### 数学代写|信息论代写information theory代考| ENGN8534

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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|Entropies for Multivariate Joint Distributions

Let $\left{p\left(x_{1}, \ldots, x_{n}\right)\right}$ be a probability distribution on $X_{1} \times \ldots \times X_{n}$ for finite $X_{i}$ ‘s. Let $S$ be a subset of $\left(X_{1} \times \ldots \times X_{n}\right)^{2}$ consisting of certain ordered pairs of ordered $n$-tuples $\left(\left(x_{1}, \ldots, x_{n}\right),\left(x_{1}^{\prime}, \ldots, x_{n}^{\prime}\right)\right)$ so the product probability measure on $S$ is:
$$\mu(S)=\sum\left{p\left(x_{1}, \ldots, x_{n}\right) p\left(x_{1}^{\prime}, \ldots, x_{n}^{\prime}\right):\left(\left(x_{1}, \ldots, x_{n}\right),\left(x_{1}^{\prime}, \ldots, x_{n}^{\prime}\right)\right) \in S\right}$$

Then all the logical entropies for this $n$-variable case are given as the product measure of certain infosets $S$. Let $I, J \subseteq N$ be subsets of the set of all variables $N=\left{X_{1}, \ldots, X_{n}\right}$ and let $x=\left(x_{1}, \ldots, x_{n}\right)$ and $x^{\prime}=\left(x_{1}^{\prime}, \ldots, x_{n}^{\prime}\right)$.

Since two ordered $n$-tuples are different if they differ in some coordinate, the joint logical entropy of all the variables is: $h\left(X_{1}, \ldots, X_{n}\right)=\mu\left(S_{\mathrm{V} N}\right)$ where:
$$\begin{gathered} S_{\vee N}=\left{\left(x, x^{\prime}\right): \vee_{i=1}^{n}\left(x_{i} \neq x_{i}^{\prime}\right)\right}=\cup\left{S_{X_{i}}: X_{i} \in N\right} \text { where } \ S_{X_{i}}=S_{x_{i} \neq x_{i}^{\prime}}=\left{\left(x, x^{\prime}\right): x_{i} \neq x_{i}^{\prime}\right} \end{gathered}$$
(where $\vee$ represents the disjunction of statements). For a non-empty $I \subseteq N$, the joint logical entropy of the variables in $I$ could be represented as $h(I)=\mu\left(S_{\vee I}\right)$ where:
$$S_{\vee I}=\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i}^{\prime}\right) \text { for } X_{i} \in I\right}=\cup\left{S_{X_{i}}: X_{i} \in I\right}$$
so that $h\left(X_{1}, \ldots, X_{n}\right)=h(N)$.
As before, the information algebra $\mathcal{I}\left(X_{1} \times \ldots \times X_{n}\right)$ is the Boolean subalgebra of $\wp\left(\left(X_{1} \times \ldots \times X_{n}\right)^{2}\right)$ generated by the basic infosets $S_{X_{i}}$ for the variables and their complements $S_{\neg X_{f}}$.

For the conditional logical entropies, let $I, J \subseteq N$ be two non-empty disjoint subsets of $N$. The idea for the conditional entropy $h(I \mid J)$ is to represent the information in the variables $I$ given by the defining condition: $\vee\left(x_{i} \neq x_{i}^{\prime}\right)$ for $X_{i} \in I$, after taking away the information in the variables $J$ which is defined by the condition: $\vee\left(x_{j} \neq x_{j}^{\prime}\right)$ for $X_{j} \in J$. “After the bar $\mid$ ” means “negate” so we negate that condition $\vee\left(x_{j} \neq x_{j}^{\prime}\right)$ for $X_{j} \in J$ and add it to the condition for $I$ to obtain the conditional logical entropy as $h(I \mid J)=h(\vee I \mid \vee J)=\mu\left(S_{\vee I \mid \vee J}\right)$ (where $\wedge$ represents the conjunction of statements):
\begin{aligned} S_{\vee I \mid \vee J}=&\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i}^{\prime}\right) \text { for } X_{i} \in I \text { and } \wedge\left(x_{j}=x_{j}^{\prime}\right) \text { for } X_{j} \in J\right} \ &=\cup\left{S_{X_{i}}: X_{i} \in I\right}-\cup\left{S_{X_{j}}: X_{j} \in J\right}=S_{\vee I}-S_{\vee J} \end{aligned}

## 数学代写|信息论代写information theory代考|An Example of Negative Mutual Information

Norman Abramson gives an example [1, pp. 130-131] where the Shannon mutual information of three variables is negative. ${ }^{3}$ William Feller gives a similar concrete example that we will use [11, Exercise 26, p. 143]. Any probability theory textbook example to show that pair-wise independence does not imply mutual independence for three or more random variables would do as well.

One fair die is thrown first and the result is recorded odd as 1 (the number of the face up mod 2) or even as 0 . Then the same is done with a second fair die so the outcome space if $U={(0,0),(0,1),(1,0),(1,1)}={0,1} \times{0,1}$ (first die on the left and second die on the right). Let $X$ be the random variable for the outcome ( 0 or 1 ) of the first throw, $Y$ for the second throw, and $Z$ for the sum $X+Y$ mod 2. Since $Z$ is a function of $X$ and $Y$, the outcome space is $U \times U=(X \times Y)^{2}$. So many Venn diagrams are illustrated rather symbolically, e.g., with circles for $h(X)$, $h(Y)$, and $h(Z)$, that it will be useful to give the actual Venn/box diagrams for this example.

The two-draw outcome space $U \times U$ can be represented as a $4 \times 4$ square with each square representing a point $\left((x, y),\left(x^{\prime}, y^{\prime}\right)\right)$ for the two trials with the dice. The points included in $h(X)$ (indicated with shading) are the pairs $\left((x, y),\left(x^{\prime}, y^{\prime}\right)\right)$ where $x \neq x^{\prime}$ and symmetrically for $h(Y)$. Since $Z=X+Y \bmod 2$, the shaded squares for $h(Z)$ are the squares $\left((x, y),\left(x^{\prime}, y^{\prime}\right)\right)$ where $x+y \neq x^{\prime}+y^{\prime} \bmod 2$ as shown in Fig. 4.2.

Since each point in $U \times U$ has the product probability $\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}$ and the logical entropies are just the sum of the probabilities of the shaded squares, we see that $h(X)=h(Y)=h(Z)=\frac{8}{16}=\frac{1}{2}$. The joint logical entropies are the sum of the probabilities for the squares that are shaded in one or the other (or both) shown in Fig. $4.3$.

## 数学代写|信息论代写information theory代考|Entropies for Countable Probability Distributions

The logical entropies for any discrete probability distributions, finite or countably infinite, can be illustrated with an logical entropy box diagram. A square with unit length sides can have the probabilities $p_{1}, p_{2} \ldots .$ marked off along the width and height so that squares $p_{i}^{2}$ and products $p_{i} p_{j}$ and $p_{j} p_{i}$ all correspond to the area of rectangles in the square. The sum of the square areas along the diagonal is $\sum_{i} p_{i}^{2}$ so the logical entropy $h(p)=1-\sum_{i} p_{i}^{2}$ is the sum of the two equal areas of rectangles on either side of the diagonal. Figure $4.7$ gives the logical entropy box diagram for the countable distribution, $p_{1}=\frac{1}{2}, p_{2}=\left(\frac{1}{2}\right)^{2}, \ldots, p_{n}=\left(\frac{1}{2}\right)^{n}, \ldots$ which sums to 1 .

Since $\sum_{i} p_{i}=1$, the logical entropy $h(p)=1-\sum_{i} p_{i}^{2}$ for countable distributions is always well-defined, strictly less than one, and interpretable as the two-draw probability of getting different indices $i \neq j$. However, the Shannon entropy for countable distributions $H(p)=\sum_{i} p_{i} \log {2}\left(\frac{1}{p{i}}\right)$ may blow up (not have a finite sum); examples are given in [23, Example 2.46, p. 30] and [6, p. 48].
For the example at hand, the sum of the probabilities squared is $\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots=$ $\left(\frac{1}{4}\right)^{1}+\left(\frac{1}{4}\right)^{2}+\left(\frac{1}{4}\right)^{3}+\ldots=\frac{1 / 4}{1-1 / 4}=\frac{1}{3}$ (the area of the diagonal squares in the box diagram of Fig. 4.7) so the logical entropy is $h(p)=1-\frac{1}{3}=\frac{2}{3}$. In the box diagram with a uniform distribution over the unit area, the logical entropy is the probability that a random point in the square lies off the boxed diagonal.

## 数学代写|信息论代写information theory代考|Entropies for Multivariate Joint Distributions

\mu(S)=\sum\left{p\left(x_{1}, \ldots, x_{n}\right) p\left(x_{1}^{\prime}, \ldots, x_{n }^{\prime}\right):\left(\left(x_{1}, \ldots, x_{n}\right),\left(x_{1}^{\prime}, \ldots, x_{ n}^{\prime}\right)\right) \in S\right}\mu(S)=\sum\left{p\left(x_{1}, \ldots, x_{n}\right) p\left(x_{1}^{\prime}, \ldots, x_{n }^{\prime}\right):\left(\left(x_{1}, \ldots, x_{n}\right),\left(x_{1}^{\prime}, \ldots, x_{ n}^{\prime}\right)\right) \in S\right}

\begin{聚集} S_{\vee N}=\left{\left(x, x^{\prime}\right): \vee_{i=1}^{n}\left(x_{i} \neq x_{i}^{\prime}\right)\right}=\cup\left{S_{X_{i}}: X_{i} \in N\right} \text { where } \ S_{X_{i }}=S_{x_{i} \neq x_{i}^{\prime}}=\left{\left(x, x^{\prime}\right): x_{i} \neq x_{i} ^{\prime}\right} \end{聚集}\begin{聚集} S_{\vee N}=\left{\left(x, x^{\prime}\right): \vee_{i=1}^{n}\left(x_{i} \neq x_{i}^{\prime}\right)\right}=\cup\left{S_{X_{i}}: X_{i} \in N\right} \text { where } \ S_{X_{i }}=S_{x_{i} \neq x_{i}^{\prime}}=\left{\left(x, x^{\prime}\right): x_{i} \neq x_{i} ^{\prime}\right} \end{聚集}
（在哪里∨表示语句的析取）。对于非空我⊆ñ, 中变量的联合逻辑熵我可以表示为H(我)=μ(小号∨我)在哪里：

S_{\vee I}=\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i}^{\prime}\right) \text { for } X_{i} \in I\right}=\cup\left{S_{X_{i}}: X_{i} \in I\right}S_{\vee I}=\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i}^{\prime}\right) \text { for } X_{i} \in I\right}=\cup\left{S_{X_{i}}: X_{i} \in I\right}

\begin{对齐} S_{\vee I \mid \vee J}=&\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i }^{\prime}\right) \text { for } X_{i} \in I \text { 和 } \wedge\left(x_{j}=x_{j}^{\prime}\right) \text { for } X_{j} \in J\right} \ &=\cup\left{S_{X_{i}}: X_{i} \in I\right}-\cup\left{S_{X_{j }}: X_{j} \in J\right}=S_{\vee I}-S_{\vee J} \end{aligned}\begin{对齐} S_{\vee I \mid \vee J}=&\left{\left(x, x^{\prime}\right): \vee\left(x_{i} \neq x_{i }^{\prime}\right) \text { for } X_{i} \in I \text { 和 } \wedge\left(x_{j}=x_{j}^{\prime}\right) \text { for } X_{j} \in J\right} \ &=\cup\left{S_{X_{i}}: X_{i} \in I\right}-\cup\left{S_{X_{j }}: X_{j} \in J\right}=S_{\vee I}-S_{\vee J} \end{aligned}

## 数学代写|信息论代写information theory代考|An Example of Negative Mutual Information

Norman Abramson 给出了一个示例 [1, pp. 130-131]，其中三个变量的香农互信息为负。3William Feller 给出了一个类似的具体示例，我们将使用该示例 [11, 练习 26, p. 143]。任何表明成对独立并不意味着三个或更多随机变量的相互独立的概率论教科书示例也可以。

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