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# 数学代写|随机过程Stochastic Porcess代考|Strong Markov Processes

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## 数学代写|随机过程代写Stochastic Porcess代考|Strong Markov Processes

Strong Markov Processes. Let $\left{\xi(t, \omega), \Im_t^s, \mathrm{P}{s, x}\right}$ be a Markov process in the phase space ${\mathscr{X}, \mathfrak{B}}$. In view of the corollary of Theorem 1 in Section 3 , we may assume (as will be done in the present section) that $\widetilde{\Xi}_t^s=\mathfrak{S}_t^s$ and $\mathfrak{B}=\mathfrak{B}^*$. A Markov process is called progressively measurable if for any $s, t, 0 \leqslant s{\mathfrak{v}} \times \mathscr{T}$-measurable function of $(s, x, t)$ on the set $0 \leqslant s \leqslant t<\infty, x \in \mathscr{X}{\mathfrak{v}}$; b) it is progressively measurable; c) for any $s \geqslant 0, t \geqslant 0$ and $f(x) \in \mathbf{B}\left(\mathfrak{B}{\mathrm{v}}\right)$ and an arbitrary Markov time $\tau$ equality
$$E_{s, x}\left{f\left(\xi_{t+\tau}\right) \mid \sigma_\tau^s\right}=E_{\tau, \xi_\tau} f\left(\xi_{t+\tau}\right)$$
is satisfied. The latter equality expresses a strong version of the “independence of the future from the past”. If we set here $\tau=u=$ const, then (4) becomes the condition which expresses the Markov property of a process (cf. Condition 4, Definition 1, Section 3).
We now clarify the meaning of the expression in the r.h.s. of (4). Set
$$g(x, s, t)=\mathrm{E}{s, x} f\left(\xi_t\right) \quad(0 \leqslant s \leqslant t) .$$ Then $$\mathrm{E}{\tau, \xi_\tau} f\left(\xi_{t+\tau}\right)=g\left(\xi_\tau, \tau, t+\tau\right)$$

## 数学代写|随机过程代写Stochastic Porcess代考|Criterion for a strong Markov property

Criterion for a strong Markov property. We now show that in many important cases a Markov process is either strong Markov or can be replaced by an equivalent strong Markov process. Here we shall use the notion of resolvent of a semi group of operators generated by the transition probabilities. This notion is of fundamental importance in the theory of homogeneous Markov processes and will be discussed in detail in Chapter II. In the present section only the definition is presented.

Let $f(x, t) \in \mathrm{B}(\mathfrak{B} \times \mathscr{T})$. Analogously to the above, we shall assume that such a function is defined on $\mathscr{X}{\mathfrak{v}} \times[0, \infty]$ and set $f(\mathfrak{v}, t)=0$; moreover we shall consider $\mathrm{B}(\mathfrak{B} \times \mathscr{T})$ as a corresponding subspace of $\mathrm{B}\left(\mathfrak{B}_v \times \mathscr{T}\right)$. Consider the function $h(x, s, t)=\mathrm{E}{s, x}(f[\xi(s+t), s+t])$. It is easy to verify that in the case when $P(s, x, t, B)$ is a measurable function of the arguments $(s, x, t), s \leqslant t$ the function $h(x, s, t)$ is $\mathfrak{B} \times(\mathscr{T})^2$-measurable. Indeed, it is sufficient to verify this claim for the functions $f(x, t)$ of the form $f(x, t)=f_1(x) g(t)$, where $g \in \mathrm{B}(\mathscr{T}), f_1 \in \mathbf{B}(\mathfrak{B})$. In this case $h(x, s, t)=g(s+t) h_1(x, s, t)$, and moreover $h_1(x, s, t)=\mathrm{E}{s, x} f_1[\xi(s+t)]$ is a $\mathfrak{B} \times(\mathscr{T})^2-$ measurable function in view of the above. Therefore, the same is also true of $h(x, s, t)$. In particular, $h(x, s, t)$ is a $\mathscr{T}$-measurable function of $t$ for fixed $(x, s)$. We set $$\left(\mathbf{R}\lambda f\right)(x, s)=\int_0^{\infty} e^{-\lambda t} \mathrm{E}{s, x} f[\xi(s+t), s+t] d t, \quad \lambda>0$$ It follows from the above that $\left(\mathbf{R}\lambda f\right)(x, s)$ is a $\mathfrak{B} \times \mathscr{T}$-measurable and bounded function for any $\lambda>0$. Therefore under a suitable assumption on the transition probabilities, the operator $\mathbf{R}_\lambda$ maps $B(\mathfrak{B} \times \mathscr{T})$ into itself.

Definition 3. A family of operators $\mathbf{R}\lambda$ is called the resolvent of the family of operators $\mathbf{T}{s t}$ (cf. Section 1).
Note that
$$\left(\mathbf{R}\lambda f\right)(x, s)=\mathrm{E}{s, x} \int_0^{\infty} e^{-\lambda t} f[\xi(s+t), s+t] d t$$

## 数学代写|随机过程代写Stochastic Porcess代考|Strong Markov Processes

$$g(x, s, t)=\mathrm{E}{s, x} f\left(\xi_t\right) \quad(0 \leqslant s \leqslant t) .$$然后 $$\mathrm{E}{\tau, \xi_\tau} f\left(\xi_{t+\tau}\right)=g\left(\xi_\tau, \tau, t+\tau\right)$$

## 数学代写|随机过程代写Stochastic Porcess代考|Criterion for a strong Markov property

$$\left(\mathbf{R}\lambda f\right)(x, s)=\mathrm{E}{s, x} \int_0^{\infty} e^{-\lambda t} f[\xi(s+t), s+t] d t$$

## MATLAB代写

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