# 数学代写|随机过程统计代写Stochastic process statistics代考|STAT4061

## 数学代写|随机过程统计代写Stochastic process statistics代考|The semigroup

A semigroup $\left(P_t\right){t \geqslant 0}$ on a Banach space $(\mathfrak{B},|\cdot|)$ is a family of linear operators $P_t: \mathfrak{B} \rightarrow \mathfrak{B}, t \geqslant 0$, which satisfies $$P_t P_s=P{t+s} \quad \text { and } \quad P_0=\mathrm{id} .$$
In this chapter we will consider the following Banach spaces:
$\mathcal{B}b\left(\mathbb{R}^d\right)$ the family of all bounded Borel measurable functions $f: \mathbb{R}^d \rightarrow \mathbb{R}$ equipped with the uniform norm $|\cdot|{\infty}$;
$\mathcal{C}{\infty}\left(\mathbb{R}^d\right)$ the family of all continuous functions $f: \mathbb{R}^d \rightarrow \mathbb{R}$ vanishing at infinity, i. e. $\lim {|x| \rightarrow \infty} u(x)=0$, equipped with the uniform norm $|\cdot|_{\infty}$.
Unless it is ambiguous, we write $\mathcal{B}b$ and $\bigodot{\infty}$ instead of $\mathcal{B}b\left(\mathbb{R}^d\right)$ and $\bigodot{\infty}\left(\mathbb{R}^d\right)$.

7.1 Lemma. Let $\left(B_t\right){t \geqslant 0}$ be a d-dimensional Brownian motion with filtration $\left(\mathcal{F}_t\right){t \geqslant 0}$. Then (7.1) defines a semigroup of operators on $\mathcal{B}_b\left(\mathbb{R}^d\right)$.

Proof. It is obvious that $P_t$ is a linear operator. Since $x \mapsto e^{-|x-y|^2 / 2 t}$ is measurable, we see that
$$P_t u(x)=\int_{\mathbb{R}^d} u(y) \mathbb{P}^x\left(B_t \in d y\right)=\frac{1}{(2 \pi t)^{d / 2}} \int_{\mathbb{R}^d} u(y) e^{-|x-y|^2 / 2 t} d y$$
is again a Borel measurable function. (For a general Markov process we must use (6.4c) with a Borel measurable $\left.g_{u, t}(\cdot)\right)$ The semigroup property (7.3) is now a direct consequence of the Markov property (6.4): For $s, t \geqslant 0$ and $u \in \mathcal{B}b\left(\mathbb{R}^d\right.$ ) \begin{aligned} P{t+s} u(x)=\mathbb{E}^x u\left(B_{t+s}\right) & \stackrel{\text { tower }}{=} \mathbb{E}^x\left[\mathbb{E}^x\left(u\left(B_{t+s}\right) \mid \mathcal{F}s\right)\right] \ & \stackrel{(6.4)}{=} \mathbb{E}^x\left[\mathbb{E}^{B_s}\left(u\left(B_t\right)\right)\right] \ &=\mathbb{E}^x\left[P_t u\left(B_s\right)\right] \ &=P_s P_t u(x) \end{aligned} For many properties of the semigroup the Banach space $\mathcal{B}_b\left(\mathbb{R}^d\right)$ is too big, and we have to consider the smaller space $\mathcal{C}{\infty}\left(\mathbb{R}^d\right)$.

## 数学代写|随机过程统计代写Stochastic process statistics代考|The generator

Let $\phi:[0, \infty) \rightarrow \mathbb{R}$ be a continuous function which satisfies the functional equation $\phi(t) \phi(s)=\phi(t+s)$ and $\phi(0)=1$. Then there is a unique $a \in \mathbb{R}$ such that $\phi(t)=e^{a t}$. Obviously,
$$a=\left.\frac{d^{+}}{d t} \phi(t)\right|{t=0}=\lim {t \rightarrow 0} \frac{\phi(t)-1}{t} .$$
Since a strongly continuous semigroup $\left(P_t\right){t \geqslant 0}$ on a Banach space $(\mathfrak{B},|\cdot|)$ also satisfies the functional equation $P_t P_s u=P{t+s} u$ it is not too wild a guess that – in an appropriate sense $-P_t u=e^{t A} u$ for some operator $A$ in $\mathfrak{B}$. In order to keep things technically simple, we will only consider Feller semigroups.
7.8 Definition. Let $\left(P_t\right){t \geqslant 0}$ be a Feller semigroup on $\mathcal{C}{\infty}\left(\mathbb{R}^d\right)$. Then
$$\begin{gathered} A u:=\lim {t \rightarrow 0} \frac{P_t u-u}{t} \quad\left(\text { the limit is taken in }\left(\mathcal{C}{\infty}\left(\mathbb{R}^d\right),|\cdot|_{\infty}\right)\right) \ \mathfrak{D}(A):=\left{u \in \mathcal{C}{\infty}\left(\mathbb{R}^d\right) \mid \exists g \in \mathcal{C}{\infty}\left(\mathbb{R}^d\right): \lim {t \rightarrow 0}\left|\frac{P_t u-u}{t}-g\right|{\infty}=0\right} \end{gathered}$$
is the (infinitesimal) generator of the semigroup $\left(P_t\right){t \geqslant 0}$. Clearly, (7.9a) defines a linear operator $A: \mathfrak{D}(A) \rightarrow \mathcal{C}{\infty}\left(\mathbb{R}^d\right)$.
7.9 Example. Let $\left(B_t\right){t \geqslant 0}, B_t=\left(B_t^1, \ldots, B_t^d\right)$, be a $d$-dimensional Brownian motion and denote by $P_t u(x)=\mathbb{E}^x u\left(B_t\right)$ the transition semigroup. Then $$\begin{gathered} \mathcal{C}{\infty}^2\left(\mathbb{R}^d\right):=\left{u \in \mathcal{C}{\infty}\left(\mathbb{R}^d\right): \partial_j u, \partial_j \partial_k u \in \mathcal{C}{\infty}\left(\mathbb{R}^d\right),\right. \ j, k=1, \ldots, d} \subset \mathfrak{D}(A) \end{gathered}$$
where
$$A=\frac{1}{2} \Delta=\frac{1}{2} \sum_{j=1}^d \partial_j^2$$
and
$$\partial_j=\partial / \partial x_j$$

## 数学代写|随机过程统计代写Stochastic process statistics代考|The semigroup

$$P_t P_s=P t+s \quad \text { and } \quad P_0=\mathrm{id} .$$

$\mathcal{C} \infty\left(\mathbb{R}^d\right)$ 所有连续函数的族 $f: \mathbb{R}^d \rightarrow \mathbb{R}$ 在无穷远处消失，即 $\lim |x| \rightarrow \infty u(x)=0$, 配备统一范数 $|\cdot|{\infty}$ 除非它是模棱两可的，我们写 $\mathcal{B} b$ 和 $\odot \infty$ 代替 $\mathcal{B} b\left(\mathbb{R}^d\right)$ 和 $\odot \infty\left(\mathbb{R}^d\right)$. $7.1$ 引理。让 $\left(B_t\right) t \geqslant 0$ 是带过滤的 $\mathrm{d}$ 维布朗运动 $\left(\mathcal{F}_t\right) t \geqslant 0$. 然后 $(7.1)$ 定义了一组运算符 $\mathcal{B}_b\left(\mathbb{R}^d\right)$. 证明。很明显， $P_t$ 是一个线性算子。自从 $x \mapsto e^{-|x-y|^2 / 2 t}$ 是可测量的，我们看到 $$P_t u(x)=\int{\mathbb{R}^d} u(y) \mathbb{P}^x\left(B_t \in d y\right)=\frac{1}{(2 \pi t)^{d / 2}} \int_{\mathbb{R}^d} u(y) e^{-|x-y|^2 / 2 t} d y$$

$P t+s u(x)=\mathbb{E}^x u\left(B_{t+s}\right) \stackrel{\text { tower }}{=} \mathbb{E}^x\left[\mathbb{E}^x\left(u\left(B_{t+s}\right) \mid \mathcal{F} s\right)\right] \quad \stackrel{(6.4)}{=} \mathbb{E}^x\left[\mathbb{E}^{B_s}\left(u\left(B_t\right)\right)\right]=\mathbb{E}^x\left[P_t u\left(B_s\right)\right]$

## 数学代写|随机过程统计代写Stochastic process statistics代考|The generator

$$a=\frac{d^{+}}{d t} \phi(t) \mid t=0=\lim t \rightarrow 0 \frac{\phi(t)-1}{t} .$$

$7.8$ 定义 $。$ 让 $\left(P_t\right) t \geqslant 0$ 是一个 Feller 半群C $\infty\left(\mathbb{R}^d\right)$. 然后

$7.9$ 示例。让 $\left(B_t\right) t \geqslant 0, B_t=\left(B_t^1, \ldots, B_t^d\right)$ ，成为 $d$ 维布朗运动并表示为 $P_t u(x)=\mathbb{E}^x u\left(B_t\right)$ 过渡半 群。然后

$$A=\frac{1}{2} \Delta=\frac{1}{2} \sum_{j=1}^d \partial_j^2$$

$$\partial_j=\partial / \partial x_j$$

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