Stochastic representation formula












2












$begingroup$


Consider the following boundary value problem in the domain $[0,T]$ x $R$ for an unknown function F.



$frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) = 0$



$F(T,x) = Phi(x)$



$Phi, mu, sigma$ are assumed to be known functions.



Derive a stochastic representation formula for this problem. Make sure it is clear at which points the functions should be evaluated.



So this is how I think you do this, but I need some help understanding the steps.



We first assume that it actually exists such stochastic representation that is the solution to the SDE



$dX_s = mu(t,X_s)ds + sigma(t,X_s)dB_s$



$X_t = x$



And the infinitesimal generator $mathcal{A}$ of X is



$mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x)$



so we can rewrite the the PDE as



$frac{partial F}{partial t} + mathcal{A}F(t,x) = 0$ (or should it be a minus sign)



$F(T,x) = Phi(x)$ $(star)$



And now we apply the Itô formula on $F(s,X_s)$ and this step I don't understand (if someone could explain it I would be very pleased, I know the Itô formula but on this problem I don't get it), but if I'm correct it we get



$F(T,X_T) = F(t,X_t) + int^T_t big(frac{partial F}{partial t}(s,X_s) + mathcal{A}F(s, X_s)big)ds + int^T_tsigma(s, X_s)frac{partial F}{partial x}(s,X_s)dB_s$
(where the ds-integral is $0$ and $F(T,X_t) = Phi(X_T)$ by assumption)



So now we take expectations on both sides and we get:



$E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t sigma(s,X_s)frac{partial F}{partial x}(s,X_s)dB_s]$



Where the integral is $0$ if $sigma(s,X_s)frac{partial F}{partial x}(s,X_s)$ is sufficiently nice.



So we then have our stochastic representation of $F(t,x) = E_{t,x}[Phi(X_T)]$



So, how do you apply the Itô formula on $(star)$? Also I'm a bit confused if it should be a minus sign at $(star)$ aswell, I think it should? Is this the Kolmogorovs backward equation? And if you instead have a initial condition in the PDE you get the foward equation? I think we should also be able to this if you add some nice function to the PDE, say $r(x)$. How would that change the derivation of this stochastic representation?










share|cite|improve this question











$endgroup$












  • $begingroup$
    It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
    $endgroup$
    – Good guy Mike
    Dec 10 '12 at 23:07










  • $begingroup$
    I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:47










  • $begingroup$
    My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:52
















2












$begingroup$


Consider the following boundary value problem in the domain $[0,T]$ x $R$ for an unknown function F.



$frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) = 0$



$F(T,x) = Phi(x)$



$Phi, mu, sigma$ are assumed to be known functions.



Derive a stochastic representation formula for this problem. Make sure it is clear at which points the functions should be evaluated.



So this is how I think you do this, but I need some help understanding the steps.



We first assume that it actually exists such stochastic representation that is the solution to the SDE



$dX_s = mu(t,X_s)ds + sigma(t,X_s)dB_s$



$X_t = x$



And the infinitesimal generator $mathcal{A}$ of X is



$mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x)$



so we can rewrite the the PDE as



$frac{partial F}{partial t} + mathcal{A}F(t,x) = 0$ (or should it be a minus sign)



$F(T,x) = Phi(x)$ $(star)$



And now we apply the Itô formula on $F(s,X_s)$ and this step I don't understand (if someone could explain it I would be very pleased, I know the Itô formula but on this problem I don't get it), but if I'm correct it we get



$F(T,X_T) = F(t,X_t) + int^T_t big(frac{partial F}{partial t}(s,X_s) + mathcal{A}F(s, X_s)big)ds + int^T_tsigma(s, X_s)frac{partial F}{partial x}(s,X_s)dB_s$
(where the ds-integral is $0$ and $F(T,X_t) = Phi(X_T)$ by assumption)



So now we take expectations on both sides and we get:



$E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t sigma(s,X_s)frac{partial F}{partial x}(s,X_s)dB_s]$



Where the integral is $0$ if $sigma(s,X_s)frac{partial F}{partial x}(s,X_s)$ is sufficiently nice.



So we then have our stochastic representation of $F(t,x) = E_{t,x}[Phi(X_T)]$



So, how do you apply the Itô formula on $(star)$? Also I'm a bit confused if it should be a minus sign at $(star)$ aswell, I think it should? Is this the Kolmogorovs backward equation? And if you instead have a initial condition in the PDE you get the foward equation? I think we should also be able to this if you add some nice function to the PDE, say $r(x)$. How would that change the derivation of this stochastic representation?










share|cite|improve this question











$endgroup$












  • $begingroup$
    It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
    $endgroup$
    – Good guy Mike
    Dec 10 '12 at 23:07










  • $begingroup$
    I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:47










  • $begingroup$
    My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:52














2












2








2





$begingroup$


Consider the following boundary value problem in the domain $[0,T]$ x $R$ for an unknown function F.



$frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) = 0$



$F(T,x) = Phi(x)$



$Phi, mu, sigma$ are assumed to be known functions.



Derive a stochastic representation formula for this problem. Make sure it is clear at which points the functions should be evaluated.



So this is how I think you do this, but I need some help understanding the steps.



We first assume that it actually exists such stochastic representation that is the solution to the SDE



$dX_s = mu(t,X_s)ds + sigma(t,X_s)dB_s$



$X_t = x$



And the infinitesimal generator $mathcal{A}$ of X is



$mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x)$



so we can rewrite the the PDE as



$frac{partial F}{partial t} + mathcal{A}F(t,x) = 0$ (or should it be a minus sign)



$F(T,x) = Phi(x)$ $(star)$



And now we apply the Itô formula on $F(s,X_s)$ and this step I don't understand (if someone could explain it I would be very pleased, I know the Itô formula but on this problem I don't get it), but if I'm correct it we get



$F(T,X_T) = F(t,X_t) + int^T_t big(frac{partial F}{partial t}(s,X_s) + mathcal{A}F(s, X_s)big)ds + int^T_tsigma(s, X_s)frac{partial F}{partial x}(s,X_s)dB_s$
(where the ds-integral is $0$ and $F(T,X_t) = Phi(X_T)$ by assumption)



So now we take expectations on both sides and we get:



$E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t sigma(s,X_s)frac{partial F}{partial x}(s,X_s)dB_s]$



Where the integral is $0$ if $sigma(s,X_s)frac{partial F}{partial x}(s,X_s)$ is sufficiently nice.



So we then have our stochastic representation of $F(t,x) = E_{t,x}[Phi(X_T)]$



So, how do you apply the Itô formula on $(star)$? Also I'm a bit confused if it should be a minus sign at $(star)$ aswell, I think it should? Is this the Kolmogorovs backward equation? And if you instead have a initial condition in the PDE you get the foward equation? I think we should also be able to this if you add some nice function to the PDE, say $r(x)$. How would that change the derivation of this stochastic representation?










share|cite|improve this question











$endgroup$




Consider the following boundary value problem in the domain $[0,T]$ x $R$ for an unknown function F.



$frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) = 0$



$F(T,x) = Phi(x)$



$Phi, mu, sigma$ are assumed to be known functions.



Derive a stochastic representation formula for this problem. Make sure it is clear at which points the functions should be evaluated.



So this is how I think you do this, but I need some help understanding the steps.



We first assume that it actually exists such stochastic representation that is the solution to the SDE



$dX_s = mu(t,X_s)ds + sigma(t,X_s)dB_s$



$X_t = x$



And the infinitesimal generator $mathcal{A}$ of X is



$mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x)$



so we can rewrite the the PDE as



$frac{partial F}{partial t} + mathcal{A}F(t,x) = 0$ (or should it be a minus sign)



$F(T,x) = Phi(x)$ $(star)$



And now we apply the Itô formula on $F(s,X_s)$ and this step I don't understand (if someone could explain it I would be very pleased, I know the Itô formula but on this problem I don't get it), but if I'm correct it we get



$F(T,X_T) = F(t,X_t) + int^T_t big(frac{partial F}{partial t}(s,X_s) + mathcal{A}F(s, X_s)big)ds + int^T_tsigma(s, X_s)frac{partial F}{partial x}(s,X_s)dB_s$
(where the ds-integral is $0$ and $F(T,X_t) = Phi(X_T)$ by assumption)



So now we take expectations on both sides and we get:



$E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t sigma(s,X_s)frac{partial F}{partial x}(s,X_s)dB_s]$



Where the integral is $0$ if $sigma(s,X_s)frac{partial F}{partial x}(s,X_s)$ is sufficiently nice.



So we then have our stochastic representation of $F(t,x) = E_{t,x}[Phi(X_T)]$



So, how do you apply the Itô formula on $(star)$? Also I'm a bit confused if it should be a minus sign at $(star)$ aswell, I think it should? Is this the Kolmogorovs backward equation? And if you instead have a initial condition in the PDE you get the foward equation? I think we should also be able to this if you add some nice function to the PDE, say $r(x)$. How would that change the derivation of this stochastic representation?







probability-theory stochastic-processes stochastic-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '12 at 11:05







Good guy Mike

















asked Dec 10 '12 at 23:04









Good guy MikeGood guy Mike

435




435












  • $begingroup$
    It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
    $endgroup$
    – Good guy Mike
    Dec 10 '12 at 23:07










  • $begingroup$
    I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:47










  • $begingroup$
    My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:52


















  • $begingroup$
    It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
    $endgroup$
    – Good guy Mike
    Dec 10 '12 at 23:07










  • $begingroup$
    I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:47










  • $begingroup$
    My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
    $endgroup$
    – Good guy Mike
    Dec 11 '12 at 12:52
















$begingroup$
It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
$endgroup$
– Good guy Mike
Dec 10 '12 at 23:07




$begingroup$
It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance!
$endgroup$
– Good guy Mike
Dec 10 '12 at 23:07












$begingroup$
I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
$endgroup$
– Good guy Mike
Dec 11 '12 at 12:47




$begingroup$
I tried adding in the PDE so that I have: $frac{partial F}{partial t}(t,x) + mu(t,x)frac{partial F}{partial x}(t,x) + frac {1}{2}sigma^2(t,x)frac{partial^2 F}{partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $frac{partial F}{partial t} + mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[Phi(X_T)] = F(t, x) + E_{t,x}[int^T_t rF(s,X_s)ds] + (dots) = {Fubini's} = F(t, x) + int^T_t rE_{t,x}[F(s,X_s)]ds + (dots)$ From here I'm not certain on how to continue.
$endgroup$
– Good guy Mike
Dec 11 '12 at 12:47












$begingroup$
My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
$endgroup$
– Good guy Mike
Dec 11 '12 at 12:52




$begingroup$
My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[Phi(X_T)]$ and that is not what I get from this.
$endgroup$
– Good guy Mike
Dec 11 '12 at 12:52










1 Answer
1






active

oldest

votes


















2












$begingroup$

I use the notation $frac{partial}{partial t}F = F_t$.



$dF = F_t,dt + F_x,dX_t + F_{xx}frac{1}{2}sigma,dt$



Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).



Put $Z(s) = e^{-int_t^s r,du}F(s, X(s))$. Then



$dZ(s) = -re^{-int_t^s r,du}Fds + e^{-int_t^s r,du}dF = e^{-int_t^s r,du}(dF - rFds)$



So, $Z(T) - Z(t) = int_t^T dZ(s)$ and



$Z(T) = Z(t) + int_t^T e^{-int_t^s r,du}(F_t + F_xmu + F_{xx}frac{1}{2}sigma -rF),ds + int_t^T e^{-int_t^s r,du}F_xsigma,dW(s)$.



The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:



$E_t[Z(T)] = E_t[Z(t)] = e^{-int_t^t r,ds}F(t, X(t)) = F(t, x)$.



On the other hand, $Z(T)$ is defined as $e^{-int_t^T r,ds}F(T, X(T)) = e^{-int_t^T r,ds}Phi(X(T))$
so we conclude:



$E_t[e^{-int_t^T r,ds}Phi(X(T))] = F(t, x)$.






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%2f255852%2fstochastic-representation-formula%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I use the notation $frac{partial}{partial t}F = F_t$.



    $dF = F_t,dt + F_x,dX_t + F_{xx}frac{1}{2}sigma,dt$



    Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).



    Put $Z(s) = e^{-int_t^s r,du}F(s, X(s))$. Then



    $dZ(s) = -re^{-int_t^s r,du}Fds + e^{-int_t^s r,du}dF = e^{-int_t^s r,du}(dF - rFds)$



    So, $Z(T) - Z(t) = int_t^T dZ(s)$ and



    $Z(T) = Z(t) + int_t^T e^{-int_t^s r,du}(F_t + F_xmu + F_{xx}frac{1}{2}sigma -rF),ds + int_t^T e^{-int_t^s r,du}F_xsigma,dW(s)$.



    The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:



    $E_t[Z(T)] = E_t[Z(t)] = e^{-int_t^t r,ds}F(t, X(t)) = F(t, x)$.



    On the other hand, $Z(T)$ is defined as $e^{-int_t^T r,ds}F(T, X(T)) = e^{-int_t^T r,ds}Phi(X(T))$
    so we conclude:



    $E_t[e^{-int_t^T r,ds}Phi(X(T))] = F(t, x)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I use the notation $frac{partial}{partial t}F = F_t$.



      $dF = F_t,dt + F_x,dX_t + F_{xx}frac{1}{2}sigma,dt$



      Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).



      Put $Z(s) = e^{-int_t^s r,du}F(s, X(s))$. Then



      $dZ(s) = -re^{-int_t^s r,du}Fds + e^{-int_t^s r,du}dF = e^{-int_t^s r,du}(dF - rFds)$



      So, $Z(T) - Z(t) = int_t^T dZ(s)$ and



      $Z(T) = Z(t) + int_t^T e^{-int_t^s r,du}(F_t + F_xmu + F_{xx}frac{1}{2}sigma -rF),ds + int_t^T e^{-int_t^s r,du}F_xsigma,dW(s)$.



      The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:



      $E_t[Z(T)] = E_t[Z(t)] = e^{-int_t^t r,ds}F(t, X(t)) = F(t, x)$.



      On the other hand, $Z(T)$ is defined as $e^{-int_t^T r,ds}F(T, X(T)) = e^{-int_t^T r,ds}Phi(X(T))$
      so we conclude:



      $E_t[e^{-int_t^T r,ds}Phi(X(T))] = F(t, x)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I use the notation $frac{partial}{partial t}F = F_t$.



        $dF = F_t,dt + F_x,dX_t + F_{xx}frac{1}{2}sigma,dt$



        Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).



        Put $Z(s) = e^{-int_t^s r,du}F(s, X(s))$. Then



        $dZ(s) = -re^{-int_t^s r,du}Fds + e^{-int_t^s r,du}dF = e^{-int_t^s r,du}(dF - rFds)$



        So, $Z(T) - Z(t) = int_t^T dZ(s)$ and



        $Z(T) = Z(t) + int_t^T e^{-int_t^s r,du}(F_t + F_xmu + F_{xx}frac{1}{2}sigma -rF),ds + int_t^T e^{-int_t^s r,du}F_xsigma,dW(s)$.



        The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:



        $E_t[Z(T)] = E_t[Z(t)] = e^{-int_t^t r,ds}F(t, X(t)) = F(t, x)$.



        On the other hand, $Z(T)$ is defined as $e^{-int_t^T r,ds}F(T, X(T)) = e^{-int_t^T r,ds}Phi(X(T))$
        so we conclude:



        $E_t[e^{-int_t^T r,ds}Phi(X(T))] = F(t, x)$.






        share|cite|improve this answer











        $endgroup$



        I use the notation $frac{partial}{partial t}F = F_t$.



        $dF = F_t,dt + F_x,dX_t + F_{xx}frac{1}{2}sigma,dt$



        Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).



        Put $Z(s) = e^{-int_t^s r,du}F(s, X(s))$. Then



        $dZ(s) = -re^{-int_t^s r,du}Fds + e^{-int_t^s r,du}dF = e^{-int_t^s r,du}(dF - rFds)$



        So, $Z(T) - Z(t) = int_t^T dZ(s)$ and



        $Z(T) = Z(t) + int_t^T e^{-int_t^s r,du}(F_t + F_xmu + F_{xx}frac{1}{2}sigma -rF),ds + int_t^T e^{-int_t^s r,du}F_xsigma,dW(s)$.



        The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:



        $E_t[Z(T)] = E_t[Z(t)] = e^{-int_t^t r,ds}F(t, X(t)) = F(t, x)$.



        On the other hand, $Z(T)$ is defined as $e^{-int_t^T r,ds}F(T, X(T)) = e^{-int_t^T r,ds}Phi(X(T))$
        so we conclude:



        $E_t[e^{-int_t^T r,ds}Phi(X(T))] = F(t, x)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 9:12

























        answered Jan 17 '14 at 21:59









        HunaphuHunaphu

        1256




        1256






























            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%2f255852%2fstochastic-representation-formula%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

            Cabo Verde

            Karlovacs län

            Gyllenstierna