# Finding the infinitesimal generator of Ornstein-Uhlenbeck process without using a theorem

Consider the 1-dimensional Ornstein-Uhlenbeck process $$X(t)=\mu+e^{-\lambda t}(x-\mu)+\frac{\sigma}{\sqrt{2\lambda}}e^{-\lambda t}W^{(e^{2\lambda t}-1)}$$

with mean $$\mu+e^{-\lambda t}(x-\mu)$$ and variance $$\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})$$. I'd like to show the infinitesimal generator of O-U process is

$$\mathcal{L}f(x)=\lambda(\mu-x)f'(x)dx+\frac{1}{2}\sigma^2f''(x)$$

My idea is first transform the equation by taking derivatives w.r.t. $$t$$,

$$dX(t)=\lambda\int_{0}^{t}(\mu-X(t))dt+\sigma dW(t),X(0)=x$$

Note that $$\int_{0}^{t}e^{\lambda s}dW(s)$$ has the same distribution as $$(2\lambda)^{-1/2}W^{(e^{2\lambda t}-1)}$$, so I can make the adjustment.

Here is a theorem in Oksendal's SDE:

(Theorem 7.5.4) Let $$f\in C^2$$, then $$f\in\mathcal{D}_A$$ and $$\mathcal{A}f=\sum_{i}b_i\frac{\partial{f}}{\partial{x_i}}+\frac{1}{2}\sum_{i,j}(\sigma\sigma)^T_{ij}\frac{\partial^2 f}{\partial x_i\partial x_j}$$.

Using this theorem, we can find the generator immediately. But I'd like to know methods without using this, since we have not covered it in our class yet. We have given the definition of infinitesimal generators:$$\mathcal{L}f=\lim_{h\to 0}\frac{\mathbb{E}[f(W^{(h)})]-f(x)}{h}$$

but using the definition, the calculation will be pretty messy.

A similar question has been asked here before: Infinitesimal Generator for Stochastic Processes, but the answer does not quite solve my problem. Can anyone provide a derivation without using the theorem? Thank you.

• "but using the definition, the calculation will be pretty messy." On the contrary, I would attack this directly using the definition. Do you know Ito's lemma?
– user711689
Commented Mar 6, 2021 at 19:16

Okay, I figured it out by doing Taylor expansion---

Write $$\mathbb{E}_x[f(X^t)]$$ as the following by Taylor expansion: $$\mathbb{E}_x[f(X^t)]\simeq\mathbb{E}_x[f(x)]+f'(x)\mathbb{E}_x[x-X(t)]+\frac{1}{2}f''(x)\mathbb{E}_x[(x-X(t))^2]$$

Compute the expectations for the second and third brackets based on expectation and variance of O-U process, $$\mathbb{E}_x[x-X(t)]=(1-e^{-\lambda t})(x-\mu)$$ $$\mathbb{E}_x[(x-X(t))^2]=x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})+(\mu+e^{-\lambda t}(x-\mu))^2-2x[\mu+e^{-\lambda t}(x-\mu)].$$

We emphasize that $$\mathbb{E}_x[f(x)]=f(x)$$.

By definition of infinitesimal generator,

$$\mathcal{L}f(x)=\lim_{t\to0}\frac{\mathbb{E}_x[f(X^t)]-f(x)}{t}=\lim_{t\to0}\frac{(1-e^{-\lambda t})(x-\mu)f'(x)+\frac{1}{2}\left (x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})+(\mu+e^{-\lambda t}(x-\mu))^2-2x[\mu+e^{-\lambda t}(x-\mu)]\right )f''(x)}{t}$$

Apply L'Hospital Rule, all exponents goes to $$1$$, and by noting that $$\frac{d}{dt}((\mu+e^{-\lambda t}(x-\mu))^2)=-2\lambda e^{-2\lambda t}(x-\mu)(\mu(e^{\lambda t}-1)+x),$$ with also some algebra, $$\mathcal{L}f(x)=\lambda(\mu-x)f'(x)+\frac{\sigma^2}{2}f''(x)$$