## 统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

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## 统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

Processes with independent increments in a separable Banach space. Let $\mathscr{X}$ be a separable Banach space, $\mathfrak{B}$ be the $\sigma$-algebra of its Borel subsets. Then ${\mathscr{X}, \mathfrak{B}}$ satisfies conditions a) and b) stated in the beginning of this Section. Thus we may consider processes with independent increments with values in $\mathscr{X}$.

Let $\xi(t)$ be a process defined on $[a, b)$. Consider the process $\overline{\xi(t)}$ independent of $\xi(t)$ with the same finite-dimensional distributions. Then $\xi(t)-\bar{\xi}(t)=\xi^(t)$ will be a symmetric process with independent increments. We shall assume that $\xi^(t)$ is a separable process. We now show that in such a case, $\xi^(t)$ has no discontinuities of the second kind. First note that for any closed convex set $S$ $$\mathrm{P}\left{\xi^(s) \in S, a \leqslant s \leqslant t\right} \geqslant 1-2 \mathrm{P}\left{\xi^(t) \notin S\right} .$$ Indeed for an arbitrary partition $a=s_0l\left(s_v\right)\right}$ has no common points with $S$ (the existence of such a functional follows from the existence of a “supporting” hyperplane which separates points outside the convex set from the convex set), then denoting by $V$ the half-space ${x: l(x)<0}$ we obtain \begin{aligned} \mathrm{P}{v \leqslant n}= & 1-\mathrm{P}\left{\xi^\left(s_k\right) \in S, k=0, \ldots, n\right} \ & \leqslant \mathrm{P}{v \leqslant n} 2 \mathrm{P}\left{\xi^(t)-\xi^\left(s_v\right) \in V \mid v \leqslant n\right} \leqslant 2 \mathrm{P}\left{\xi^(t) \notin S\right} . \end{aligned} In view of the separability of the process $\xi^(t)$ equation (18) follows from (19). We now construct closed convex compact sets $K_{m, n}$ such that
$$\mathrm{P}\left{\xi^\left(b-\frac{1}{n}\right) \notin K_{m, n}\right} \leqslant \frac{1}{m} .$$ Then $$\mathrm{P}\left{\xi^(s) \in K_{m, n}, a \leqslant s \leqslant b-\frac{1}{n}\right} \geqslant 1-\frac{2}{m} .$$

## 统计代写|随机过程代写stochastic process代考|Some properties of sample functions

Some properties of sample functions. The process $\xi(t)$ is called a step process if it has only a finite number of jumps on each interval $[a, b-\varepsilon]$ and if $\xi(t)$ is constant between the jumps.

Theorem 6. In order for a stochastically continuous process $\xi(t)$ defined on $[a, b)$ and taking values in a separable Banach space $\mathscr{X}$ to be a step process it is necessary and sufficient that its characteristic functional $\varphi_t(l)$ be of the form
$$\varphi_t(l)=\varphi_a(l) \exp \left{\int\left(e^{i l(x)}-1\right) \Pi_t(d x)\right}$$
where $\Pi_t(A)$ for all $t \in[a, b)$ is a finite measure and is a continuous and monotone function of $t$ for $A \in \mathfrak{B}$.

Proof. Let $\xi(t)$ be a step process. Let $\Delta_{\varepsilon}={x:|x|>\varepsilon}$. Then $\xi(t)-\xi(a)=\lim {\varepsilon \rightarrow 0} \xi{\Delta_{\varepsilon}}(t)$ and moreover for $t<b$ there exists the $\lim {\varepsilon \rightarrow 0} v_t\left(\Delta{\varepsilon}\right)=v_t$, where $v_t$ is the total number of jumps of the process $\xi(s)$ up to the moment $t$. Being the limit of Poisson variables, the variable $v_t$ will also have a Poisson distribution. Moreover
$$\mathrm{E} v_t=\lim {\varepsilon \rightarrow 0} \mathrm{E} v_t\left(\Delta{\varepsilon}\right)=\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)$$
Hence, $\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)<\infty$. Set $\lim {\varepsilon \rightarrow 0} \Pi_t\left(\Delta{\varepsilon}\right)=\Pi_t(\mathscr{X})$. Then $\Pi_t(A)$ is defined and is finite for all $A \in \mathfrak{B}$. Furthermore
$$\mathrm{E} e^{i l\left(\zeta_{\Delta_{\varepsilon}}(t)\right)}=\exp \left{\int_{\Delta_{\varepsilon}}\left(e^{i l(x)}-1\right) \Pi_t(d x)\right} .$$

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Processes with independent increments in a separable Banach space

$$\int \alpha^{\infty} 2 \exp \left(-t^2 / 2\right) \mathrm{d} t \leq \frac{1}{\alpha} \int_\alpha^{\infty} 2 t \exp \left(-t^2 / 2\right) \mathrm{d} t=\mathrm{P}(A) / \alpha \leq L \mathrm{P}(A) \alpha$$

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