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












0














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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  • What’s the integral of x with respect to x?
    – Fede Poncio
    2 days ago










  • $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    – AddSup
    2 days ago
















0














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
























  • What’s the integral of x with respect to x?
    – Fede Poncio
    2 days ago










  • $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    – AddSup
    2 days ago














0












0








0







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















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













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share|cite|improve this question








edited 2 days ago









J.G.

23.2k22137




23.2k22137










asked 2 days ago









tosik

1264




1264












  • What’s the integral of x with respect to x?
    – Fede Poncio
    2 days ago










  • $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    – AddSup
    2 days ago


















  • What’s the integral of x with respect to x?
    – Fede Poncio
    2 days ago










  • $frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
    – AddSup
    2 days ago
















What’s the integral of x with respect to x?
– Fede Poncio
2 days ago




What’s the integral of x with respect to x?
– Fede Poncio
2 days ago












$frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
– AddSup
2 days ago




$frac{d}{dt}left(frac{1}{2}g^2(t)right)=g(t)g'(t)$.
– AddSup
2 days ago










2 Answers
2






active

oldest

votes


















1














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





















  • You are right, I confused the integrals, thank you.
    – tosik
    2 days ago



















1














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 $f=g^2/2$.






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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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





















    • You are right, I confused the integrals, thank you.
      – tosik
      2 days ago
















    1














    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





















    • You are right, I confused the integrals, thank you.
      – tosik
      2 days ago














    1












    1








    1






    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












    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 2 days ago









    Clive Newstead

    50.7k474133




    50.7k474133












    • You are right, I confused the integrals, thank you.
      – tosik
      2 days ago


















    • You are right, I confused the integrals, thank you.
      – tosik
      2 days ago
















    You are right, I confused the integrals, thank you.
    – tosik
    2 days ago




    You are right, I confused the integrals, thank you.
    – tosik
    2 days ago











    1














    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


























      1














      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
























        1












        1








        1






        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












        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 2 days ago









        J.G.

        23.2k22137




        23.2k22137






























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