Steps of a Markov chain subordinated to a Poisson process












2












$begingroup$


Let





  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a time-homogeneous Markov chain on $(Omega,mathcal A,operatorname P)$

  • $D([0,1]):=left{f:[0,1]tomathbb Rmid ftext{ is càdlàg and left-continuous at }1right}$


  • $X^{(n)}$ be a $D([0,1])$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $$X^{(n)}_t=Y^{(n)}_{lfloor ntrfloor};;;text{for all }tin[0,1)$$


  • $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $ninmathbb N$ and $$Z^{(n)}_t:=begin{cases}Y^{(n)}_{N_{nt}}&text{for }tin[0,1)\ Z^{(n)}_{1-}end{cases}$$



I've read that first $(n-1)wedge N_{n-}$ steps of $X^{(n)}$ and $Z^{(n)}$ coincide, but appear at different times ($frac kn$ vs the $k$th jump time of $left(N_{nt}right)_{tge0}$). What's exactly meant and how can we prove it rigorously?




I've got a vague intuition what's meant: Assuming that $N$ is right-continuous (in practice, we can always find a right-continuous modification), $N$ is almost surely nondecreasing and makes almost surely jumps of size $1$. So, $tmapsto N_{nt}$ somehow behaves like $tmapstolfloor ntrfloor$, but the time-scale is stretched. How can we formulate this rigorously?



What's confusing me most is that it's written that only the first $(n-1)wedge N_{n-}$ steps coincide. Where does the $N_{n-}=lim_{tto n-}N_t$ come from? Shouldn't the first $n-1$ steps coincide?






probability-theory stochastic-processes markov-chains markov-process poisson-process







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$endgroup$

















    2












    $begingroup$


    Let





    • $(Omega,mathcal A,operatorname P)$ be a probability space


    • $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a time-homogeneous Markov chain on $(Omega,mathcal A,operatorname P)$

    • $D([0,1]):=left{f:[0,1]tomathbb Rmid ftext{ is càdlàg and left-continuous at }1right}$


    • $X^{(n)}$ be a $D([0,1])$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $$X^{(n)}_t=Y^{(n)}_{lfloor ntrfloor};;;text{for all }tin[0,1)$$


    • $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $ninmathbb N$ and $$Z^{(n)}_t:=begin{cases}Y^{(n)}_{N_{nt}}&text{for }tin[0,1)\ Z^{(n)}_{1-}end{cases}$$



    I've read that first $(n-1)wedge N_{n-}$ steps of $X^{(n)}$ and $Z^{(n)}$ coincide, but appear at different times ($frac kn$ vs the $k$th jump time of $left(N_{nt}right)_{tge0}$). What's exactly meant and how can we prove it rigorously?




    I've got a vague intuition what's meant: Assuming that $N$ is right-continuous (in practice, we can always find a right-continuous modification), $N$ is almost surely nondecreasing and makes almost surely jumps of size $1$. So, $tmapsto N_{nt}$ somehow behaves like $tmapstolfloor ntrfloor$, but the time-scale is stretched. How can we formulate this rigorously?



    What's confusing me most is that it's written that only the first $(n-1)wedge N_{n-}$ steps coincide. Where does the $N_{n-}=lim_{tto n-}N_t$ come from? Shouldn't the first $n-1$ steps coincide?






    probability-theory stochastic-processes markov-chains markov-process poisson-process







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a time-homogeneous Markov chain on $(Omega,mathcal A,operatorname P)$

      • $D([0,1]):=left{f:[0,1]tomathbb Rmid ftext{ is càdlàg and left-continuous at }1right}$


      • $X^{(n)}$ be a $D([0,1])$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $$X^{(n)}_t=Y^{(n)}_{lfloor ntrfloor};;;text{for all }tin[0,1)$$


      • $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $ninmathbb N$ and $$Z^{(n)}_t:=begin{cases}Y^{(n)}_{N_{nt}}&text{for }tin[0,1)\ Z^{(n)}_{1-}end{cases}$$



      I've read that first $(n-1)wedge N_{n-}$ steps of $X^{(n)}$ and $Z^{(n)}$ coincide, but appear at different times ($frac kn$ vs the $k$th jump time of $left(N_{nt}right)_{tge0}$). What's exactly meant and how can we prove it rigorously?




      I've got a vague intuition what's meant: Assuming that $N$ is right-continuous (in practice, we can always find a right-continuous modification), $N$ is almost surely nondecreasing and makes almost surely jumps of size $1$. So, $tmapsto N_{nt}$ somehow behaves like $tmapstolfloor ntrfloor$, but the time-scale is stretched. How can we formulate this rigorously?



      What's confusing me most is that it's written that only the first $(n-1)wedge N_{n-}$ steps coincide. Where does the $N_{n-}=lim_{tto n-}N_t$ come from? Shouldn't the first $n-1$ steps coincide?






      probability-theory stochastic-processes markov-chains markov-process poisson-process







      share|cite|improve this question











      $endgroup$




      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a time-homogeneous Markov chain on $(Omega,mathcal A,operatorname P)$

      • $D([0,1]):=left{f:[0,1]tomathbb Rmid ftext{ is càdlàg and left-continuous at }1right}$


      • $X^{(n)}$ be a $D([0,1])$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $$X^{(n)}_t=Y^{(n)}_{lfloor ntrfloor};;;text{for all }tin[0,1)$$


      • $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $ninmathbb N$ and $$Z^{(n)}_t:=begin{cases}Y^{(n)}_{N_{nt}}&text{for }tin[0,1)\ Z^{(n)}_{1-}end{cases}$$



      I've read that first $(n-1)wedge N_{n-}$ steps of $X^{(n)}$ and $Z^{(n)}$ coincide, but appear at different times ($frac kn$ vs the $k$th jump time of $left(N_{nt}right)_{tge0}$). What's exactly meant and how can we prove it rigorously?




      I've got a vague intuition what's meant: Assuming that $N$ is right-continuous (in practice, we can always find a right-continuous modification), $N$ is almost surely nondecreasing and makes almost surely jumps of size $1$. So, $tmapsto N_{nt}$ somehow behaves like $tmapstolfloor ntrfloor$, but the time-scale is stretched. How can we formulate this rigorously?



      What's confusing me most is that it's written that only the first $(n-1)wedge N_{n-}$ steps coincide. Where does the $N_{n-}=lim_{tto n-}N_t$ come from? Shouldn't the first $n-1$ steps coincide?






      probability-theory stochastic-processes markov-chains markov-process poisson-process




      probability-theory stochastic-processes markov-chains markov-process poisson-process






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 4 '18 at 20:14







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      asked Dec 4 '18 at 18:00









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          $begingroup$

          It's easy to see that $$X^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{n-1}right}$$ and $$Z^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{N_{n-}}right}$$ for all $ninmathbb N$. In order to make these processes coincide, we can apply a random time change $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ with $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ and $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ for $ninmathbb N$.






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            $begingroup$

            It's easy to see that $$X^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{n-1}right}$$ and $$Z^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{N_{n-}}right}$$ for all $ninmathbb N$. In order to make these processes coincide, we can apply a random time change $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ with $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ and $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ for $ninmathbb N$.






            share|cite|improve this answer









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              $begingroup$

              It's easy to see that $$X^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{n-1}right}$$ and $$Z^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{N_{n-}}right}$$ for all $ninmathbb N$. In order to make these processes coincide, we can apply a random time change $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ with $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ and $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ for $ninmathbb N$.






              share|cite|improve this answer









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                0












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                $begingroup$

                It's easy to see that $$X^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{n-1}right}$$ and $$Z^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{N_{n-}}right}$$ for all $ninmathbb N$. In order to make these processes coincide, we can apply a random time change $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ with $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ and $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ for $ninmathbb N$.






                share|cite|improve this answer









                $endgroup$



                It's easy to see that $$X^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{n-1}right}$$ and $$Z^{(n)}_{[0,:1)}=left{Y^{(n)}_0,ldots,Y^{(n)}_{N_{n-}}right}$$ for all $ninmathbb N$. In order to make these processes coincide, we can apply a random time change $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ with $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ and $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ for $ninmathbb N$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 20:17









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