Find the generator of Markov Process

Question

Homework question:

Consider the Markov process $X_t=B_t-t^2+t$ where $B_t$ is the Brownian motion. Find the generator $Q$ of this process.

I am completely confused how to find the generator for this.
Any help please?

Answer 1

I think the trick is usually to find an Ito diffusion that is relevant and augment it by $t$; then you can use the results for Ito diffusions.

So for example $\mathrm{d}Y_t = \mathrm{d}B_t + \mathrm{d}t$ is an Ito diffusion and we have $Y_t = B_t + t$. So then lets look at

$$Z_t = \left( \begin{array}{c} Y_t \\ t \\ \end{array} \right)$$ We know the generator of the above because it's a two dimensional Ito diffusion.

$$ A f(y,t) = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial y}+\frac{1}{2} \frac{\partial^2 f}{\partial y^2}$$

Then we have

$$ Q_{t_0} g(x) = \lim_{s\downarrow t_0} \frac{E^x_{t_0}[g(X_s)]-g(x)}{s-t_0} = \left(\lim_{h \downarrow 0}\frac{E^{y,t}[g(Y_h-(t+h)^2)]-g(y-t^2)}{h}\right)_{y = x+t_0^2,t=t_0}$$ $$ =\left(A g(y-t^2)\right)_{y = x+t_0^2,t=t_0}$$ $$ =-2 t_0 g'(x) + g'(x) + \frac{1}{2}g''(x)$$

Answer 2

First of all, we note that

$$X_t := B_t-t^2+t = \int_0^t \, dB_s + \int_0^t (-2s+1) \, ds.$$

Therefore, we find by Itô's formula for any $f \in C^2$

$$\begin{align*} f(X_t)-f(X_{t_0}) &= \int_{t_0}^t f'(X_s) \, dX_s + \frac{1}{2} \int_{t_0}^t f''(X_s) \, d\langle X \rangle_s \\ &= \int_{t_0}^t f(X_s) \, dB_s + \int_{t_0}^t f'(X_s) (-2s+1) \, ds + \frac{1}{2} \int_{t_0}^t f''(X_s) \, ds. \end{align*}$$

As $\int_{t_0}^t f(X_s) \, dB_s$ is a stochastic integral, hence a martingale, we see that

$$\begin{align*} &\frac{1}{t-t_0} \bigg(\mathbb{E}^{t_0,x}f(X_t)-f(x)\bigg) \\ &= \frac{1}{t-t_0} \int_{t_0}^t \mathbb{E}^{t_0,x}(f'(X_s)) (-2s+1) \, ds + \frac{1}{2(t-t_0)} \int_0^t \mathbb{E}^{t_0,x}(f''(X_s)) \, ds. \end{align*}$$

Since both $s \mapsto \mathbb{E}^{t_0,x}(f'(X_s))$ and $s \mapsto \mathbb{E}^{t_0,x}(f''(X_s))$ are continuous functions, the fundamental theorem of calculus yields

$$\begin{align*} \lim_{t \to t_0} \frac{1}{t-t_0} \int_{t_0}^{t} \mathbb{E}^{t_0,x}(f'(X_s)) (-2s+1) \, ds &= \mathbb{E}^{t_0,x}(f'(X_{t_0})) (-2 \cdot t_0+1) \\ &= f'(x)(-2t_0+1) \\ \lim_{t \to t_0} \frac{1}{2t-t_0} \int_{t_0}^t \mathbb{E}^{t_0,x}(f''(X_s))&= \frac{1}{2} \mathbb{E}^{t_0,x}(f''(X_{t_0})) \\ &= \frac{1}{2} f''(x) \end{align*}$$

Hence,

$$Q_{t_0}f(x) = \lim_{t \downarrow t_0} \frac{1}{t-t_0} \bigg(\mathbb{E}^{t_0,x}f(X_t)-f(x)\bigg) \to f'(x)(-2t_0+1)+ \frac{1}{2} f''(x).$$