### 数学代写|信息论代写information theory代考| ECE4042

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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

## 数学代写|信息论代写information theory代考|Metrical Logical Entropy = (Twice) Variance

Since logical entropy is a measure of distinctions, differences, distinguishability, and diversity or heterogeneity, there should be a relationship with the notion of variance in statistics. Given a probability distribution $p=\left(p_{1}, \ldots, p_{n}\right)$, the logical entropy is the two-draw probability of getting a different index $h(p)=\sum_{i \neq j} p_{i} p_{j}=1-$ $\sum_{i} p_{i}^{2}$. The probabilities are typically supplied from the inverse-image of a random variable $X: U \rightarrow \mathbb{R}$ on an outcome set or sample space $U$ where the possible values of $x_{1}, \ldots, x_{n}$ have probabilities $p_{i}=\operatorname{Pr}\left(X=x_{i}\right)$ for $i=1, \ldots, n$. If the statistical data being modeled is categorical, then the distinct numbers $x_{i}$ assigned to different categories have no metrical significance, and thus cannot be meaningfully added or multiplied. But the situation can still be modeled to compute a variance by using a random vector $X=\left(X_{1}, \ldots, X_{n}\right)$ where on each trial, one of the random variables has value 1 and the rest 0 with the probabilities $\operatorname{Pr}\left(X_{i}=1\right)=p_{i}$ for $i=1, \ldots, n$. Then the expectation of the random vector is $E(X)=\left(p_{1}, \ldots, p_{n}\right)$. The variance added over the components of the vector is:
\begin{aligned} \operatorname{Var}(X) &=\sum_{i} p_{i}\left[\left(1-p_{i}\right)^{2}+\sum_{j \neq i}\left(0-p_{j}\right)^{2}\right] \ &=\sum_{i} p_{i}\left[1-2 p_{i}+p_{i}^{2}+\sum_{j \neq i}\left(0-p_{j}\right)^{2}\right] \ &=\sum_{i} p_{i}\left[1-2 p_{i}+\sum_{j} p_{j}^{2}\right] \ &=1-2 \sum_{i} p_{i}^{2}+\sum_{j} p_{j}^{2}=1-\sum_{i} p_{i}^{2}=h(p) \end{aligned}
Variance of random vector for categorical data $=$ logical entropy. [8]
Another way to model this situation is $n$ Bernoulli trials with variable probabilities $p_{i}$ for success $X_{i}=1$ and $1-p_{i}=q_{i}$ for failure $X_{i}=0$ on the $i$ th trial where the expectations are $E\left(X_{i}\right)=p_{i}$ and those probabilities make up a distribution, i.e., $\sum_{i=1}^{n} p_{i}=1$. The random variable for the number of successes is: $S=\sum_{i=1}^{n} X_{i}$ with the expectation $E(S)=\sum_{i} E\left(X_{i}\right)=\sum_{i} p_{i}=1$. The variance of $X_{i}$ is:
\begin{aligned} \operatorname{Var}\left(X_{i}\right) &=E\left[\left(X_{i}-p_{i}\right)\left(X_{i}-p_{i}\right)\right]=\left(1-p_{i}\right)^{2} p_{i}+\left(0-p_{i}\right)^{2} q_{i} \ &=q_{i}^{2} p_{i}+p_{i}^{2} q_{i}=p_{i} q_{i}\left(p_{i}+q_{i}\right)=p_{i}\left(1-p_{i}\right) \end{aligned}

## 数学代写|信息论代写information theory代考|Boltzmann and Shannon Entropies

In Boltzmannian statistical mechanics, when $n$ particles can be in $m$ different distinguishable “states” (e.g., levels of energy), then a configuration or macrostate $\left(n_{1}, \ldots, n_{m}\right)$ is given by the occupation numbers $n_{i}$ of particles in the $i$ th state where $n_{1}+\ldots+n_{m}=n$. Then the total number of possible microstates for that configuration is the multinomial coefficient $\left(\begin{array}{c}n \ n_{1}, \ldots, n_{m}\end{array}\right)=\frac{n !}{n_{1} ! \ldots n_{m} !}$. If all the $m^{n}$ possible states are equiprobable, then the larger the multinomial coefficient, the higher probability that a system will be in that state. The system will spontaneously propagate to higher probability states subject to the constraint that the sum of the energies of the particles is a constant, the total energy in the system. Hence to find the characteristics of the equilibrium towards which a system will spontaneously evolve, one needs to maximize the multinomial coefficient subject to the energy constraint. The natural logarithm of the multinomial coefficient is a monotonic transformation so maximizing the one is equivalent to maximizing the other. And taking the natural log of the multinomial coefficient yields an additive quantity to be associated with the extensive quantity of entropy in thermodynamics.

The Boltzmann entropy ${ }^{5} \ln \left(\frac{n !}{n_{1} ! \ldots n_{m} !}\right)$ is usually represented as related to the Shannon entropy via the two-term Stirling approximation. It is important to note that the Boltzmann entropy is already a Shannon entropy for the probability distribution of $p=\left(1 / \frac{n !}{n_{1} ! \ldots n_{m} !}, \ldots, 1 / \frac{n !}{n_{1} ! \ldots n_{m} !}\right)$ where all the microstates for the configuration $\left(n_{1}, \ldots, n_{m}\right)$ are equiprobable:
$$\begin{gathered} H_{e}\left(1 / \frac{n !}{n_{1} ! \ldots n_{m} !}, \ldots, 1 / \frac{n !}{n_{1} ! \ldots n_{m} !}\right)=\sum_{i=1}^{\frac{n !}{n_{1} \cdots n_{m}}} \frac{1}{\frac{n !}{n ! ! \ldots n_{m} !}} \ln \left(\frac{1}{1 / \frac{n !}{n_{1} ! \ldots n_{m} !}}\right) \ \quad=\frac{n !}{n_{1} ! \ldots n_{m} !} \frac{1}{\frac{n !}{n ! \ldots n_{m} !}} \ln \left(\frac{n !}{n_{1} ! \ldots n_{m} !}\right)=\ln \left(\frac{n !}{n_{1} ! \ldots n_{m} !}\right) \end{gathered}$$
There is no approximation involved. Taking the Shannon entropy to base 2 , the usual interpretation is that it is the average number of yes-or-no questions it takes to determine the specific microstate given the macrostate. The logical entropy for that probability distribution is: $h\left(1 / \frac{n !}{n_{1} ! \ldots n_{m} !}, \ldots, 1 / \frac{n !}{n_{1} ! \ldots n_{m} !}\right)=1-\frac{1}{\frac{n !}{n_{1} ! \ldots n_{m} !}}$

## 数学代写|信息论代写information theory代考|MaxEntropies for Discrete Distributions

Given a function $X: U \rightarrow \mathbb{R}$ with metrical real values $\left(x_{1}, \ldots, x_{n}\right)$, what is the probability distribution $p=\left(p_{1}, \ldots, p_{n}\right)$ such that $\sum_{i=1}^{n} p_{i} x_{i}=m$ for a given mean or average $m$ such that logical entropy is a maximum? Since there are inequality constraints $p_{i} \geq 0$ on the variables and since logical entropy is quadratic in the variables, this is a quadratic programming problem. But we can first approach it as a classical optimization problem (equality constraints and no non-negativity constraints on the variables) and then derive the conditions on $m$ so that all the probabilities will be non-negative anyway.
The Lagrangian for the maximization problem is:
$$\mathcal{L}\left(p_{1}, \ldots, p_{n}\right)=1-\sum_{i} p_{i}^{2}-\lambda\left(1-\sum p_{i}\right)+\tau\left(m-\sum p_{i} x_{i}\right)$$
so the first-order conditions are:
$$\partial \mathcal{L} / \partial p_{j}=-2 p_{j}+\lambda-\tau x_{j}=0 \text { so } p_{j}=\frac{1}{2}\left(\lambda-\tau x_{j}\right) .$$
The calculations are simplified if we define $\mu=\sum_{i} x_{i} / n$ and $\operatorname{Var}(X)=E\left(X^{2}\right)-$ $\mu^{2}=\sum_{i} x_{i}^{2} / n-\mu^{2}=\sigma^{2}$. Using the first constraint:
$$1=\sum p_{i}=\frac{1}{2}(n \lambda-\tau n \mu)=\frac{n}{2}(\lambda-\tau \mu),$$
and using the second constraint:
$$\begin{gathered} m=\sum p_{i} x_{i}=\sum_{i} x_{i} \frac{1}{2}\left(\lambda-\tau x_{i}\right)=\lambda \frac{n}{2} \mu-\tau \frac{1}{2} \sum x_{i}^{2} \ =\frac{n}{2} \lambda \mu-\frac{n}{2} \tau\left[\operatorname{Var}(X)+\mu^{2}\right]=\mu\left(\frac{n}{2}(\lambda-\tau \mu)\right)-\frac{n}{2} \operatorname{Var}(X) \tau=\mu-\frac{n}{2} \operatorname{Var}(X) \tau \end{gathered}$$
so we have two equations that can be used to solve for the Lagrange multipliers $\lambda$ and $\tau$. Solving for $\tau$ in the second constraint:
$$\tau=\frac{2}{n} \frac{(\mu-m)}{\operatorname{Var}(X)}$$

## 数学代写|信息论代写information theory代考|Boltzmann and Shannon Entropies

H和(1/n!n1!…n米!,…,1/n!n1!…n米!)=∑一世=1n!n1⋯n米1n!n!!…n米!ln⁡(11/n!n1!…n米!) =n!n1!…n米!1n!n!…n米!ln⁡(n!n1!…n米!)=ln⁡(n!n1!…n米!)

## 数学代写|信息论代写information theory代考|MaxEntropies for Discrete Distributions

∂大号/∂pj=−2pj+λ−τXj=0 所以 pj=12(λ−τXj).

1=∑p一世=12(nλ−τnμ)=n2(λ−τμ),

τ=2n(μ−米)曾是⁡(X)

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