Antiderivative of $g(x)dg(x)$












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I am reading a book by Shreve "Stochastic Calculus for Finance II" and after computing a stochastic integral $int_{0}^{T}W(t)dW(t)$ where $W(t)$ is a Brownian motion he compares it to the integral
$$int_{0}^{T}g(t)dg(t) = int_{0}^{T}g(t)g^prime (t)dt = 0.5g^2(T),$$
where $g(t)$ is a differentiable function with $g(0)=0$. I don't get the fact that $int g(t)g^prime (t)dt = 0.5g^2(t)$. For me the right hand side is equal to $int g(t)dt$, without the $g^prime (t)$ term. Thanks in advance.






calculus integration self-learning stochastic-calculus







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  • $begingroup$
    What’s the integral of x with respect to x?
    $endgroup$
    – Fede Poncio
    Jan 3 at 20:07










  • $begingroup$
    $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    $endgroup$
    – AddSup
    Jan 4 at 8:31
















0












$begingroup$


I am reading a book by Shreve "Stochastic Calculus for Finance II" and after computing a stochastic integral $int_{0}^{T}W(t)dW(t)$ where $W(t)$ is a Brownian motion he compares it to the integral
$$int_{0}^{T}g(t)dg(t) = int_{0}^{T}g(t)g^prime (t)dt = 0.5g^2(T),$$
where $g(t)$ is a differentiable function with $g(0)=0$. I don't get the fact that $int g(t)g^prime (t)dt = 0.5g^2(t)$. For me the right hand side is equal to $int g(t)dt$, without the $g^prime (t)$ term. Thanks in advance.






calculus integration self-learning stochastic-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What’s the integral of x with respect to x?
    $endgroup$
    – Fede Poncio
    Jan 3 at 20:07










  • $begingroup$
    $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    $endgroup$
    – AddSup
    Jan 4 at 8:31














0












0








0





$begingroup$


I am reading a book by Shreve "Stochastic Calculus for Finance II" and after computing a stochastic integral $int_{0}^{T}W(t)dW(t)$ where $W(t)$ is a Brownian motion he compares it to the integral
$$int_{0}^{T}g(t)dg(t) = int_{0}^{T}g(t)g^prime (t)dt = 0.5g^2(T),$$
where $g(t)$ is a differentiable function with $g(0)=0$. I don't get the fact that $int g(t)g^prime (t)dt = 0.5g^2(t)$. For me the right hand side is equal to $int g(t)dt$, without the $g^prime (t)$ term. Thanks in advance.






calculus integration self-learning stochastic-calculus







share|cite|improve this question











$endgroup$




I am reading a book by Shreve "Stochastic Calculus for Finance II" and after computing a stochastic integral $int_{0}^{T}W(t)dW(t)$ where $W(t)$ is a Brownian motion he compares it to the integral
$$int_{0}^{T}g(t)dg(t) = int_{0}^{T}g(t)g^prime (t)dt = 0.5g^2(T),$$
where $g(t)$ is a differentiable function with $g(0)=0$. I don't get the fact that $int g(t)g^prime (t)dt = 0.5g^2(t)$. For me the right hand side is equal to $int g(t)dt$, without the $g^prime (t)$ term. Thanks in advance.






calculus integration self-learning stochastic-calculus




calculus integration self-learning stochastic-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 3 at 20:06









J.G.

32.6k23250




32.6k23250










asked Jan 3 at 19:58









tosiktosik

1504




1504












  • $begingroup$
    What’s the integral of x with respect to x?
    $endgroup$
    – Fede Poncio
    Jan 3 at 20:07










  • $begingroup$
    $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    $endgroup$
    – AddSup
    Jan 4 at 8:31


















  • $begingroup$
    What’s the integral of x with respect to x?
    $endgroup$
    – Fede Poncio
    Jan 3 at 20:07










  • $begingroup$
    $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    $endgroup$
    – AddSup
    Jan 4 at 8:31
















$begingroup$
What’s the integral of x with respect to x?
$endgroup$
– Fede Poncio
Jan 3 at 20:07




$begingroup$
What’s the integral of x with respect to x?
$endgroup$
– Fede Poncio
Jan 3 at 20:07












$begingroup$
$frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
$endgroup$
– AddSup
Jan 4 at 8:31




$begingroup$
$frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
$endgroup$
– AddSup
Jan 4 at 8:31










2 Answers
2






active

oldest

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1












$begingroup$

You seem to be confusing $int g(t), dt$ with $int g(t), dg(t)$. It might help to make the substitution explicitly.



Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives
$$int_{0}^{T} g(t), dg(t) = int_{t=0}^{t=T} y, dy = left[ dfrac{y^2}{2} right]_{t=0}^{t=T} = left[ dfrac{g(t)^2}{2} right]_0^T = dfrac{g(T)^2}{2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are right, I confused the integrals, thank you.
    $endgroup$
    – tosik
    Jan 3 at 20:16



















1












$begingroup$

Just as $int g^prime dt=g+C$, $int f^prime(g)g^prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.






share|cite|improve this answer









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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You seem to be confusing $int g(t), dt$ with $int g(t), dg(t)$. It might help to make the substitution explicitly.



    Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives
    $$int_{0}^{T} g(t), dg(t) = int_{t=0}^{t=T} y, dy = left[ dfrac{y^2}{2} right]_{t=0}^{t=T} = left[ dfrac{g(t)^2}{2} right]_0^T = dfrac{g(T)^2}{2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You are right, I confused the integrals, thank you.
      $endgroup$
      – tosik
      Jan 3 at 20:16
















    1












    $begingroup$

    You seem to be confusing $int g(t), dt$ with $int g(t), dg(t)$. It might help to make the substitution explicitly.



    Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives
    $$int_{0}^{T} g(t), dg(t) = int_{t=0}^{t=T} y, dy = left[ dfrac{y^2}{2} right]_{t=0}^{t=T} = left[ dfrac{g(t)^2}{2} right]_0^T = dfrac{g(T)^2}{2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You are right, I confused the integrals, thank you.
      $endgroup$
      – tosik
      Jan 3 at 20:16














    1












    1








    1





    $begingroup$

    You seem to be confusing $int g(t), dt$ with $int g(t), dg(t)$. It might help to make the substitution explicitly.



    Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives
    $$int_{0}^{T} g(t), dg(t) = int_{t=0}^{t=T} y, dy = left[ dfrac{y^2}{2} right]_{t=0}^{t=T} = left[ dfrac{g(t)^2}{2} right]_0^T = dfrac{g(T)^2}{2}$$






    share|cite|improve this answer









    $endgroup$



    You seem to be confusing $int g(t), dt$ with $int g(t), dg(t)$. It might help to make the substitution explicitly.



    Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives
    $$int_{0}^{T} g(t), dg(t) = int_{t=0}^{t=T} y, dy = left[ dfrac{y^2}{2} right]_{t=0}^{t=T} = left[ dfrac{g(t)^2}{2} right]_0^T = dfrac{g(T)^2}{2}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 3 at 20:10









    Clive NewsteadClive Newstead

    52k474136




    52k474136












    • $begingroup$
      You are right, I confused the integrals, thank you.
      $endgroup$
      – tosik
      Jan 3 at 20:16


















    • $begingroup$
      You are right, I confused the integrals, thank you.
      $endgroup$
      – tosik
      Jan 3 at 20:16
















    $begingroup$
    You are right, I confused the integrals, thank you.
    $endgroup$
    – tosik
    Jan 3 at 20:16




    $begingroup$
    You are right, I confused the integrals, thank you.
    $endgroup$
    – tosik
    Jan 3 at 20:16











    1












    $begingroup$

    Just as $int g^prime dt=g+C$, $int f^prime(g)g^prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just as $int g^prime dt=g+C$, $int f^prime(g)g^prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just as $int g^prime dt=g+C$, $int f^prime(g)g^prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.






        share|cite|improve this answer









        $endgroup$



        Just as $int g^prime dt=g+C$, $int f^prime(g)g^prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 20:07









        J.G.J.G.

        32.6k23250




        32.6k23250






























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