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I came across a generalized Black-Scholes equation formulation in this paper.

Let me highlight the basic idea below. Consider a random field $W(t,T)$ where for a fixed $T$, $W$ is a Brownian motion and for a fixed $t$, $W$ is a continuous function. $W$ satisfies the following:

  • $dW(t,u)dW(t,v)=c(u,v)dt$
  • $dW(t,u)dW(t,u)=dt$

The author's define the dynamics of some asset as follows:

$$\frac{dS(t)}{S(t)}=\mu(t)dt+\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du$$

The second term on the right side intrigues me. The paper doesn't really get into explaining the doing calculus with such processes so I just want to ask the following.

If I define $dg(t)=\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du$ then does it follow that

$$dg(t)dg(s)=\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du \int_{T_1}^{T_2}\sigma(t,v)dW(t,v)dv\\=\int_{T_1}^{T_2}\int_{T_1}^{T_2}\sigma(t,u)\sigma(t,v)c(u,v)dtdudv$$

Do we have the Ito isometry?

$$E\left[\left(\int_0^t\int_{T_1}^{T_2}\sigma(s,u)dW(s,u)du\right)^2\right]=E\left[\int_0^t\int_{T_1}^{T_2}\int_{T_1}^{T_2}\sigma(s,u)\sigma(s,v)c(u,v)dudvds\right]$$

I was not able to find any literature on such integrals. If anyone can suggest any references that would also help a lot. Thanks!

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  • $\begingroup$ Let's see what happens if we set some things constant that you haven't ruled out could be constant. If $\sigma\equiv 1$ and $W(s,u)=W_s$ (so $W$ does not depend on $u$) then $c\equiv 1$ and it looks like your Ito isometry does not hold: $$t^2 (T_2-T_1)=E\left(\left(t(W_{T_2}-W_{T_1})\right)^2\right)=E\left(\left(\int_0^t\int_{T_1}^{T_2} dW_s du\right)^2\right) \ne E\left(\int\int\int du\,dv\,ds\right)= t(T_2-T_1)^2$$ $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Jul 7, 2020 at 6:39
  • $\begingroup$ @BjørnKjos-Hanssen Thank you $\endgroup$
    – Heisenberg
    Commented Jul 7, 2020 at 19:06

1 Answer 1

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The notations in the question are ambiguous (Bjørn Kjos-Hanssen showed that the other interpretation cannot be correct). I assume that the expression of interest is given by $$ g(t) = \int_0^t \Sigma(s)\,dW(s)\;, $$ where $W$ is a $C(\mathbb{R})$-valued Wiener process with covariance $\hat c$ (at time $1$) and $\Sigma(s) \in C(\mathbb{R})^*$ is the finite measure given by $\Sigma(s) f = \int_{T_1}^{T_2}\sigma(s,u)f(u)\,du$. Here, the covariance $\hat c$ is the bilinear map on $C(\mathbb{R})^*$ such that, for measures $\mu$ and $\nu$, $$ \hat c(\mu,\nu) = \int c(u,v)\, \mu(du)\,\nu(dv)\;. $$ Itô isometry then indeed reads $$ \mathbb{E} g(t)^2 = \int_0^t \mathbb{E} \hat c(\Sigma(s),\Sigma(s))\,ds \;, $$ assuming of course that $\Sigma$ is adapted and square integrable. Regarding references, any book on SPDEs would do, for example "Stochastic Equations in Infinite Dimensions" by Da Prato & Zabczyk or Section 3 of my lecture notes.

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  • $\begingroup$ Thank you this helps a lot. This is probably a very basic questions but I don't fully understand how you interpret $c$. On the left side of the second equation you have $c$ evaluated at two measures and on the right side $c$ is evaluated at to real numbers. Why is this possible? $\endgroup$
    – Heisenberg
    Commented Jul 7, 2020 at 19:22
  • $\begingroup$ Edited for clarity. $\endgroup$
    – Martin Hairer
    Commented Jul 7, 2020 at 19:43
  • $\begingroup$ @MartinHairer Usually there is some linear operator $Q$ with finite trace in $C(\mathbb{R})$ so that $W_t-W_s\sim N(0,(t-s)Q)$. What is $Q$ in terms of $c$? $\endgroup$
    – user81883
    Commented Jul 11, 2020 at 18:35
  • $\begingroup$ Maybe @MartinHairer can correct me but I think if we were to think of a Hilbert space valued Weiner process, $Qf(\cdot)=\int c(s,\cdot) f(s)ds$, $f\in H$ and $Q\in L(H).$ $\endgroup$
    – Heisenberg
    Commented Jul 11, 2020 at 23:48

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