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The Ornstein-Uhlenbeck Semigroup

Part 7 of the series on Malliavin calculus

By L.Llamazares-Elias
Posted on November 16, 2022

Three line summary

  • There is a natural extension of the Laplacian to the Wiener space.

  • The generator of the Laplacian is the Ornstein-Uhlenbeck semigroup.

  • The Ornstein-Uhlenbeck semigroup in finite dimensions is the generator of the Ornstein-Uhlenbeck process, from which it derives its name.

    The Laplacian of a random variable

    First, we give some finite-dimensional motivation. Suppose that $f\in C\U c^\infty({\mathbb R}^d\to{\mathbb R}^d)$ and $g\in C\U c^\infty({\mathbb R}^d)$. Then an integration by parts shows that the adjoint of the gradient in $L^2({\mathbb R}^d)$ is minus the divergence. That is,

    $$\int\U \mathbb{R}^df(x) \cdot \nabla g(x) dx=-\int\U \mathbb{R}^d\nabla\cdot f(x) \nabla g(x) dx.$$

    Then, we define the Laplacian as minus the adjoint of the gradient $\nabla$ composed with the gradient

    $$\Delta := -\nabla^\circ \nabla .$$

    Which gives the familiar

    $$\Delta\U \mathbb{R}^d =\nabla\cdot \nabla=\partial\U 1^2+\ldots\partial \U d^2.$$

Of course, this is all well and good when the domain of $f,g$ is a finite-dimensional space. Otherwise, there is no Lebesgue measure. We now move to what is our base case in our series of blog posts and consider a probability space $(\Omega,\mathbb{P},\mathcal{F}\U t)$ where $\mathcal{F}\U t$ is generated by a Wiener process $W\U t$. Then, as we have seen previously the Skorohod integral $\delta$ is the adjoint of the Malliavin derivative $D$ so we would like to define

$$\Delta := -\delta \circ D.$$

On what kind of random variables can we define this? Well let us take $X=\sum\U {n=0}^{\infty} I\U n(f\U n)$ with a rapidly decaying chaos expansion, then

$$\Delta X=-\delta (DX)=-\delta \left(\sum\U {n=1}^\infty nI\U {n-1}(f\U n(\cdot ,t))\right)=-\sum\U {n=1}^\infty nI\U {n}(f\U n).$$

All we require for this expression to make sense is that- the right-hand side is in $L^2(\Omega)$. That is, by Ito’s $n$-th isometry, that

$$\sum\U {n=0}^\infty n^2 \norm{f\U n}\U {L^2(I\U n)}< \infty.$$

Is this a space we’ve dealt with before? Well if we recall our old spaces $\mathbb{D}^{k,p}$. Then we have that

$$\begin{gathered} \int\U {I^2}\norm{D\U {t,s}X}\U {L^2(\Omega)}^2 ds d t=\int\U {I^2}\norm{\sum\U {n=2}^\infty n(n-1)I\U {n-2}(f\U n(\cdot ,s,t))}\U {L^2(\Omega)}^2\\=\int\U {I^2}\sum\U {n=2}^\infty n^2(n-1)^2(n-2)!\norm{f\U n(\cdot ,s,t)}\U {L^2(I\U {n-2})}^2=\sum\U {n=2}^\infty n(n-1)n!\norm{f\U n(\cdot ,s,t)}\U {L^2(I\U n)}^2. \end{gathered}$$

Where analogous calculations go through if we have more derivatives to get the terms $n(n-1)\cdots (n-(k-1))$. This shows that

$$\mathbb{D}^{k,p}:=\left\{X\in L^2(\Omega):\quad \norm{X}\U {\mathbb{D}^{k,2}}=\sum\U {n=0}^\infty n^kn! \norm{f\U n}\U {L^p(I\U n)}< \infty\right\} .$$

Thus, the domain of $\Delta$ is exactly $\mathbb{D}^{2,2}$. This is quite pleasing as, as we have observed earlier, the spaces $\mathbb{D}^{k,p}$ mimic the Sobolev spaces $W^{k,p}$, when $p=2$ this resemblance is quite strong as we have that the norm on $H^k:=W^{k,2}$ is

$$\norm{f}\U {H^k{\mathbb R}^d)}=\int\U \mathbb{R}^d\left\langle\xi \right\rangle^k \hat{f}(\xi )^2d\xi .$$

Which is formally equal to the one just derived for $\mathbb{D}^{k,2}.$ It is very interesting to observe that, directly from the definition, we obtain a basis of eigenvalues of $\Delta$. Let us define

$$H\U n:=\{X\in L^2(\Omega): X=I\U n(f\U n),\quad \text{for some } f\U n \in L^2\U S(I^n) \} .$$

That is, $H\U n$ are the random variables that only have the $n$-th term in their chaos expansion to be non-zero. Then by the chaos expansion theorem, we know that

$$L^2(\Omega)=\overline{\oplus\U {n=0}^\infty H\U n}.$$

And by construction of the Laplacian, $\Delta e\U n=n e\U n$ for every $e\U n \in H\U n$. In fact, by the uniqueness of the chaos expansion, the elements of $H\U n$ for some $n \in \mathbb{N}$ are the unique eigenvectors of $\Delta .$

The Ornstein-Uhlenbeck semigroup

As it turns out, $\Delta$ defines a semigroup

Definition 1. The Ornstein-Uhlenbeck semigroup is the family of operators $\Phi(t):L^2(\Omega)\to L^2(\Omega)$

$$\Phi(t)X:=\sum\U {n=0}^{\infty} e^{-nt}I\U n(f\U n), \quad\forall t\in I.$$

The term $e^{-nt}$ is quite reminiscent of the semigroup for the heat equation

$$e^{t\Delta }u\U 0:=\int\U \mathbb{R}^de^{-4 \pi^2 \xi^2t}\widehat{u\U 0}(\xi ) d\xi,$$

and will cause an analogous smoothing effect by making the terms in the chaos expansion to decrease faster. To see that $\Phi$ defines a semigroup first note that, by the linearity of the iterated integrals,

$$\Phi(t)X:=\sum\U {n=0}^{\infty} I\U n(e^{-nt}f\U n).$$

So as a result

$$\Phi(t+s)X=\sum\U {n=0}^{\infty} e^{-nt}I\U n(e^{-ns}f\U n)=\sum\U {n=0}^{\infty} \Phi(t)\Phi(s)X.$$

Which shows that $\Phi(t+s)=\Phi(t)\circ \Phi(s)$. Finally, note that

$$\frac{\Phi(t)X-X}{t}=\sum\U {n=0}^{\infty} \left(\frac{e^{-nt}-1}{nt} \right)nI\U n(f\U n)\to -\sum\U {n=0}^{\infty} nI\U n(f\U n)=\Delta X \in L^2(\Omega) .$$

Where the commutation under the integral sign (with the counting measure) is justified as $(e^{-nt}-1)/(nt)$ is uniformly bounded in $n$. There’s an explicit formula for $\Phi(t)$. Proposition 1 (Mehler’s formula). Let $(\Omega,\mathcal{F}\U t,\gamma )$ be the Wiener space, then

$$\Phi(t)X(\omega)=\int\U {\Omega}X\left(e^{-t}\omega+\sqrt{1-e^{-2t}}\eta\right) \gamma (\eta)\in L^2(\Omega).$$

The proof is technical and can be found in Nualart’s book 1 on page 74. Let us try to understand the formula and also the reason for the name of the semigroup. We consider as at the beginning of this post the finite-dimensional case but now with some Gaussian measure $\mu$

$$\mu (A):=\int\U {A}e^{-\frac{\norm{x}^2}{2} } dx.$$

Then, integration by parts shows that

$$\begin{aligned} \int\U \mathbb{R}^df(x) \cdot \nabla g(x) d\mu(x) & =-\int\U \mathbb{R}^d\nabla\cdot \left(e^{-\frac{\norm{x}^2}{2} }f(x)\right) \nabla g(x) dx \\&=\int\U \mathbb{R}^d(x\cdot f(x)-\nabla\cdot f(x)) d\mu (x). \end{aligned}$$

That is, the adjoint of the gradient in $L^2({\mathbb R}^d,\mu )$ is $x\cdot -\nabla\cdot$. Notice that we get the extra term that corresponds to multiplication by $x\cdot$.As a result, the Laplacian on $L^2({\mathbb R}^d, \mu )$ is given by

$$\Delta\U \mu =\nabla\cdot \nabla-x\cdot \nabla .$$

Furthermore, by Itô’s formula, $\Delta\U \mu$ is the generator of the SDE

$$dX(t)=-X(t)d t+ \sqrt{2}dW(t)$$

Let us write $X\U x$ for the solution to the above SDE with initial data $x \in {\mathbb R}^d$ That is, if we define

$$P\U tX(x):=E[\varphi(X(t))],$$

then

$$\partial \U tP\U tX(x)=\Delta\U \mu P\U tX(x) .$$

The process $X$ that solves the SDE above is known as the Ornstein-Uhlenbeck process and, by the theory of linear SDEs, is given by

$$X\U x(t)=e^{-t}x+\sqrt{2} \int\U {0}^te^{s-t} dW(t).$$

Since

$$\sqrt{2} \int\U {0}^te^{s-t} dW(t)\sim \sqrt{2} e^{-t}\mathcal{N}\left(0,\norm{e^\cdot }^2\U {L^2([0,t])}\right)=\sqrt{1-e^{-2t}}\mathcal{N}(0,1)$$

We deduce that for each fixed $t$ we can find a measure $\gamma \sim \mathcal{N}(0,1)$ with

$$X(t)=e^{-t}x+\sqrt{1-e^{-2t}}\gamma .$$

We then get that

$$P\U t \varphi(x)=\mathbb{E}\left[\varphi\left(e^{-t}x+\sqrt{1-e^{-2t}}\gamma \right)\right]$$

And by taking $\varphi=Id$ we recover Mehler’s formula. This correspondence is expanded on in chapter $7$ of Hairer’s notes 2.

Tags: Malliavin calculus
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