Why $mathbb P(X_t=Y_ttext{ for all }t)=1$ if $mathbb P(X_t=Y_t)=1$ for all $t$?












3












$begingroup$


Let $X_t$ and $Y_t$ two continuous stochastic process on $(Omega ,mathcal F,mathbb P)$ s.t. $X_t=Y_t$ a.s. for all $t$. Why $mathbb P{X_t=Y_ttext{ for all }t}=1$ ?



The solution goes as : We have that $X_r=Y_r$ a.s. for all $rinmathbb Q$. Therefore, $mathbb P{sup_{sinmathbb R}|X_s-Y_s|>0}=0$. The claim follow.



I really don't understand the argument. And for me $mathbb P(X_t=Y_t)=1$ for all $t$ is really the same as $mathbb P(X_t=Y_ttext{ for all }t)=1$. I don't see the difference.






probability measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
    $endgroup$
    – Calvin Khor
    Dec 3 '18 at 13:23


















3












$begingroup$


Let $X_t$ and $Y_t$ two continuous stochastic process on $(Omega ,mathcal F,mathbb P)$ s.t. $X_t=Y_t$ a.s. for all $t$. Why $mathbb P{X_t=Y_ttext{ for all }t}=1$ ?



The solution goes as : We have that $X_r=Y_r$ a.s. for all $rinmathbb Q$. Therefore, $mathbb P{sup_{sinmathbb R}|X_s-Y_s|>0}=0$. The claim follow.



I really don't understand the argument. And for me $mathbb P(X_t=Y_t)=1$ for all $t$ is really the same as $mathbb P(X_t=Y_ttext{ for all }t)=1$. I don't see the difference.






probability measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
    $endgroup$
    – Calvin Khor
    Dec 3 '18 at 13:23
















3












3








3





$begingroup$


Let $X_t$ and $Y_t$ two continuous stochastic process on $(Omega ,mathcal F,mathbb P)$ s.t. $X_t=Y_t$ a.s. for all $t$. Why $mathbb P{X_t=Y_ttext{ for all }t}=1$ ?



The solution goes as : We have that $X_r=Y_r$ a.s. for all $rinmathbb Q$. Therefore, $mathbb P{sup_{sinmathbb R}|X_s-Y_s|>0}=0$. The claim follow.



I really don't understand the argument. And for me $mathbb P(X_t=Y_t)=1$ for all $t$ is really the same as $mathbb P(X_t=Y_ttext{ for all }t)=1$. I don't see the difference.






probability measure-theory







share|cite|improve this question









$endgroup$




Let $X_t$ and $Y_t$ two continuous stochastic process on $(Omega ,mathcal F,mathbb P)$ s.t. $X_t=Y_t$ a.s. for all $t$. Why $mathbb P{X_t=Y_ttext{ for all }t}=1$ ?



The solution goes as : We have that $X_r=Y_r$ a.s. for all $rinmathbb Q$. Therefore, $mathbb P{sup_{sinmathbb R}|X_s-Y_s|>0}=0$. The claim follow.



I really don't understand the argument. And for me $mathbb P(X_t=Y_t)=1$ for all $t$ is really the same as $mathbb P(X_t=Y_ttext{ for all }t)=1$. I don't see the difference.






probability measure-theory




probability measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 13:12









NewMathNewMath

4059




4059












  • $begingroup$
    First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
    $endgroup$
    – Calvin Khor
    Dec 3 '18 at 13:23




















  • $begingroup$
    First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
    $endgroup$
    – Calvin Khor
    Dec 3 '18 at 13:23


















$begingroup$
First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
$endgroup$
– Calvin Khor
Dec 3 '18 at 13:23






$begingroup$
First one is for each $t$, ${ omega : X_t(omega) = Y_t(omega) }$, second one is $bigcap_{tin mathbb R} { omega : X_t(omega) = Y_t(omega) }$
$endgroup$
– Calvin Khor
Dec 3 '18 at 13:23












2 Answers
2






active

oldest

votes


















2












$begingroup$

The only thing you can say for $t$ fixed is $$mathbb P{X_t=Y_ttext{ for all }t}leq mathbb P{X_t=Y_t}.$$
You should pay more attention to the answers I gave you here. It's the same problem here.



What you must show is that there is $N$ with $mathbb P(N)=0$ s.t. for all $omega notin Omega $ and all $t$, $X_t(omega )=Y_t(omega ).$




  • If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $qinmathbb Q$. Therefore, for all $qinmathbb Q$, there is $N_q$ with $mathbb P(N_q)=0$ s.t. for all $omega notin N_q$ $X_q(omega )=Y_q(omega )$.


  • Set $N=bigcup_{qinmathbb Q}N_q$. It's s.t. $mathbb P(N)=0$ and for all $omega notin N$ and all $qinmathbb Q$, $X_q(omega )=Y_q(omega )$.


  • Let $tin mathbb R$ and let $(q_n)_n$ a sequence of $mathbb Q$ s.t. $q_nto t$. If $omega notin N$, then, by continuity
    $$X_t=lim_{nto infty }X_{q_n}=lim_{nto infty }Y_{q_n}=Y_t.$$



Therefore $mathbb P{X_t=Y_ttext{ for all $t$}}=1$.




For a counter example for the equality : Let $mathbb P$ the Lebesgue measure on $Omega =[0,1]$. Let $tin [0,1]$, $X_t=0$ for all $omega $ and $Y_t(omega )=boldsymbol 1_{{t}}(omega )$.



Then $mathbb P{X_t=Y_t}=1$ for all $t$, but $$mathbb P{X_t=Y_ttext{ for all }t}=0,$$
since there is no $omega $ s.t. $X_t(omega )=Y_t(omega )$ for all $t$ because for $omega in Omega $ fixed, when $t=omega $ then $X_t(omega )=0$ but $Y_t(omega )=1$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Since $X_t, Y_t$ are continuous, we have of course
    $$
    mathbb Pleft{sup_{sinmathbb R}|X_s-Y_s|=0right}
    =
    mathbb Pleft{sup_{sinmathbb Q}|X_s-Y_s|=0right}
    =
    mathbb Pleft(bigcap_{sinmathbb Q}{|X_s-Y_s|=0}right)
    $$
    a countable intersection of measurable sets of measure $1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
      $endgroup$
      – Did
      Dec 3 '18 at 14:06











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    2 Answers
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    2 Answers
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    oldest

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    2












    $begingroup$

    The only thing you can say for $t$ fixed is $$mathbb P{X_t=Y_ttext{ for all }t}leq mathbb P{X_t=Y_t}.$$
    You should pay more attention to the answers I gave you here. It's the same problem here.



    What you must show is that there is $N$ with $mathbb P(N)=0$ s.t. for all $omega notin Omega $ and all $t$, $X_t(omega )=Y_t(omega ).$




    • If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $qinmathbb Q$. Therefore, for all $qinmathbb Q$, there is $N_q$ with $mathbb P(N_q)=0$ s.t. for all $omega notin N_q$ $X_q(omega )=Y_q(omega )$.


    • Set $N=bigcup_{qinmathbb Q}N_q$. It's s.t. $mathbb P(N)=0$ and for all $omega notin N$ and all $qinmathbb Q$, $X_q(omega )=Y_q(omega )$.


    • Let $tin mathbb R$ and let $(q_n)_n$ a sequence of $mathbb Q$ s.t. $q_nto t$. If $omega notin N$, then, by continuity
      $$X_t=lim_{nto infty }X_{q_n}=lim_{nto infty }Y_{q_n}=Y_t.$$



    Therefore $mathbb P{X_t=Y_ttext{ for all $t$}}=1$.




    For a counter example for the equality : Let $mathbb P$ the Lebesgue measure on $Omega =[0,1]$. Let $tin [0,1]$, $X_t=0$ for all $omega $ and $Y_t(omega )=boldsymbol 1_{{t}}(omega )$.



    Then $mathbb P{X_t=Y_t}=1$ for all $t$, but $$mathbb P{X_t=Y_ttext{ for all }t}=0,$$
    since there is no $omega $ s.t. $X_t(omega )=Y_t(omega )$ for all $t$ because for $omega in Omega $ fixed, when $t=omega $ then $X_t(omega )=0$ but $Y_t(omega )=1$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      The only thing you can say for $t$ fixed is $$mathbb P{X_t=Y_ttext{ for all }t}leq mathbb P{X_t=Y_t}.$$
      You should pay more attention to the answers I gave you here. It's the same problem here.



      What you must show is that there is $N$ with $mathbb P(N)=0$ s.t. for all $omega notin Omega $ and all $t$, $X_t(omega )=Y_t(omega ).$




      • If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $qinmathbb Q$. Therefore, for all $qinmathbb Q$, there is $N_q$ with $mathbb P(N_q)=0$ s.t. for all $omega notin N_q$ $X_q(omega )=Y_q(omega )$.


      • Set $N=bigcup_{qinmathbb Q}N_q$. It's s.t. $mathbb P(N)=0$ and for all $omega notin N$ and all $qinmathbb Q$, $X_q(omega )=Y_q(omega )$.


      • Let $tin mathbb R$ and let $(q_n)_n$ a sequence of $mathbb Q$ s.t. $q_nto t$. If $omega notin N$, then, by continuity
        $$X_t=lim_{nto infty }X_{q_n}=lim_{nto infty }Y_{q_n}=Y_t.$$



      Therefore $mathbb P{X_t=Y_ttext{ for all $t$}}=1$.




      For a counter example for the equality : Let $mathbb P$ the Lebesgue measure on $Omega =[0,1]$. Let $tin [0,1]$, $X_t=0$ for all $omega $ and $Y_t(omega )=boldsymbol 1_{{t}}(omega )$.



      Then $mathbb P{X_t=Y_t}=1$ for all $t$, but $$mathbb P{X_t=Y_ttext{ for all }t}=0,$$
      since there is no $omega $ s.t. $X_t(omega )=Y_t(omega )$ for all $t$ because for $omega in Omega $ fixed, when $t=omega $ then $X_t(omega )=0$ but $Y_t(omega )=1$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        The only thing you can say for $t$ fixed is $$mathbb P{X_t=Y_ttext{ for all }t}leq mathbb P{X_t=Y_t}.$$
        You should pay more attention to the answers I gave you here. It's the same problem here.



        What you must show is that there is $N$ with $mathbb P(N)=0$ s.t. for all $omega notin Omega $ and all $t$, $X_t(omega )=Y_t(omega ).$




        • If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $qinmathbb Q$. Therefore, for all $qinmathbb Q$, there is $N_q$ with $mathbb P(N_q)=0$ s.t. for all $omega notin N_q$ $X_q(omega )=Y_q(omega )$.


        • Set $N=bigcup_{qinmathbb Q}N_q$. It's s.t. $mathbb P(N)=0$ and for all $omega notin N$ and all $qinmathbb Q$, $X_q(omega )=Y_q(omega )$.


        • Let $tin mathbb R$ and let $(q_n)_n$ a sequence of $mathbb Q$ s.t. $q_nto t$. If $omega notin N$, then, by continuity
          $$X_t=lim_{nto infty }X_{q_n}=lim_{nto infty }Y_{q_n}=Y_t.$$



        Therefore $mathbb P{X_t=Y_ttext{ for all $t$}}=1$.




        For a counter example for the equality : Let $mathbb P$ the Lebesgue measure on $Omega =[0,1]$. Let $tin [0,1]$, $X_t=0$ for all $omega $ and $Y_t(omega )=boldsymbol 1_{{t}}(omega )$.



        Then $mathbb P{X_t=Y_t}=1$ for all $t$, but $$mathbb P{X_t=Y_ttext{ for all }t}=0,$$
        since there is no $omega $ s.t. $X_t(omega )=Y_t(omega )$ for all $t$ because for $omega in Omega $ fixed, when $t=omega $ then $X_t(omega )=0$ but $Y_t(omega )=1$.






        share|cite|improve this answer











        $endgroup$



        The only thing you can say for $t$ fixed is $$mathbb P{X_t=Y_ttext{ for all }t}leq mathbb P{X_t=Y_t}.$$
        You should pay more attention to the answers I gave you here. It's the same problem here.



        What you must show is that there is $N$ with $mathbb P(N)=0$ s.t. for all $omega notin Omega $ and all $t$, $X_t(omega )=Y_t(omega ).$




        • If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $qinmathbb Q$. Therefore, for all $qinmathbb Q$, there is $N_q$ with $mathbb P(N_q)=0$ s.t. for all $omega notin N_q$ $X_q(omega )=Y_q(omega )$.


        • Set $N=bigcup_{qinmathbb Q}N_q$. It's s.t. $mathbb P(N)=0$ and for all $omega notin N$ and all $qinmathbb Q$, $X_q(omega )=Y_q(omega )$.


        • Let $tin mathbb R$ and let $(q_n)_n$ a sequence of $mathbb Q$ s.t. $q_nto t$. If $omega notin N$, then, by continuity
          $$X_t=lim_{nto infty }X_{q_n}=lim_{nto infty }Y_{q_n}=Y_t.$$



        Therefore $mathbb P{X_t=Y_ttext{ for all $t$}}=1$.




        For a counter example for the equality : Let $mathbb P$ the Lebesgue measure on $Omega =[0,1]$. Let $tin [0,1]$, $X_t=0$ for all $omega $ and $Y_t(omega )=boldsymbol 1_{{t}}(omega )$.



        Then $mathbb P{X_t=Y_t}=1$ for all $t$, but $$mathbb P{X_t=Y_ttext{ for all }t}=0,$$
        since there is no $omega $ s.t. $X_t(omega )=Y_t(omega )$ for all $t$ because for $omega in Omega $ fixed, when $t=omega $ then $X_t(omega )=0$ but $Y_t(omega )=1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 13:31

























        answered Dec 3 '18 at 13:24









        SurbSurb

        37.5k94375




        37.5k94375























            1












            $begingroup$

            Since $X_t, Y_t$ are continuous, we have of course
            $$
            mathbb Pleft{sup_{sinmathbb R}|X_s-Y_s|=0right}
            =
            mathbb Pleft{sup_{sinmathbb Q}|X_s-Y_s|=0right}
            =
            mathbb Pleft(bigcap_{sinmathbb Q}{|X_s-Y_s|=0}right)
            $$
            a countable intersection of measurable sets of measure $1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
              $endgroup$
              – Did
              Dec 3 '18 at 14:06
















            1












            $begingroup$

            Since $X_t, Y_t$ are continuous, we have of course
            $$
            mathbb Pleft{sup_{sinmathbb R}|X_s-Y_s|=0right}
            =
            mathbb Pleft{sup_{sinmathbb Q}|X_s-Y_s|=0right}
            =
            mathbb Pleft(bigcap_{sinmathbb Q}{|X_s-Y_s|=0}right)
            $$
            a countable intersection of measurable sets of measure $1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
              $endgroup$
              – Did
              Dec 3 '18 at 14:06














            1












            1








            1





            $begingroup$

            Since $X_t, Y_t$ are continuous, we have of course
            $$
            mathbb Pleft{sup_{sinmathbb R}|X_s-Y_s|=0right}
            =
            mathbb Pleft{sup_{sinmathbb Q}|X_s-Y_s|=0right}
            =
            mathbb Pleft(bigcap_{sinmathbb Q}{|X_s-Y_s|=0}right)
            $$
            a countable intersection of measurable sets of measure $1$.






            share|cite|improve this answer











            $endgroup$



            Since $X_t, Y_t$ are continuous, we have of course
            $$
            mathbb Pleft{sup_{sinmathbb R}|X_s-Y_s|=0right}
            =
            mathbb Pleft{sup_{sinmathbb Q}|X_s-Y_s|=0right}
            =
            mathbb Pleft(bigcap_{sinmathbb Q}{|X_s-Y_s|=0}right)
            $$
            a countable intersection of measurable sets of measure $1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 '18 at 14:03

























            answered Dec 3 '18 at 13:57









            GEdgarGEdgar

            62.2k267168




            62.2k267168












            • $begingroup$
              Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
              $endgroup$
              – Did
              Dec 3 '18 at 14:06


















            • $begingroup$
              Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
              $endgroup$
              – Did
              Dec 3 '18 at 14:06
















            $begingroup$
            Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
            $endgroup$
            – Did
            Dec 3 '18 at 14:06




            $begingroup$
            Actually, more is true, namely that, almost surely, $$sup_{sinmathbb R}|X_s-Y_s|=sup_{sinmathbb Q}|X_s-Y_s|$$
            $endgroup$
            – Did
            Dec 3 '18 at 14:06


















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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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