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












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












$begingroup$


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




share|cite|improve this question








edited Jan 3 at 20:06









J.G.

25.3k22539




25.3k22539










asked Jan 3 at 19:58









tosiktosik

1264




1264












  • $begingroup$
    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

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












$begingroup$

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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060938%2fantiderivative-of-gxdgx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    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

    51.5k474135




    51.5k474135












    • $begingroup$
      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.

        25.3k22539




        25.3k22539






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060938%2fantiderivative-of-gxdgx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Questions related to Moebius Transform of Characteristic Function of the Primes

            Image
            2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
            Read more

            List of scandals in India

            Image
            List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more