      Why is it true?






      probability convergence




      probability convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 8:28









      Bremen000Bremen000

      429210




      429210






















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          $begingroup$

          For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.



          Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.






          share|cite|improve this answer









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            $begingroup$

            For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.



            Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.



              Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.



                Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.






                share|cite|improve this answer









                $endgroup$



                For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.



                Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 8:31









                Kavi Rama MurthyKavi Rama Murthy

                59k42161




                59k42161






























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