I need help.

Defining the parametric stochastic integral

$$ F_t = \int_t^T\xi(t,s)g(s)ds $$


with $\xi$ a generic stochastic process such that $d\xi(t,s) = \mu(t,s)dt + \sigma(t,s)dW_t$, I'm trying to prove that

$\\\\$ $$ dF_t = - g(t)\xi(t,t)dt + \int_t^Td\xi(t,s)g(s)ds$$

My first attempt was as follows :

$$ \xi(t,s) = \xi(0,s) + \int_0^t\mu(u,s)du + \int_0^t \sigma(u,s)dW_u $$

and so

\begin{eqnarray*} F_t &=& \int_t^T\xi(0,s)g(s)ds + \int_t^T\int_0^t\mu(u,s)g(s)duds + \int_t^T\int_0^t\sigma(u,s)g(s)dW_uds\\ &=& \int_t^T\xi(s,s)g(s)ds - \int_t^T\int_t^s\mu(u,s)g(s)duds - \int_t^T\int_t^s\sigma(u,s)g(s)dW_uds \end{eqnarray*}


Assuming suitable conditions to apply the stochastic Fubini theorem, we get


\begin{eqnarray*} F_t = \int_t^T\xi(s,s)g(s)ds - \int_t^T\alpha(u,T)du - \int_t^T\beta(u,T)dW_u \end{eqnarray*}


\begin{eqnarray*} \alpha(u,T) = \int_u^T\mu(u,s)g(s)ds \quad \quad \text{and} \quad \quad \beta(u,T) = \int_u^T\sigma(u,s)g(s)ds \end{eqnarray*}

Applying Ito's lemma, we find


\begin{eqnarray*} dF_t &=& -\xi(t,t)g(t)dt + \alpha(t,T)dt + \beta(t,T)dW_t\\ &=& -\xi(t,t)g(t)dt + \int_t^T\left(\mu(t,s)dt + \sigma(t,s)dW_t\right)g(s)ds\\ &=& -\xi(t,t)g(t)dt + \int_t^Td\xi(t,s)g(s)ds \end{eqnarray*}

Now, I have two questions :

  • Is my proof correct ?
  • Is there a more clever and faster answer ?

Thank you in advance for your answer.


1 Answer 1


I am having trouble to understand your notation $$ \int_t^Td\xi(t,s)g(s)\,ds\,. $$ What is the meaning of this when you switch from the differential form $dF_t$ to the integral form $$ F_t=F_0-\int_0^tg(s)\,\xi(s,s)\,ds\,+\quad? $$ Surely, in the deterministic case when $\sigma\equiv 0\,$ we have by ordinary calculus $$ \frac{dF}{dt}=-\xi(t,t)\,g(t)+\int_t^T\frac{\partial}{\partial t}\xi(t,s)\,g(s)\,ds\,, $$ or, in integral form $$\tag{0} F_t=F_0-\int_0^tg(s)\,\xi(s,s)\,ds+\int_0^t\int_u^T\frac{\partial}{\partial u}\xi(u,s)\,g(s)\,ds\,du\,. $$ To get to the bottom of the stochastic case I consider only the case $\mu\equiv 0,\sigma\not\equiv 0,g\equiv1$ to simplify notation.

From $\xi(t,s)=\xi(0,s)+\int_0^t\sigma(u,s)\,dW_u$ we get (using stochastic Fubini) \begin{align} F_t&=\int_t^T\xi(t,s)\,ds=\int_t^T\xi(0,s)\,ds+\int_t^T\left(\int_0^t\sigma(u,s)\,dW_u\right)\,ds\\ &=\int_t^T\xi(0,s)\,ds+\int_0^t\int_t^T\sigma(u,s)\,ds\,dW_u\,. \end{align} By Ito's formula, $$\tag{1} dF_t=-\xi(0,t)\,dt+\left(\int_t^T\sigma(t,s)\,ds\right)dW_t-\left(\int_0^t\sigma(u,t)\,dW_u\right)\,dt\,. $$ The last term in (1) can be combined with the first term and gives \begin{align}\tag{2} dF_t&=-\xi(t,t)\,dt+\left(\int_t^T\sigma(t,s)\,ds\right)dW_t\,. \end{align} In integral form, (2) is $$\tag{3}\boxed{ F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^t\int_u^T\sigma(u,s)\,ds\,dW_u\,.} $$ By stochastic Fubini, this is $$\tag{4} F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^T\int_0^{s\wedge t}\sigma(u,s)\,dW_u\,ds\,. $$ Using $$ d\xi(t,s)=\sigma(t,s)\,dW_t $$ one could write (4) as $$\tag{5}\boxed{ F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^T\big\{\xi(s\wedge t,s)-\xi(0,s)\big\} \,ds\,.} $$ It is fairly easy to see that (0) can also be written in the same form. In other words, (5) is the form that comprises the deterministic and the stochastic case.

  • $\begingroup$ If you switch $dW_t$ and $ds$ in (2), you get the anwser to the initial question, no ? $\endgroup$
    – Deros
    Sep 7, 2021 at 12:06
  • $\begingroup$ In (2) you can't switch $dW_t$ and $ds$. To do so you need to put this equation into integral form and use Fubini. That's what I have done. $\endgroup$
    – Kurt G.
    Sep 7, 2021 at 12:39
  • $\begingroup$ Thanks for answering me, but I don't understand your question "What is the meaning ...". If you switch $dWt$ and $ds$ in (2), with your dynamic for $\xi$ you get exactly $$dF_t = -\xi(t,t)dt + \int_t^T\left(\sigma(t,s)dW_t\right)ds = -\xi(t,t)dt + \int_t^Td\xi(t,s)ds$$ so the meaning is just your equation (3). For my part, I am having trouble with your first equation (1). Where does the last $dt$ factor come from ? I mean, if $G_t = \int_0^tg(u,t)dW_u$, you just write $dG_t = g(t,t)dW_t + \int_0^t(\partial_tg(u,t)dt)dW_u$. It's may be obvious, but I have never seen this kind of generalization $\endgroup$
    – Deros
    Sep 7, 2021 at 14:05
  • $\begingroup$ Sorry but why we can't switch ? $\endgroup$
    – Deros
    Sep 7, 2021 at 14:05
  • $\begingroup$ We will get to the point where I can explain to you why we can't switch. Before that, the term $\int_0^t(\partial_tg(u,t)\,dt)\,dW_u$ in your $dG_t$ looks strange to me. I prefer to 'pull' $dt$ out of this integral and write this as $(\int_0^t(\partial_tg(u,t)\,dW_u)\,dt\,.$ There you have the $dt$ factor in my eq. (3). In general I think is is a source of much confusion to do hocus pocus with those equations in differential form. We know that in Ito calculus the only rigorous meaning is always the integral form. $\endgroup$
    – Kurt G.
    Sep 8, 2021 at 8:08

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