# Question in a theorem about diffusion process

$$\mathbf{Definition}$$: A continuous $$n$$-dimensional Markov process with transition probability function $$p(s, x, t, A)$$ is called a diffusion process if:
(i) for any $$\epsilon>0, t \geqslant 0, x \in R^{n}$$, $$\lim _{h \downarrow 0} \frac{1}{h} \int_{|y-x|>\epsilon} p(t, x, t+h, d y)=0 \quad (1)$$ (ii) there exist an $$n$$-vector $$b(x, t)$$ and an $$n \times n$$ matrix $$a(x, t)$$ such that for any $$\epsilon>0, t \geqslant 0, x \in R^{n}$$, $$\lim _{h \downarrow 0} \frac{1}{h} \int_{|y-x|<\epsilon}\left(y_{i}-x_{i}\right) p(t, x, t+h, d y)=b_{i}(x, t)\quad(1 \leqslant i \leqslant n) \quad (2)$$ $$\lim _{h \downarrow 0} \frac{1}{h} \int_{|y-x|<\epsilon}\left(y_{i}-x_{i}\right)\left(y_{j}-x_{i}\right) p(t, x, t+h, d y)=a_{i j}(x, t)\quad (1 \leqslant i, j \leqslant n) \quad (3)$$ where $$b=\left(b_{1}, \ldots, b_{n}\right), a=\left(a_{i j}\right) .$$ The vector $$b$$ is called the drift coefficient and the matrix $$a$$ is called the diffusion coefficient.

$$\mathbf{Lemma}$$ The following conditions imply the conditions (i), (ii):
(i*) for some $$\delta>0, t \geqslant 0, x \in R^{n}$$ $$\lim _{h \downarrow 0} \frac{1}{h} \int_{R^{n}}|x-y|^{2+\delta} p(t, x, t+h, d y)=0\quad(1')$$ (ii*) for any $$t \geqslant 0, x \in R^{n}$$, $$\lim _{h \downarrow 0} \frac{1}{h} \int_{R^{n}}\left(y_{i}-x_{i}\right) p(t, x, t+h, d y)=b_{i}(x, t) \quad(1 \leqslant i \leqslant n) \quad(2')$$ $$\lim _{h \downarrow 0} \frac{1}{h} \int_{R^{n}}\left(y_{i}-x_{i}\right)\left(y_{j}-x_{t}\right) p(t, x, t+h, d y)=a_{i j}(x, t) \quad (1 \leqslant i, j \leqslant n)\quad(3') .$$

$$\mathbf{Theorem}$$ Let $$b(x, t), \sigma(x, t)$$ be measurable continuous in $$(x, t) \in R^{n} \times[0, \infty)$$ and satisfy Lipschitz and linear growth condition. $$\xi_0$$ is independent of $$\mathcal{F}(w(t),t\ge 0)$$ and $$E|\xi_0|^2<\infty$$. Then the solution of $$d\xi(t)=b(t,\xi(t))dt+\sigma(t,\xi(t))dw(t)$$ $$\xi(0)=\xi_0 \quad a.s.$$ is a diffusion process with drift $$b(x, t)$$ and diffusion matrix $$a(x, t)=\sigma(x, t) \sigma^{*}(x, t)$$.

In the proof of the theorem here, I am confused about how to show that the limit in (2) and (3) exists. Here we are given a lemma so that we can check (1')-(3') instead of checking the condition (1)-(3) from definition. However, in the proof of Theorem, author does not check condition (1'). So I wonder how the assumption (1') is satisfied here? (Apparently (1) and (1') is very different. Is the reason behind related to the second moment of $$\xi(0)$$ is finite?

Supplement: In the proof of the theorem, I notice that we have such inequality $$E(|\xi(t)-\xi_0|^4)\le C(t-0)^2$$ so, does it mean that (1') is satisfied since 4th moment is finite implies that we can choose $$\delta=2$$?

• Yes, it implies (1') for $\delta=2$. In fact, if $w(t)$ means a Brownian motion, then by Burkholder-Davis-Gundy inequality, $\xi$ is 1/2-Hölder continuous under the $L_{p}$ norm for all $p\geq2$, hence (1') naturally holds with any $\delta>0$.
– Q9y5
Jul 3 at 14:36

Yes, $$\delta=2$$ here so that 1’ is satisfied.