I am reading Normal Approximations with Malliavin Calculus: From Stein's Method to Universality by Stein and Peccati and got stuck on problem 1.7.4.

We define the OU-semigroup of operators to act on $f\in \mathcal S$ (Schwartz space) by the following:

$$P_t f(x)=\int_{-\infty}^{\infty} f(e^{-t}x+\sqrt{1-e^{-2t}}y)d\gamma(y)$$

where $\gamma$ is the standard normal measure on $\Bbb R$.

We define the OU-process as the solution to

$$dX_t=\sqrt 2 dB_t-X_t dt, \ X_0=x$$

I've shown the solution to this is


Prove that $P_tf(x)=E(f(X_t))$ for $f\in \mathcal S$

I applied Ito's fomula to $f(X_t)$ and I got:

\begin{align*}df(X_t)&=f'(X_t)dX_t+\frac{1}{2}f''(X_t)2dt\\ &=f'(X_t)(\sqrt 2 dB_t-X_t dt)+f''(X_t)dt\\ &=(-f'(X_t)X_t+f''(X_t))dt+\sqrt{2}f'(X_t) dB_t\end{align*}

Noting that the generator $L$ of $P_t$ acts on $f$ by $Lf(x)=-xf'(x)+f''(x)$ and that $f(X_0)=x$ we have

$$f(X_t)=f(x)+\int_0^t L f(X_s) ds+\int_0^t \sqrt{2} f'(X_s) dB_s$$

Taking expectations of both sides and using Fubini/martingale property of Ito integral gives:

$$E(f(X_t))=f(x)+\int_0^t E(Lf(X_s)) ds$$

Here I am stuck. We also know that $L=\frac{d}{ds}\Bigg|_{s=0} P_s$ and $L=-\delta D$ if those are of use.


1 Answer 1


Since the integrand

$$s \mapsto e^{-(t-s)}$$

is deterministic, the stochastic integral

$$ \int_0^t e^{-(t-s)} \, dB_s$$

is Gaussian, and so is $X_t$. It is not difficult to see that $$\mathbb{E}(X_t)=e^{-t} x \qquad \text{and} \qquad \text{var}(X_t) = 2 (1-e^{-2t}),$$

and therefore we conclude that $X_t \sim N(e^{-t} x, 2(1-e^{-2t}))$. Expressing $\mathbb{E}f(X_t)$ in terms of the density $p_t$ of $X_t$, $$\mathbb{E}f(X_t) = \int_{\mathbb{R}} f(y) p_t(y) \, dy,$$ it follows now easily that $\mathbb{E}f(X_t) = P_t f(x)$.

  • $\begingroup$ Of course, it's not so bad. Thanks. $\endgroup$
    – user223391
    Jun 6, 2018 at 13:38

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