物理代写|统计力学代写Statistical mechanics代考|PHYS3034

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

物理代写|统计力学代写Statistical mechanics代考|Definition of a Measure

Given a set $\Omega$, a measure $\mu$ is a map from subsets of $\Omega$ to the real numbers that will give the “size” of that set. The simplest example is when $\Omega$ is finite or countable, $\Omega=\left{x_{1}, x_{2}, \ldots\right}$ and we have a sequence of numbers $p_{i} \geq 0, i=1, \ldots$; then, the measure $\mu$ is defined on subsets $A \subset \Omega$ by:
$$\mu(A)=\sum_{i ; x_{i} \in A} p_{i}$$
with $\mu(A)=\infty$ if the sum diverges.
However, it is in general not possible to define a “natural” measure on all subsets of uncountable sets. For a simple example, consider the set $[0,1[$ with addition modulo 1. Define an equivalence relation $x \equiv y$ if $x-y \in \mathbb{Q}$. The set $[0,1[$ is thus an uncountable union of equivalence classes, each of which is countable, since $\mathbb{Q}$ is countable and [0, 1[ is not. Let $E$ be a set composed of one element taken from each equivalence class (we need the axiom of choice to prove that such a set exists but let us assume that), let $q_{n}$ be an enumeration of the rational numbers in $[0,1[$ and let $E_{n}=E+q_{n}$. The sets $E_{n}$ ‘s are two by two disjoint (by definition of equivalence classes: if $E_{n} \cap E_{m} \neq \emptyset$ for $n \neq m$ then there exists $x, y \in E$, with $x+q_{n}=y+q_{m}$, and that means that $x \equiv y$, which contradicts the definition of $E)$ and $\cup_{n} E_{n}=[0,1[$.
Now if we want the sets $E_{n}$ to be measurable and if we want to define a translation invariant measure on $[0,1[$ (with addition modulo 1) satisfying (2.A.1) (e.g. the Lebesgue measure defined after proposition 2.4), then we run into a contradiction since $\mu\left(E_{n}\right)=\mu(E), \forall n$, by translation invariance, and $\mu\left(\cup_{n} E_{n}\right)=\mu([0,1[)=1$ : the infinite sum of identical terms in (2.A.1) (which extends (2.2.1) to infinite sums) cannot equal $1 .$

A more sophisticated example of non-measurable sets, called the Banach-Tarski paradox, relies on the construction of a subtle partition of the unit ball in $\mathbb{R}^{3}$ into ten disjoint subsets $A_{1}, \ldots, A_{10}$ (this construction again uses the axiom of choice), such that there exist ten rotations $R_{1}, \ldots, R_{10}$ with the property that $R_{1} A_{1}, \ldots, R_{5} A_{5}$ form a partition of the unit ball and $R_{6} A_{6}, \ldots, R_{10} A_{10}$ form also a partition of the unit ball. Thus, by partitioning adequately one unit ball and rotating without deformation the elements of the partition, one can construct two balls of the same size. This would be a paradox if the sets $A_{1}, \ldots, A_{10}$ were measurable, because the Lebesgue measure is invariant under rotations and then we would have, since the $A_{i}$ ‘s are disjoint:
\begin{aligned} 1 &=\mu_{\mathrm{Leb}}\left(\cup_{i=1}^{10} A_{i}\right)=\sum_{i=1}^{10} \mu_{\mathrm{Leb}}\left(A_{i}\right) \ &=\sum_{i=1}^{10} \mu_{\mathrm{Leb}}\left(R_{i} A_{i}\right)=\mu_{\mathrm{Leb}}\left(\cup_{i=1}^{5} R_{i} A_{i}\right)+\mu_{\mathrm{Leb}}\left(\cup_{i=6}^{10} R_{i} A_{i}\right)=2 \end{aligned}
This proves that the sets $A_{1}, \ldots, A_{10}$ are not measurable.

物理代写|统计力学代写Statistical mechanics代考|Constructions of Measures

The family of Borel sets, as well as other $\sigma$-algebras used here, is actually very large and it would be quite cumbersome to define explicitly the value of the map $\mu$ on every of those sets. Luckily, there exists extension theorems that guarantee that, if one defines $\mu$ on a much smaller class of sets, then it can be extended to the whole $\sigma$-algebra.

Definition $2.3$ A semi-algebra $\mathcal{S}$ is a family of subsets of a set $\Omega$ such that:

1. $\emptyset \in \mathcal{S}$.
2. $\forall A, B \in \mathcal{S}$, we have $A \cap B \in \mathcal{S}$ (the family is closed under pairwise intersections).
3. $\forall A, B \in \mathcal{S}$, there exist disjoint sets $C_{i} \in \mathcal{S}, i=1,2, \ldots, n$, such that $A \backslash B=\bigcup_{i=1}^{n} C_{i}$ (relative complements of elements of $\mathcal{S}$ can be written as finite disjoint unions of elements of $\mathcal{S}$ ).

Proposition 2.4 Extensions of measures: If a map defined on a semi-algebra of sets $\mathcal{S}$ satisfies the properties in Definition $2.2$, and if $\Omega$ can be written as $\Omega=\cup_{i \in \mathbb{N}} A_{i}$, with $A_{i} \in \mathcal{S}, \mu\left(A_{i}\right)<\infty, \forall i \in \mathbb{N}$, then that map can be extended in a unique way to a measure defined on the $\sigma$-algebra generated by $\mathcal{S}$ (i.e. the smallest $\sigma$-algebra containing $\mathcal{S}$ ).
For a proof, see e.g. Royden [278, Sect. 12.2].
It is easy to check that the set of intervals in $\mathbb{R}$ or of rectangles in $\mathbb{R}^{n}$ (sets of the form $I_{1} \times \cdots \times I_{n}$, where $I_{k}$ are intervals in $\mathbb{R}$ ) are semi-algebras (exercise). Thus, it is sufficient to define a measure on the intervals of $\mathbb{R}$ to have it defined on the Borel subsets of $\mathbb{R}$.

If we take the measure of an interval to be its length, or the measure of a rectangle in $\mathbb{R}^{n}$ to be its volume, one obtains by extension the Lebesgue measure $\mu_{\mathrm{L}} \mathrm{eb}$, which is thus uniquely defined. ${ }^{25}$ For the Lebesgue measure of a set $E$, we will write $\mu$ Leb $(E)$ or, when there is no ambiguity, $|E|$, which also denotes the cardinality of the set $E$ for finite sets.

物理代写|统计力学代写Statistical mechanics代考|Integration

The idea of Riemann integration is to approximate the integral, that is, the area under a curve (for a positive valued function) by the sum of the areas of little vertical rectangles whose upper side lies just under that curve or just above it.

However that method of integration has two limitations: the set of functions that can be integrated with that method is restricted: for example, one cannot integrate à la Riemann the indicator function of the rational numbers, since the height of the only rectangles under the graph of that function is 0 and the height of the only rectangles above that graph is 1 , although intuitively, since the set of rational numbers is of measure 0 , that integral should exist and be also equal to 0 . Moreover, if a sequence of Riemann integrable functions $F_{n}(x) \rightarrow F(x), \forall x$, as $n \rightarrow \infty$, the conditions under which one can write the obviously desirable equation $\int F(x) d x=\lim {n \rightarrow \infty} \int F{n}(x) d x$ are not simple.

The idea of Lebesgue integration solves those problems. Consider for simplicity a bounded map $F: \mathbb{R} \rightarrow \mathbb{R}^{+}$of bounded support (i.e. that vanishes outside a finite interval). Instead of dividing the domain of definition of that function into small intervals, as one does in the theory of Riemann integration, one divides the image of the function into small intervals and one seeks to approximate the integral by integrating functions of the form:
$$F_{n}=\sum_{m=0}^{\infty} \frac{m}{n} 1\left(A_{m}\right),$$
with $A_{m}=F^{-1}\left(\left[\frac{m}{n}, \frac{m+1}{n}\left[\right.\right.\right.$ and where the sum $\sum_{m=0}^{\infty}$ is finite for any bounded $F$, since then $F^{-1}\left(\left[\frac{m}{n}, \frac{m+1}{n}[)=\emptyset\right.\right.$ for $m$ large enough. The integral of $F_{n}$ is naturally defined as $\sum_{m=0}^{\infty} \frac{m}{n} \mu\left(A_{m}\right)$.

For that expression to make sense, it is enough for the measure of sets of the form $F^{-1}\left(\left[\frac{m}{n}, \frac{m+1}{n}\right.\right.$ [ to exists. It is convenient to introduce a more general notion.

物理代写|统计力学代写Statistical mechanics代考|Definition of a Measure

μ(一个)=∑一世;X一世∈一个p一世

1=μ大号和b(∪一世=110一个一世)=∑一世=110μ大号和b(一个一世) =∑一世=110μ大号和b(R一世一个一世)=μ大号和b(∪一世=15R一世一个一世)+μ大号和b(∪一世=610R一世一个一世)=2

物理代写|统计力学代写Statistical mechanics代考|Constructions of Measures

Borel 系列以及其他σ-这里使用的代数实际上非常大，明确定义映射的值会很麻烦μ在每一组上。幸运的是，存在扩展定理保证，如果一个定义μ在更小的集合类上，然后它可以扩展到整个σ-代数。

1. ∅∈小号.
2. ∀一个,乙∈小号， 我们有一个∩乙∈小号（家庭在成对交叉点下是封闭的）。
3. ∀一个,乙∈小号, 存在不相交集C一世∈小号,一世=1,2,…,n, 这样一个∖乙=⋃一世=1nC一世（元素的相对补充小号可以写成元素的有限不相交并集小号 ).

物理代写|统计力学代写Statistical mechanics代考|Integration

Lebesgue 积分的想法解决了这些问题。为简单起见考虑有界地图F:R→R+有界支持（即在有限区间外消失）。不像在黎曼积分理论中那样将该函数的定义域划分为小区间，而是将函数的图像划分为小区间，并通过对以下形式的函数进行积分来近似积分：

Fn=∑米=0∞米n1(一个米),

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MATLAB代写

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