Ito's formula proof - why can we assume $u(t,omega), v(t,omega)$ are elementary?












1












$begingroup$


My question is about a simplification made in the proof of Ito's formula.



In the proof of Ito's formula in my textbook it says that since $int_0^T f dBs$ is defined as



$$int_0^T f dB_s = lim_{nto infty} int_0^T f_n dB_s$$



(with limit in probability) where $f_n$ is a sequence of step-functions such that



$$int_0^t (f-f_n)^2 ds to 0$$



in probability, then if $X_t$ is an Ito process of the form



$$X_t = X_0 + int_0^t u(s,omega)ds + int_0^t v(s,omega)dB_s$$



then to prove Ito's formula, we can assume $u$ and $v$ are elementary functions. ie, they have the form



$$f(s,omega) = sum_j e_j (omega) 1{t in [t_j, t_{j+1})}$$



So I understand that if $X_t$ is defined by



$$X_t^n = int_0^t f_n dB_s$$



where $f_n$ is a step function. Then clearly $f_n$ is an elementary function. So the fact we can assume $u,v$ are elementary is due the fact $X_t$ is the limit of integrals of step functions. Where I am confused is how this limit can be interchanged with the function $g in C^2$. That is, why is it the case that



$$g(X_t) = g(lim_n X_t^n) = lim_n g(X_t^n)$$



I suspect it is because:




  1. $u,v$ are assumed to be almost surely integrable (definition of Ito process)


  2. $g$ is a continuous mapping



And so it follows by some form of the continuous mapping theorem? But I am still unsure of the exact justification.



I am guessing this is the reason we need $g(t,x)$ to be twice continuously differentiable? So that we can take the limit out of all:



$$g, frac{partial g}{partial t}, frac{partial g}{partial x}, frac{partial^2 g}{partial x^2}$$



Using the continuous mapping theorem? Is this correct?






probability probability-theory stochastic-processes stochastic-calculus stochastic-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
    $endgroup$
    – AddSup
    Dec 19 '18 at 13:05










  • $begingroup$
    Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
    $endgroup$
    – Xiaomi
    Dec 19 '18 at 13:09
















1












$begingroup$


My question is about a simplification made in the proof of Ito's formula.



In the proof of Ito's formula in my textbook it says that since $int_0^T f dBs$ is defined as



$$int_0^T f dB_s = lim_{nto infty} int_0^T f_n dB_s$$



(with limit in probability) where $f_n$ is a sequence of step-functions such that



$$int_0^t (f-f_n)^2 ds to 0$$



in probability, then if $X_t$ is an Ito process of the form



$$X_t = X_0 + int_0^t u(s,omega)ds + int_0^t v(s,omega)dB_s$$



then to prove Ito's formula, we can assume $u$ and $v$ are elementary functions. ie, they have the form



$$f(s,omega) = sum_j e_j (omega) 1{t in [t_j, t_{j+1})}$$



So I understand that if $X_t$ is defined by



$$X_t^n = int_0^t f_n dB_s$$



where $f_n$ is a step function. Then clearly $f_n$ is an elementary function. So the fact we can assume $u,v$ are elementary is due the fact $X_t$ is the limit of integrals of step functions. Where I am confused is how this limit can be interchanged with the function $g in C^2$. That is, why is it the case that



$$g(X_t) = g(lim_n X_t^n) = lim_n g(X_t^n)$$



I suspect it is because:




  1. $u,v$ are assumed to be almost surely integrable (definition of Ito process)


  2. $g$ is a continuous mapping



And so it follows by some form of the continuous mapping theorem? But I am still unsure of the exact justification.



I am guessing this is the reason we need $g(t,x)$ to be twice continuously differentiable? So that we can take the limit out of all:



$$g, frac{partial g}{partial t}, frac{partial g}{partial x}, frac{partial^2 g}{partial x^2}$$



Using the continuous mapping theorem? Is this correct?






probability probability-theory stochastic-processes stochastic-calculus stochastic-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
    $endgroup$
    – AddSup
    Dec 19 '18 at 13:05










  • $begingroup$
    Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
    $endgroup$
    – Xiaomi
    Dec 19 '18 at 13:09














1












1








1





$begingroup$


My question is about a simplification made in the proof of Ito's formula.



In the proof of Ito's formula in my textbook it says that since $int_0^T f dBs$ is defined as



$$int_0^T f dB_s = lim_{nto infty} int_0^T f_n dB_s$$



(with limit in probability) where $f_n$ is a sequence of step-functions such that



$$int_0^t (f-f_n)^2 ds to 0$$



in probability, then if $X_t$ is an Ito process of the form



$$X_t = X_0 + int_0^t u(s,omega)ds + int_0^t v(s,omega)dB_s$$



then to prove Ito's formula, we can assume $u$ and $v$ are elementary functions. ie, they have the form



$$f(s,omega) = sum_j e_j (omega) 1{t in [t_j, t_{j+1})}$$



So I understand that if $X_t$ is defined by



$$X_t^n = int_0^t f_n dB_s$$



where $f_n$ is a step function. Then clearly $f_n$ is an elementary function. So the fact we can assume $u,v$ are elementary is due the fact $X_t$ is the limit of integrals of step functions. Where I am confused is how this limit can be interchanged with the function $g in C^2$. That is, why is it the case that



$$g(X_t) = g(lim_n X_t^n) = lim_n g(X_t^n)$$



I suspect it is because:




  1. $u,v$ are assumed to be almost surely integrable (definition of Ito process)


  2. $g$ is a continuous mapping



And so it follows by some form of the continuous mapping theorem? But I am still unsure of the exact justification.



I am guessing this is the reason we need $g(t,x)$ to be twice continuously differentiable? So that we can take the limit out of all:



$$g, frac{partial g}{partial t}, frac{partial g}{partial x}, frac{partial^2 g}{partial x^2}$$



Using the continuous mapping theorem? Is this correct?






probability probability-theory stochastic-processes stochastic-calculus stochastic-integrals







share|cite|improve this question











$endgroup$




My question is about a simplification made in the proof of Ito's formula.



In the proof of Ito's formula in my textbook it says that since $int_0^T f dBs$ is defined as



$$int_0^T f dB_s = lim_{nto infty} int_0^T f_n dB_s$$



(with limit in probability) where $f_n$ is a sequence of step-functions such that



$$int_0^t (f-f_n)^2 ds to 0$$



in probability, then if $X_t$ is an Ito process of the form



$$X_t = X_0 + int_0^t u(s,omega)ds + int_0^t v(s,omega)dB_s$$



then to prove Ito's formula, we can assume $u$ and $v$ are elementary functions. ie, they have the form



$$f(s,omega) = sum_j e_j (omega) 1{t in [t_j, t_{j+1})}$$



So I understand that if $X_t$ is defined by



$$X_t^n = int_0^t f_n dB_s$$



where $f_n$ is a step function. Then clearly $f_n$ is an elementary function. So the fact we can assume $u,v$ are elementary is due the fact $X_t$ is the limit of integrals of step functions. Where I am confused is how this limit can be interchanged with the function $g in C^2$. That is, why is it the case that



$$g(X_t) = g(lim_n X_t^n) = lim_n g(X_t^n)$$



I suspect it is because:




  1. $u,v$ are assumed to be almost surely integrable (definition of Ito process)


  2. $g$ is a continuous mapping



And so it follows by some form of the continuous mapping theorem? But I am still unsure of the exact justification.



I am guessing this is the reason we need $g(t,x)$ to be twice continuously differentiable? So that we can take the limit out of all:



$$g, frac{partial g}{partial t}, frac{partial g}{partial x}, frac{partial^2 g}{partial x^2}$$



Using the continuous mapping theorem? Is this correct?






probability probability-theory stochastic-processes stochastic-calculus stochastic-integrals




probability probability-theory stochastic-processes stochastic-calculus stochastic-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 10:03







Xiaomi

















asked Dec 19 '18 at 9:44









XiaomiXiaomi

1,066115




1,066115












  • $begingroup$
    Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
    $endgroup$
    – AddSup
    Dec 19 '18 at 13:05










  • $begingroup$
    Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
    $endgroup$
    – Xiaomi
    Dec 19 '18 at 13:09


















  • $begingroup$
    Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
    $endgroup$
    – AddSup
    Dec 19 '18 at 13:05










  • $begingroup$
    Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
    $endgroup$
    – Xiaomi
    Dec 19 '18 at 13:09
















$begingroup$
Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
$endgroup$
– AddSup
Dec 19 '18 at 13:05




$begingroup$
Hi Xiaomi, does your textbook define stochastic integrals as limits in probability (as opposed to limits in $L^2$)?
$endgroup$
– AddSup
Dec 19 '18 at 13:05












$begingroup$
Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
$endgroup$
– Xiaomi
Dec 19 '18 at 13:09




$begingroup$
Yes. Limit in $L^2$ is used for the introduction, but then switched to limit in probability to relax the assumptions on the class of functions.
$endgroup$
– Xiaomi
Dec 19 '18 at 13:09










1 Answer
1






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oldest

votes


















1












$begingroup$

If $X_t $ is of the form
$$
X_t = int_0^t f(s,omega)dB_s(omega),quad tgeq 0,
$$ then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes ${f_n}$ such that
$$
E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0
$$ for all $T>0$. Let $X_{n,t} = int_0^t f_n(s,omega)dB_s(omega)$. Then by martingale maximal inequality, we have
$$
E[sup_{tin [0,T]}|X_t -X_{n,t}|^2]leq C E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0.
$$ This implies that $X_{n.t} to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $gin C^2_0$ via localization method. Then,
$$
g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
$$ converge in probability to
$$
g(X_t), g'(X_t), g''(X_t)
$$ locally uniformly.






share|cite|improve this answer









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    $begingroup$

    If $X_t $ is of the form
    $$
    X_t = int_0^t f(s,omega)dB_s(omega),quad tgeq 0,
    $$     then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes ${f_n}$ such that
    $$
    E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0
    $$     for all $T>0$. Let $X_{n,t} = int_0^t f_n(s,omega)dB_s(omega)$. Then by martingale maximal inequality, we have
    $$
    E[sup_{tin [0,T]}|X_t -X_{n,t}|^2]leq C E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0.
    $$     This implies that $X_{n.t} to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $gin C^2_0$ via localization method. Then,
    $$
    g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
    $$     converge in probability to
    $$
    g(X_t), g'(X_t), g''(X_t)
    $$     locally uniformly.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $X_t $ is of the form
      $$
      X_t = int_0^t f(s,omega)dB_s(omega),quad tgeq 0,
      $$       then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes ${f_n}$ such that
      $$
      E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0
      $$       for all $T>0$. Let $X_{n,t} = int_0^t f_n(s,omega)dB_s(omega)$. Then by martingale maximal inequality, we have
      $$
      E[sup_{tin [0,T]}|X_t -X_{n,t}|^2]leq C E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0.
      $$       This implies that $X_{n.t} to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $gin C^2_0$ via localization method. Then,
      $$
      g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
      $$       converge in probability to
      $$
      g(X_t), g'(X_t), g''(X_t)
      $$       locally uniformly.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $X_t $ is of the form
        $$
        X_t = int_0^t f(s,omega)dB_s(omega),quad tgeq 0,
        $$         then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes ${f_n}$ such that
        $$
        E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0
        $$         for all $T>0$. Let $X_{n,t} = int_0^t f_n(s,omega)dB_s(omega)$. Then by martingale maximal inequality, we have
        $$
        E[sup_{tin [0,T]}|X_t -X_{n,t}|^2]leq C E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0.
        $$         This implies that $X_{n.t} to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $gin C^2_0$ via localization method. Then,
        $$
        g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
        $$         converge in probability to
        $$
        g(X_t), g'(X_t), g''(X_t)
        $$         locally uniformly.






        share|cite|improve this answer









        $endgroup$



        If $X_t $ is of the form
        $$
        X_t = int_0^t f(s,omega)dB_s(omega),quad tgeq 0,
        $$         then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes ${f_n}$ such that
        $$
        E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0
        $$         for all $T>0$. Let $X_{n,t} = int_0^t f_n(s,omega)dB_s(omega)$. Then by martingale maximal inequality, we have
        $$
        E[sup_{tin [0,T]}|X_t -X_{n,t}|^2]leq C E[int_0^T |f_n(s,cdot)-f(s,cdot)|^2ds]to 0.
        $$         This implies that $X_{n.t} to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $gin C^2_0$ via localization method. Then,
        $$
        g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
        $$         converge in probability to
        $$
        g(X_t), g'(X_t), g''(X_t)
        $$         locally uniformly.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 10:43









        SongSong

        16.2k1739




        16.2k1739






























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