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The Chaos Expansion

Iterated Itô integrals and their consequences

By L. Llamazares-Elias
Posted on May 26, 2022

This is the first post of many to come on this blog, thanks for joining! This is also the first in a series on Malliavin calculus. This chain is based on the book by Oksendal and Nunno 1 and inspired by a reading group on the topic. Some technical details are only sketched out and not detailed and full. Full proofs can be found in the aformentioned book.

Three line summary

  • Square-integrable deterministic functions of multiple variables can be iteratively Itô integrated to get a square-integrable random function.

  • Iterated integration satisfies an isometry.

  • Every square-integrable function can be uniquely written as the a sum of iterated integrals.

Why this is important

The chaos expansion gives us a way to represent random variables as a sum of functions in $L^2$. This representation can be used to define the Skorohod integral and later on the Malliavin derivative.

Iterated integrals

We will write $\mathcal{F}\U t$ for the completion of the natural filtration generated by $W(t)$, on some measure space $\Omega$ and consider an interval $I=\zl 0,T\zr $ or $I=\zl 0,+\infty \zr$. In this second case we say that $T=\infty.$ We also recall the notation

$$\mathcal{M}\U t^2:=\{f\in L^2(\zl 0,t\zr \times\Omega)\text{ that are continuous martingales}\}.$$

We will work with the spaces $L^2(I^n), L^2\U S(I^n),L^2(S\U n)$ of respectively square-integrable functions, symmetric square-integrable functions, and finally square-integrable functions on

$$S\U n:=\{0\leq t\U 1\leq\ldots\leq t\U n\leq T\}.$$

Since these functions don’t involve a random variable we call them deterministic. Given $f\in L^2(I^n)$ we denote its symmetrization by

$$f\U S:=\frac{1}{n!}\sum\U {\sigma} f(t\U {\sigma\U 1},\ldots,t\U {\sigma\U n}).$$

Where $\sigma$ is summed over all the permutations of ${1,\ldots,n}$. Note that if $f$ is already symmetric then $f=f\U S$. Given a deterministic function we can transform it into a random one by taking its Itô integral. We will want to do this multiple times, this is permitted by the following lemma.

Lemma 0. Given $f\in L^2(S\U n)$ and any $m<n$, the integral

$$\int\U {0}^{t\U m}\cdots\int\U {0}^{t\U 2} f(t\U 1,\ldots,t\U n)dW(t\U 1)\ldots dW(t\U {m-1}).$$

is in $\mathbb{L}^2(I\times\Omega)$ as a function of $t\U m$.

Proof. The proof can be seen by verifying the above property for smooth functions $f \in C\U c^\infty(S\U n)$ and taking limits as $C\U c^\infty(S\U n)$ is dense in $L^2(S\U n)$ and measurability is preserved by limits. ◻

We can now define, the following

Definition 1. Let $f\in L^2(S\U n)$, then we define the n-fold Itô integral as

$$J\U n(f):=\int\U I\cdots \int\U {0}^{t\U 3}\int\U {0}^{t\U 2} f(t\U 1,\ldots,t\U n)dW(t\U 1) dW(t\U 2)\ldots dW(t\U n).$$

To notice the subtleties involved in this definition, we note that one example of an integral that would be ill-defined however is if for $f\in L^2(I^n)$ we defined

$${\color{red}J\U n(f)}:=\int\U I\cdots \int\U I\int\U I f(t\U 1,\ldots,t\U n)dW(t\U 1) dW(t\U 2)\ldots dW(t\U n).$$

In this case, we have that the first integral is $\mathcal{F}\U T$ adapted and not $\mathcal{F}\U {t\U 2}$ adapted so we cannot continue integrating! As a result, we instead give the following definition for symmetric functions.

Definition 2. Let $f\in L^2\U S(I^n)$, then we define

$$I\U n(f):=\int\U I\cdots \int\U I\int\U I f(t\U 1,\ldots,t\U n)dW(t\U 1) dW(t\U 2)\ldots dW(t\U n):=n! J\U n(f).$$

Where it is important to note that the second equality is by definition and where the rescaling factor is motivated by the fact that, by counting permutations, if $I$ is finite then $\mu(I^n)=n! \mu(S\U n)$. Based on Itô’s isometry we can obtain a similar result for the just defined iterated integrals.

Proposition 1 (Itô’s n-th isometry). Let $f,g\in L^2(S^n)$ then

$$\left\langle J\U n(f),J\U m(g)\right\rangle\U {L^2(\Omega)}=\left\langle f,g\right\rangle\U {L^2(S\U n)}\delta\U {nm}$$.

Proof. The proof is an application of Itô’s (product) isometry to move the expectation into the iterated integrals. If the number of integrals is unequal ($n\neq m$) you get the expectation of an Itô integral which is zero. Otherwise, you just get the (deterministic) inner product. ◻

As a result we also get by a calculation that if $f,g\in L^2(I^n)$ then

$$\left\langle I\U n(f),I\U m(g)\right\rangle\U {L^2(\Omega)}=n!\left\langle f,g\right\rangle\U {L^2(I^n)}\delta\U {nm}.$$

The chaos expansion

Lemma 1 (Itô’s representation theorem). Let $\xi\in L^2(\Omega,\mathcal{F}\U T)$, then there exists a unique $X(t)\in \mathbb{L}^2(I)$ such that

$$\xi=\mathbb{E}\zl \xi\zr +\int\U {I} X(t)dW(t).$$

Furthermore it holds that $\norm{X}\U {L^2(I\times\Omega)}\leq \norm{\xi}\U {L^2(\Omega)}$.

Theorem 1 (Chaos expansion). Let $\xi\in L^2(\Omega,\mathcal{F}\U T)$, then there exists a unique sequence of functions $g\U n \in L^2(S\U n),f\U n \in L^2\U S(I^n)$ such that

$$\xi=\sum\U {n=0}^{\infty} J\U n (g\U n);\quad \xi=\sum\U {n=0}^{\infty} I \U n (f\U n).$$

Proof. The proof is a bit technical but we sketch the main idea which is to iteratively apply Itô’s representation theorem to get

$$\begin{gathered} \xi=\mathbb{E}\zl \xi\zr +\int\U I X\U 1(t\U 1)dW(t\U 1)=\mathbb{E}\zl \xi\zr +\int\U {I} \mathbb{E}\zl X\U 1(t\U 1)\zr dW(t\U 1)\\+ \int\U {I}\int\U {0}^{t\U 1} X\U 2(t\U 1,t\U 2) dW(t\U 2)dW(t\U 1)= \ldots=\sum\U {n=0}^{N} J\U n(f\U n)+ \int\U {S\U {N+1}} X\U {N+1} dW^{\otimes(N+1)}. \end{gathered}$$

Where $g\U 0=\mathbb{E}\zl \xi\zr , g\U 1(t\U 1) =\mathbb{E}\zl X\U 1\zr (t\U 1)$ and so on. By an iteration we deduce that, if we write $\phi\U {N+1}$ for the last integral term, then

$$\mathbb{E}\zl \phi\U {N+1}\zr \leq\ldots\leq \mathbb{E}\zl \xi^2\zr .$$

Furthermore, by Itô’s $n$-th isometry we have that the terms in the above sum are orthogonal so

$$\norm{\xi}\U {L^2(\Omega)}=\sum\U {n=0}^{N} \norm{J\U n(g\U n)}\U {L^2(\Omega)} +\norm{\phi\U {N+1}}\U {L^2(\Omega)}.$$

Since the last term is bounded, we obtain a bound on the $L^2(\Omega)$ norm of the sum uniform in $n$ so the sum converges. As a result, so does $\phi\U N$ to some $\phi\U \infty$. It suffices to see that $\phi\U \infty=0$. This can be proved using that, by Itô’s $n$-th isometry, $\phi\U {N+1}$ is orthogonal to $J\U n(h)$ for any $h\in L^2(\Omega)$ and $n\leq N$. As a result, the limit $\phi\U \infty$ is orthogonal to $J\U n(h)$ for all $n$ and by a density argument (there are a lot of these functions $J\U n(h)$!) we obtain $\phi\U \infty=0$. This proves the first sum of the theorem.

It remains to prove the second part, but this follows by extending the $g\U n$ by $0$ on $I^n\setminus S\U n$ and taking $f\U n$ to be the symmetrization of this extension. ◻

Tags: Malliavin calculus
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