### 物理代写|统计力学代写Statistical mechanics代考|PHYS3934

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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|Approximation of Integrals

Since measurable functions can be quite irregular (think for example of the indicator function of the rational numbers), it is convenient to approximate them by regular functions. Let $F: \Omega \rightarrow \mathbb{R}$ be integrable: $\int_{\Omega}|F(x)| d \mu(x)<\infty$ and assume that $\Omega$ is a Borel subset of $\mathbb{R}^{n}$ for some $n$. Then, $\forall \epsilon>0, \exists$ a continuous function $G: \Omega \rightarrow \mathbb{R}$ so that:
$$\int_{\Omega}|F(x)-G(x)| d \mu(x) \leq \epsilon$$
The function $G$ can chosen to be $C^{\infty}$ and to vanish outside a bounded set. $G$ can also be chosen as a function of the form $\sum_{m=1}^{N} c_{m} \mathbb{1}{A{m}}$ with $A_{m}$ rectangles in $\Omega$ and $N<\infty$ (see e.g. [197, Chap. 7] for the proofs).

If $\Omega=\times_{i \in \mathcal{Z}} \Omega_{i}$ is a product space, $\boldsymbol{\mu}$ a product measure on that space (see (2.A.6)) and $F: \Omega \rightarrow \mathbb{R}$ an integrable function, $\int_{\Omega}|F(\mathbf{x})| d \mu(\mathbf{x})<\infty$, then, $\forall \epsilon>0, \exists N<$ $\infty, \exists G: \mathbf{\Omega} \rightarrow \mathbb{R}$ which is a function only of the variables $\left(x_{-N}, \ldots, x_{N}\right)$, so that:
$$\int_{\Omega}|F(\mathbf{x})-G(\mathbf{x})| d \boldsymbol{\mu}(\mathbf{x}) \leq \epsilon$$
Bounds similar to (2.A.11) hold for square integrable $F$ ‘s, $\int_{\Omega}|F(x)|^{2} d \mu(x)<\infty$, with $|F(x)-G(x)|$ replaced by $|F(x)-G(x)|^{2}$ and similarly for (2.A.12).

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

Consider a map $T: \Omega \rightarrow \Omega$ from $\Omega$ into itself. Later, specially in Chap. 4 , we will think of $T$ as a dynamical transformation on a space of physical states $\Omega$. An important notion is the one of measures that are invariant under such transformations ${ }^{28}$ :

Definition $2.7$ Let $(\Omega, \Sigma, \mu)$ be a measure space and $T: \Omega \rightarrow \Omega$ a map from $\Omega$ into itself. One says that $\mu$ is invariant under $T$ or, equivalently, that the map $T$ preserves the measure $\mu$, if, $\forall A \in \Sigma$,
$$\mu\left(T^{-1} A\right)=\mu(A)$$
where
$$T^{-1} A={x \mid T x \in A}$$

Remark $2.8$ If the map $T$ is invertible, (2.A.13) is obviously equivalent to: $\forall A \in \Sigma$,
$$\mu(T A)=\mu(A)$$
The reader might wonder why one uses definition (2.A.13) and not (2.A.14). The logic is that, considering $T$ to be a transformation acting on a set of states $\Omega$, one wants to compare the measure of a subset of states $A$, with the measure of the set of initial conditions that are mapped onto that subset by the transformation $T$ and that set is $T^{-1} A$. Equation (2.A.13) says that those two probabilities are equal and that expresses the fact that the map $T$ preserves the measure $\mu$.
Remark 2.9 Property (2.A.13) is equivalent to: $\forall F \in L^{1}(\Omega, d \mu)$,
$$\int_{\Omega} F(T x) d \mu(x)=\int_{\Omega} F(x) d \mu(x)$$
For a function of the form $F(x)=1_{A}(x)$, the equivalence of (2.A.13) and (2.A.15) is immediate. To prove that (2.A.13) implies (2.A.15) for more general functions $F \in L^{1}(\Omega)$, one uses (2.A.9). Since (2.A.15) holds for each term of the sum in the right hand side of (2.A.9), it holds also in the limit and that proves (2.A.15).

## 物理代写|统计力学代写Statistical mechanics代考|Probability Densities, Marginal and Conditional

A measure $\nu$ on $(\Omega, \Sigma)$ is absolutely continuous with respect to another measure $\mu$ on $(\Omega, \Sigma)$ if, $\forall A \in \Sigma$,
$$\mu(A)=0 \rightarrow \nu(A)=0$$
In that situation, the Radon-Nikodym theorem implies that one can write, $\forall A \in \Sigma$,
$$\nu(A)=\int_{A} F(x) d \mu(x)$$
for a $\mu$-integrable function $F: \Omega \rightarrow[0, \infty[$, see e.g. [278, Chap. 11] for a proof.
The function $F(x)$ is the probability density of $\nu$ relative to $\mu$ and one writes: $\frac{d v}{d \mu}=F$. We are often interested in situation where $\mu$ is the Lebesgue measure and then, one write $(2 . \mathrm{A} .19)$ as: $\nu(A)=\int_{A} F(x) d x$.

Given a measure space $(\Omega, \Sigma, \mu)$, the marginal probability distribution of a random variable $f: \Omega \rightarrow \mathbb{R}$ is the measure $\nu_{f}$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ given by:
$$\nu_{f}(A)=\mu(f(x) \in A)=\mu\left(f^{-1}(A)\right)$$
For these $\nu_{f}$ and $\mu, \frac{d v_{f}}{d \mu}$ is the probability density of the random variable $f$.
Two random variable $f_{1}, f_{2}$ are independent if, $\forall A, B \in \mathcal{B}(\mathbb{R})$,
$$\mu\left(f_{1}(x) \in A, f_{2}(x) \in B\right)=\mu\left(f_{1}(x) \in A\right) \mu\left(f_{2}(x) \in B\right)$$
This definition can be extended to any finite collection of random variables; an infinite collection of random variables is independent if any finite sub-collection of random variables is.
If $\mu(\Omega)=1$, and if $f_{1}, f_{2}$ are independent, then:
$$\int_{\Omega} f_{1}(x) f_{2}(x) d \mu(x)=\int_{\Omega} f_{1}(x) d \mu(x) \int_{\Omega} f_{2}(x) d \mu(x)$$
We introduced in Sect. $2.2 .2$ the notion of conditional probability $P(A \mid B)$ of an event $A$, given some event $B$. For a discrete random variable $X$ (taking a finite or countable number of values), one defines the conditional probability distribution of a random variable $Y$, given that $X=x$ by:
$$P(Y \in A \mid X=x)=\frac{P(Y \in A, X=x)}{P(X=x)}$$

## 物理代写|统计力学代写Statistical mechanics代考|Approximation of Integrals

∫Ω|F(X)−G(X)|dμ(X)≤ε

∫Ω|F(X)−G(X)|dμ(X)≤ε

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

μ(吨−1一个)=μ(一个)

μ(吨一个)=μ(一个)

∫ΩF(吨X)dμ(X)=∫ΩF(X)dμ(X)

## 物理代写|统计力学代写Statistical mechanics代考|Probability Densities, Marginal and Conditional

μ(一个)=0→ν(一个)=0

ν(一个)=∫一个F(X)dμ(X)

νF(一个)=μ(F(X)∈一个)=μ(F−1(一个))

μ(F1(X)∈一个,F2(X)∈乙)=μ(F1(X)∈一个)μ(F2(X)∈乙)

∫ΩF1(X)F2(X)dμ(X)=∫ΩF1(X)dμ(X)∫ΩF2(X)dμ(X)

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## MATLAB代写

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