统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Measurable space

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统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Measurable space

Borel $\sigma$-algebra $\beta\left(R^{n}\right)$. Let
$$R^{n}=\underbrace{R \times R \times \ldots \times R}{n}=\left{\left(x{1}, x_{2}, \ldots, x_{n}\right): x_{1} \in R, \ldots, x_{n} \in R\right}$$
be a direct (Cartesian) product of $n$ exemplars (copies) of the number scale $R$.
We define the system:
$$J^{(n)}=\left{I_{1} \times I_{2} \times \ldots \times I_{n}: I_{i}=\left(a_{i}, b_{i}\right], a_{i}<b_{i}, i=1,2, \ldots, n\right}$$
The system $J^{(n)}$ forms an algebra.
The smallest $\sigma$-algebra $\sigma\left(J^{(n)}\right)$ that contains $J^{(n)}$ is called a Borel $\sigma$-algebra on $R^{n}$ and will be denoted by $\beta\left(R^{n}\right)$ :
$$\beta\left(R^{n}\right)=\sigma\left(J^{(n)}\right)$$
Elements of $\sigma$-algebra $\beta\left(R^{n}\right)$ are called the $n$ dimensional Borel sets (or Borel sets on $R^{n}$ ).

We will show that the definition of a Borel $\sigma$-algebra $\beta\left(R^{n}\right)$ could be obtained differently.

Indeed, along with rectangles $I^{(n)}=I_{1} \times I_{2} \times \ldots \times I_{n}$ we consider rectangles $B^{(n)}=B_{1} \times B_{2} \times \ldots \times B_{n}$ with Borel sides $\left(B_{k}\right.$ is a Borel set on a number scale, standing in the $k$-th place in the direct product $\underbrace{R \times R \times \ldots \times R}{n}$. The smallest $\sigma$-algebra containing all possible rectangles with Borel sides is denoted by $$\beta^{(n)}=\underbrace{\beta(R) \otimes \beta(R) \otimes . . \otimes \beta(R)}{n}$$
and is called a direct product of $\sigma$-algebras $\beta(R)$ :
$$\beta^{(n)}=\beta(R) \otimes \beta(R) \otimes \ldots \otimes \beta(R)=\left{B_{1} \times B_{2} \times \ldots \times B_{n}: B_{i} \in \beta(R)\right}$$
We show that in fact $\beta^{(n)}=\beta\left(R^{n}\right)$, in other words, the smallest $\sigma$-algebras, generated by the rectangles $I^{(n)}=I_{1} \times I_{2} \times \ldots \times I_{n}$, coinside with the $\sigma$-algebras, generated by the wider class of rectangles $B^{(n)}=B_{1} \times B_{2} \times \ldots \times B_{n}$ with Borel sides.
To prove this statement, i.e., the equality $\beta^{(n)}=\beta\left(R^{n}\right)$, we will first prove an auxiliary lemma.

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Measurable space

Borel $\sigma$-algebra $\beta\left(R^{\infty}\right)$. This $\sigma$-algebra plays a significant role in the probability theory, since it serves as a basis for constructing probabilistic models of experiments with an infinite number of outcomes.
Let
$$R^{\infty}=\left{x=\left(x_{1}, x_{2}, \ldots\right):-\infty<x_{k}<\infty, k=1,2, \ldots\right}=R \times R \times \ldots$$
Let’s denote by $I_{k}, B_{k}$ (respectively) the intervals $\left(a_{k}, b_{k}\right]$ and Borel sets of the $k$-th number scale (with the coordinates $x_{k}$ ).
Let’s consider cylindrical sets
\begin{aligned} &J\left(I_{1} \times \ldots \times I_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\right): x_{1} \in I_{1}, x_{2} \in I_{2}, \ldots, x_{n} \in I_{n}\right} \ &J\left(B_{1} \times \ldots \times B_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\right): x_{1} \in B_{1}, x_{2} \in B_{2}, \ldots, x_{n} \in B_{n}\right} \ &J\left(B^{(n)}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\right),\left(x_{1}, x_{2}, \ldots x_{n}\right) \in B^{(n)}, \quad B^{(n)}=B_{1} \times \ldots \times B_{n}\right} \end{aligned}
We can consider each of the cylinders $J\left(B_{1} \times \ldots \times B_{n}\right)$ or $J\left(B^{(n)}\right)$ as a cylinders with bases in $R^{n+1}, R^{n+2}, \ldots$ :
\begin{aligned} J\left(B_{1} \times \ldots \times B_{n}\right) &=J\left(B_{1} \times \ldots \times B_{n} \times R\right) \ J\left(B^{(n)}\right) &=J\left(B^{(n)} \times R\right) \end{aligned}
etc.
It follows that both systems of cylinders $J\left(B_{1} \times \ldots \times B_{n}\right)$ and $J\left(B^{(n)}\right)$ form algebras.

It is not difficult to verify that sets composed of unions of disjoint cylinders $J\left(I_{1} \times \ldots \times I_{n}\right)$ also form an algebra.

We denote by $\beta\left(R^{\infty}\right), \beta_{1}\left(R^{\infty}\right)$ and $\beta_{2}\left(R^{\infty}\right)$ the smallest algebras containing all the sets of the forms (1), (2), (3), respectively.
It is clear that
$$\beta\left(R^{\infty}\right) \subseteq \beta_{1}\left(R^{\infty}\right) \subseteq \beta_{2}\left(R^{\infty}\right)$$
Let’s show that in fact these three $\sigma$-algebras coincide.

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Tasks for independent work

1. Let’s introduce on the number scale $R=(-\infty,+\infty)$ the metrics:
$$\text { for } x, y \in R, \rho_{1}(x, y)=\frac{|x-y|}{1+|x-y|} \text {. }$$
a) Prove that $\rho_{1}(x, y)$ is really a metrics.
b) Let’s denote by $\beta_{0}(R)$ the smallest $\sigma$-algebra, generated by the open sets
$$S_{\rho}\left(x^{0}\right)=\left{x \in R: \rho_{1}\left(x, x^{0}\right)<\rho, \rho>0, x^{0} \in R\right} .$$
Prove that $\beta_{0}(R)=\beta(R)$.
2. Let $\beta_{0}\left(R^{n}\right)$ be the smallest $\sigma$-algebra, generated by the open sets
$$S_{\rho}\left(x^{0}\right)=\left{x \in R^{n}: \rho_{n}\left(x, x^{0}\right)<\rho, x^{0} \in R^{n}, \rho>0\right}$$
in metrics
$$\begin{gathered} \rho_{n}\left(x, x^{0}\right)=\sum_{k=1}^{n} 2^{-k} \rho_{1}\left(x_{k}, x_{k}^{0}\right), \quad x=\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in R^{n}, \quad x^{0}=\left(x_{1}^{0}, x_{2}^{0}, \ldots, x_{n}^{0}\right) \in R^{n}, \ \rho_{1}\left(x_{k}, x_{k}^{0}\right)=\frac{\left|x_{k}-x_{k}^{0}\right|}{1+\left|x_{k}-x_{k}^{0}\right|} . \end{gathered}$$
Prove that $\beta_{0}\left(R^{n}\right)=\beta\left(R^{n}\right)$.
3. For all points $x, x_{0} \in R^{\infty}$ the distance between them is defined by the formula:
$$\rho_{\infty}\left(x, x^{0}\right)=\sum_{k=1}^{\infty} 2^{-k} \frac{\left|x_{k}-x_{k}^{0}\right|}{1+\left|x_{k}-x_{k}^{0}\right|} .$$
Let $\beta_{0}\left(R^{\infty}\right)$ be the smallest $\sigma$-algebra, generated by the open sets
$$S_{\rho}\left(x^{0}\right)=\left{x \in R^{\infty}: \rho_{\infty}\left(x, x^{0}\right)<\rho, x^{0} \in R^{\infty}, \rho>0\right} .$$

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Measurable space

$$R^{n}=\underbrace{R \times R \times \ldots \times R} {n}=\left{\left(x {1}, x_{2}, \ldots, x_{n }\right): x_{1} \in R, \ldots, x_{n} \in R\right} b和一种d一世r和C吨(C一种r吨和s一世一种n)pr这d在C吨这Fn和X和米pl一种rs(C这p一世和s)这F吨H和n在米b和rsC一种l和R.在和d和F一世n和吨H和s是s吨和米: J^{(n)}=\left{I_{1} \times I_{2} \times \ldots \times I_{n}: I_{i}=\left(a_{i}, b_{i}\对], a_{i}<b_{i}, i=1,2, \ldots, n\right} 吨H和s是s吨和米Ĵ(n)F这r米s一种n一种lG和br一种.吨H和s米一种ll和s吨σ−一种lG和br一种σ(Ĵ(n))吨H一种吨C这n吨一种一世nsĴ(n)一世sC一种ll和d一种乙这r和lσ−一种lG和br一种这nRn一种nd在一世llb和d和n这吨和db是b(Rn): \beta\left(R^{n}\right)=\sigma\left(J^{(n)}\right)$$

\beta^{(n)}=\beta(R) \otimes \beta(R) \otimes \ldots \otimes \beta(R)=\left{B_{1} \times B_{2} \times \ldots \times B_{n}: B_{i} \in \beta(R)\right}\beta^{(n)}=\beta(R) \otimes \beta(R) \otimes \ldots \otimes \beta(R)=\left{B_{1} \times B_{2} \times \ldots \times B_{n}: B_{i} \in \beta(R)\right}

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Measurable space

R^{\infty}=\left{x=\left(x_{1}, x_{2}, \ldots\right):-\infty<x_{k}<\infty, k=1,2, \ ldots\right}=R \times R \times \ldotsR^{\infty}=\left{x=\left(x_{1}, x_{2}, \ldots\right):-\infty<x_{k}<\infty, k=1,2, \ ldots\right}=R \times R \times \ldots

\begin{对齐} &J\left(I_{1} \times \ldots \times I_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\右): x_{1} \in I_{1}, x_{2} \in I_{2}, \ldots, x_{n} \in I_{n}\right} \ &J\left(B_{1} \times \ldots \times B_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\right): x_{1} \in B_{1} , x_{2} \in B_{2}, \ldots, x_{n} \in B_{n}\right} \ &J\left(B^{(n)}\right)=\left{x: x =\left(x_{1}, x_{2}, \ldots\right),\left(x_{1}, x_{2}, \ldots x_{n}\right) \in B^{(n) }, \quad B^{(n)}=B_{1} \times \ldots \times B_{n}\right} \end{aligned}\begin{对齐} &J\left(I_{1} \times \ldots \times I_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\右): x_{1} \in I_{1}, x_{2} \in I_{2}, \ldots, x_{n} \in I_{n}\right} \ &J\left(B_{1} \times \ldots \times B_{n}\right)=\left{x: x=\left(x_{1}, x_{2}, \ldots\right): x_{1} \in B_{1} , x_{2} \in B_{2}, \ldots, x_{n} \in B_{n}\right} \ &J\left(B^{(n)}\right)=\left{x: x =\left(x_{1}, x_{2}, \ldots\right),\left(x_{1}, x_{2}, \ldots x_{n}\right) \in B^{(n) }, \quad B^{(n)}=B_{1} \times \ldots \times B_{n}\right} \end{aligned}

Ĵ(乙1×…×乙n)=Ĵ(乙1×…×乙n×R) Ĵ(乙(n))=Ĵ(乙(n)×R)

b(R∞)⊆b1(R∞)⊆b2(R∞)

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Tasks for independent work

1. 让我们在数字尺度上介绍R=(−∞,+∞)指标：
为了 X,是∈R,ρ1(X,是)=|X−是|1+|X−是|.
a) 证明ρ1(X,是)真的是一个指标。
b) 让我们用b0(R)最小的σ-代数，由开集生成
S_{\rho}\left(x^{0}\right)=\left{x \in R: \rho_{1}\left(x, x^{0}\right)<\rho, \rho> 0, x^{0} \in R\right} 。S_{\rho}\left(x^{0}\right)=\left{x \in R: \rho_{1}\left(x, x^{0}\right)<\rho, \rho> 0, x^{0} \in R\right} 。
证明b0(R)=b(R).
2. 让b0(Rn)做最小的σ-代数，由开集生成
S_{\rho}\left(x^{0}\right)=\left{x \in R^{n}: \rho_{n}\left(x, x^{0}\right)<\rho , x^{0} \in R^{n}, \rho>0\right}S_{\rho}\left(x^{0}\right)=\left{x \in R^{n}: \rho_{n}\left(x, x^{0}\right)<\rho , x^{0} \in R^{n}, \rho>0\right}
在指标中
ρn(X,X0)=∑ķ=1n2−ķρ1(Xķ,Xķ0),X=(X1,X2,…,Xn)∈Rn,X0=(X10,X20,…,Xn0)∈Rn, ρ1(Xķ,Xķ0)=|Xķ−Xķ0|1+|Xķ−Xķ0|.
证明b0(Rn)=b(Rn).
3. 对于所有点X,X0∈R∞它们之间的距离由以下公式定义：
ρ∞(X,X0)=∑ķ=1∞2−ķ|Xķ−Xķ0|1+|Xķ−Xķ0|.
让b0(R∞)做最小的σ-代数，由开集生成
S_{\rho}\left(x^{0}\right)=\left{x \in R^{\infty}: \rho_{\infty}\left(x, x^{0}\right)< \rho, x^{0} \in R^{\infty}, \rho>0\right} 。

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