# Parametric Stochastic Integral

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

$$\\\\$$

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

with

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

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 ?

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.
• If you switch $dW_t$ and $ds$ in (2), you get the anwser to the initial question, no ? Sep 7, 2021 at 12:06
• 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. Sep 7, 2021 at 12:39
• 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 Sep 7, 2021 at 14:05
• 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. Sep 8, 2021 at 8:08