### 物理代写|统计力学代写Statistical mechanics代考|PHYS3006

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|Cantor Sets and Measures

As an aside, we mention a family of sets that are interesting from a topological and measure-theoretic viewpoint: the Cantor sets.

The original or “middle-third” Cantor set can be obtained by removing successively the “middle-third” intervals: let $C_{0}=[0,1], C_{1}=[0,1] \backslash(] \frac{1}{3}, \frac{2}{3}[)=\left[0, \frac{1}{3}\right] \cup$ $\left[\frac{1}{3}, 1\right], \quad C_{2}=C_{1} \backslash(] \frac{1}{9}, \frac{2}{9}[\cup] \frac{7}{9}, \frac{8}{9}[)=\left[0, \frac{1}{9}\right] \cup\left[\frac{2}{9}, \frac{1}{3}\right] \cup\left[\frac{2}{3}, \frac{7}{9}\right] \cup\left[\frac{8}{9}, 1\right]$, and so on. Then $C=n_{n=0}^{\infty} C_{n}$.

Actually, $C=\left{x \mid x=\sum_{n=1}^{\infty} \frac{a_{n}}{3^{n}}, a_{n} \neq 1, \forall n\right}$. To see this, observe that, if we write the ternary expansion of a number $x \in[0,1], x=\sum_{n=1}^{\infty} \frac{a_{n}}{3 \pi}$, then the numbers in $] \frac{1}{3}, \frac{2}{3}\left[\right.$ correspond to $a_{1}=1$ those in $] \frac{1}{9}, \frac{2}{9}[\cup] \frac{7}{9}, \frac{8}{9}\left[\right.$ correspond to $a_{2}=1$ and so on. So that the numbers in $C$ are exactly those $x \in[0,1]$ whose ternary expansion does not contain the symbol 1 .
The Cantor set $C$ has the following properties:

1. $C$ is of zero measure since the sum of the lengths of the removed intervals is: $\sum_{n=1}^{\infty} \frac{2^{n-1}}{3 n}=1$.
2. $C$ is uncountable: the map $g: C \rightarrow[0,1], \quad g(x)=\sum_{n=1}^{\infty} \frac{a_{n}}{2^{n+1}}$, where $x=\sum_{n=1}^{\infty} \frac{a_{n}}{3 n}, a_{n} \neq 1, \forall n$, is a surjection from $C$ to $[0,1]$, hence the cardinality of $C$ must be at least the one of $[0,1]$ (and also at most that cardinality, since $C \subset[0,1])$.
3. $C$ is a perfect set, which, by definition, means that it is closed (since it is the complement of a union of open intervals) and that each element of $C$ is a limit of other elements of $C$ : if $x=\sum_{n=1}^{\infty} \frac{a_{n}}{3^{n}}, a_{n} \neq 1$, $\forall n$, then $x=\lim {k \rightarrow \infty} x{k}$, where $x_{k}=\sum_{n=1}^{\infty} \frac{a_{s}^{k}}{3^{k}}$, with $a_{n}^{k}=a_{n}, \forall n \neq k$, and $a_{k}^{k}=2-a_{k}$.
4. $C$ is nowhere dense which, for a closed set, means that its interior is empty: in any neighborhood of $x \in C$, there are points not in $C$. Given $x=\sum_{n=1}^{\infty} \frac{a_{n}}{3^{n}}, a_{n} \neq 1$,

$\forall n$ and $\epsilon>0$, choose $k$ so that $3^{-k}<\epsilon$, and let $y=\sum_{n=1}^{\infty} \frac{b_{n}}{3^{n}}$, with $b_{n}=a_{n}$, $\forall n \neq k$, and $b_{k}=1$. Then, $y \notin C$ and $|x-y| \leq \epsilon$.

1. $C$ is totally disconnected, i.e. its only connected components are singletons.

## 物理代写|统计力学代写Statistical mechanics代考|Proofs of the Law of Large Numbers

Let us first prove $(2.3 .8)$. We need to estimate the probability of:
$$\left|\frac{1}{N} \sum_{i=1}^{N} f_{i}\left(x_{i}\right)-\mathbb{E}(f(x))\right| \geq \epsilon$$
We may assume that $\mathbb{E}(f(x))=0$, by redefining $f$. We will use Markov’s (or Chebyshev’s) inequality: for a random variable $F$ on $(\Omega, \Sigma, \mu)$, and $a>0$,

$$\mu(1(F \geq a)) \leq \frac{\mathbb{E}\left(F^{2}\right)}{a^{2}}$$
which follows trivially from $a \mathbb{1}(F \geq a) \leq F$.
Apply this to $F=\left|\frac{1}{N} \sum_{i=1}^{N} f_{i}\left(x_{i}\right)\right|, \mu=\mu$ as in (2.3.8) and $a=\epsilon$. We get:
$$\boldsymbol{\mu}\left(\left|\frac{1}{N} \sum_{i=1}^{N} f_{i}\left(x_{i}\right)\right| \geq \epsilon\right) \leq \frac{\mathbb{E}\left(\left(\sum_{i=1}^{N} f_{i}\left(x_{i}\right)\right)^{2}\right)}{\epsilon^{2} N^{2}}$$
We have $\mathbb{E}\left(\left(\sum_{i=1}^{N} f_{i}\left(x_{i}\right)\right)^{2}\right)=\sum_{i, j=1}^{N} \mathbb{E}\left(f_{i}\left(x_{i}\right) f_{j}\left(x_{j}\right)\right)=\sum_{i=1}^{N} \mathbb{E}\left(f_{i}\left(x_{i}\right)^{2}\right)=$ $N \mathbb{E}\left(f(x)^{2}\right)$, since $\mathbb{E}\left(f_{i}\left(x_{i}\right) f_{j}\left(x_{j}\right)\right)=\mathbb{E}\left(f_{i}\left(x_{i}\right)\right) \mathbb{E}\left(f_{j}\left(x_{j}\right)\right)=0$ for $i \neq j$ because the variables $x_{i}, x_{j}$ are independent and, by assumption, $\mathbb{E}\left(f_{i}\left(x_{i}\right)\right)=0, \forall i$.
Inserting this in the right hand side of (2.B.2), we get:
$$\boldsymbol{\mu}\left(\left|\frac{1}{N} \sum_{i=1}^{N} f_{i}\left(x_{i}\right)\right| \geq \epsilon\right) \leq \frac{C}{\epsilon^{2} N}$$
with $C=\mathbb{E}\left(f(x)^{2}\right.$ ) (we assumed $f$ to be bounded), which implies
$$\boldsymbol{\mu}\left(G_{N}(\epsilon)\right) \geq 1-\frac{C}{\epsilon^{2} N}$$
which proves (2.3.8).
To prove (2.3.12), we proceed similarly. Applying (2.B.1) to $\left|F_{\alpha}\right|$, for $\alpha=$ $1, \ldots, k$, where $F_{\alpha}=\frac{1}{N}\left(\sum_{i=1}^{N} \mathbb{1}\left(x_{i} \in A_{\alpha}\right)-\mathbb{E}\left(\sum_{i=1}^{N} \mathbb{1}\left(x_{i} \in A_{\alpha}\right)\right)\right)=\frac{1}{N} \sum_{i=1}^{N}$ $\mathbb{1}\left(x_{i} \in A_{\alpha}\right)-P_{\alpha}$, (since $\left.\mathbb{E}\left(\mathbb{1}\left(x_{i} \in A_{\alpha}\right)\right)=P_{\alpha}\right)$, we get:
$$\boldsymbol{\mu}\left(\left|F_{a}\right| \geq \epsilon\right) \leq \frac{\mathbb{E}\left(F_{\alpha}^{2}\right)}{\epsilon^{2}}$$

## 物理代写|统计力学代写Statistical mechanics代考|Newton’s Laws

In classical mechanics one starts by defining frames of reference and by distinguishing between inertial and non inertial frames.

A frame of reference is simply a coordinate system that ascribes a set of numbers specifying the positions of all the particles in the system under consideration at any given time. We will use here cartesian coordinates. If these positions change in time, those coordinates, as functions of time, describe the trajectories of the particles.
An inertial frame of reference is one in which a particle which is not subjected to any force moves along a straight line at constant velocity. Instead of saying “which is not subjected to any force” one could say “is infinitely distant from any other particle.” If we describe the motion of such a particle in a frame of reference attached to a merry-go-round or to an accelerating rocket, that motion will no longer appear to be on a straight line or to move at constant speed. Such frames are called non inertial.

It is obvious from the definition given here that we are dealing with an idealization, which is nevertheless approximately realized in many situations: the fact that the Earth rotates around the Sun and around itself makes a frame of reference attached to the Earth, strictly speaking, non inertial; but it can nevertheless be considered inertial for most experiments performed in laboratories.

A basic principle of mechanics is the equivalence of all inertial frames of reference, also called Galilean invariance: the laws of motion take the same form in all inertial frames of reference and the transformations between such frames consist of (constant in time) rotations and translations on a straight line at constant velocity of the origin of coordinates. This invariance implies conservation laws for total momentum, total angular momentum and energy (checking the first conservation laws will be left as exercises). ${ }^{1}$

Here we will always work in a fixed inertial frame, so we will not be concerned with Galilean invariance. Moreover, we will not discuss conservations laws apart form the conservation of energy. Newton’s first law, says, in modern terminology, that there exist inertial reference frames; since we decided to work in one such frame, we will not discuss it further.

## 物理代写|统计力学代写Statistical mechanics代考|Cantor Sets and Measures

1. C是零度量，因为已删除间隔的长度之和为：∑n=1∞2n−13n=1.
2. C不可数：地图G:C→[0,1],G(X)=∑n=1∞一个n2n+1， 在哪里X=∑n=1∞一个n3n,一个n≠1,∀n, 是从C至[0,1]，因此基数C必须至少是其中之一[0,1]（而且最多也就是那个基数，因为C⊂[0,1]).
3. C是一个完美集，根据定义，这意味着它是闭集（因为它是开区间并集的补集）并且C是其他元素的限制C： 如果X=∑n=1∞一个n3n,一个n≠1, ∀n， 然后X=林ķ→∞Xķ， 在哪里Xķ=∑n=1∞一个sķ3ķ， 和一个nķ=一个n,∀n≠ķ， 和一个ķķ=2−一个ķ.
4. C是无处稠密的，对于一个闭集，意味着它的内部是空的：在任何邻域X∈C, 有点不在C. 给定X=∑n=1∞一个n3n,一个n≠1,

∀n和ε>0， 选择ķ以便3−ķ<ε， 然后让是=∑n=1∞bn3n， 和bn=一个n,∀n≠ķ， 和bķ=1. 然后，是∉C和|X−是|≤ε.

1. C是完全断开的，即它唯一连接的组件是单例。

## 物理代写|统计力学代写Statistical mechanics代考|Proofs of the Law of Large Numbers

|1ñ∑一世=1ñF一世(X一世)−和(F(X))|≥ε

μ(1(F≥一个))≤和(F2)一个2

μ(|1ñ∑一世=1ñF一世(X一世)|≥ε)≤和((∑一世=1ñF一世(X一世))2)ε2ñ2

μ(|1ñ∑一世=1ñF一世(X一世)|≥ε)≤Cε2ñ

μ(Gñ(ε))≥1−Cε2ñ

μ(|F一个|≥ε)≤和(F一个2)ε2

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