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$\mathbf{Definition}$: 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 $\delta=2$?

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    $\begingroup$ 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 $\delta>0$. $\endgroup$
    – Q9y5
    Jul 3 at 14:36

1 Answer 1

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

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jul 3 at 15:08

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