Posted on Categories:Statistical inference, 统计代写, 统计代考, 统计推断

# 统计代写|统计推断代考Statistical Inference代写|Counting

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## 统计代写|统计推断代考Statistical Inference代写|Counting

The elementary process of counting can become quite sophisticated when placed in the hands of a statistician. Most often, methods of counting are used in order to construct probability assignments on finite sample spaces, although they can be used to answer other questions also.

Example 1.2.12 (Lottery-I) For a number of years the New York state lottery operated according to the following scheme. From the numbers $1,2, \ldots, 44$, a person may pick any six for her ticket. The winning number is then decided by randomly selecting six numbers from the forty-four. To be able to calculate the probability of winning we first must count how many different groups of six numbers can be chosen from the forty-four.

Example 1.2.13 (Tournament) In a single-elimination tournament, such as the U.S. Open tennis tournament, players advance only if they win (in contrast to doubleelimination or round-robin tournaments). If we have 16 entrants, we might be interested in the number of paths a particular player can take to victory, where a path is taken to mean a sequence of opponents.

Counting problems, in general, sound complicated, and often we must do our counting subject to many restrictions. The way to solve such problems is to break them down into a series of simple tasks that are easy to count, and employ known rules of combining tasks. The following theorem is a first step in such a process and is sometimes known as the Fundamental Theorem of Counting.

Theorem 1.2.14 If a job consists of $k$ separate tasks, the ith of which can be done in $n_i$ ways, $i=1, \ldots, k$, then the entire $j o b$ can be done in $n_1 \times n_2 \times \cdots \times n_k$ ways.

Proof: It suffices to prove the theorem for $k=2$ (see Exercise 1.15). The proof is just a matter of careful counting. The first task can be done in $n_1$ ways, and for each of these ways we have $n_2$ choices for the second task. Thus, we can do the job in
$$\underbrace{\left(1 \times n_2\right)+\left(1 \times n_2\right)+\cdots+\left(1 \times n_2\right)}_{n_1 \text { terms }}=n_1 \times n_2$$
ways, establishing the theorem for $k=2$.

## 统计代写|统计推断代考Statistical Inference代写|Enumerating Outcomes

The counting techniques of the previous section are useful when the sample space $S$ is a finite set and all the outcomes in $S$ are equally likely. Then probabilities of events can be calculated by simply counting the number of outcomes in the event. To see this, suppose that $S=\left{s_1, \ldots, s_N\right}$ is a finite sample space. Saying that all the outcomes are equally likely means that $P\left(\left{s_i\right}\right)=1 / N$ for every outcome $s_i$. Then, using Axiom 3 from Definition 1.2.4, we have, for any event $A$,
$$P(A)=\sum_{s_i \in A} P\left(\left{s_i\right}\right)=\sum_{s_i \in A} \frac{1}{N}=\frac{# \text { of elements in } A}{# \text { of elements in } S} .$$
For large sample spaces, the counting techniques might be used to calculate both the numerator and denominator of this expression.

Example 1.2.18 (Poker) Consider choosing a five-card poker hand from a standard deck of 52 playing cards. Obviously, we are sampling without replacement from the deck. But to specify the possible outcomes (possible hands), we must decide whether we think of the hand as being dealt sequentially (ordered) or all at once (unordered). If we wish to calculate probabilities for events that depend on the order, such as the probability of an ace in the first two cards, then we must use the ordered outcomes. But if our events do not depend on the order, we can use the unordered outcomes. For this example we use the unordered outcomes, so the sample space consists of all the five-card hands that can be chosen from the 52-card deck. There are $\left(\begin{array}{c}52 \ 5\end{array}\right)=2,598,960$ possible hands. If the deck is well shuffled and the cards are randomly dealt, it is reasonable to assign probability $1 / 2,598,960$ to each possible hand.

We now calculate some probabilities by counting outcomes in events. What is the probability of having four aces? How many different hands are there with four aces? If we specify that four of the cards are aces, then there are 48 different ways of specifying the fifth card. Thus,
$$P(\text { four aces })=\frac{48}{2,598,960}$$
less than 1 chance in 50,000 . Only slightly more complicated counting, using Theorem 1.2.14, allows us to calculate the probability of having four of a kind. There are 13 ways to specify which denomination there will be four of. After we specify these four cards, there are 48 ways of specifying the fifth. Thus, the total number of hands with four of a kind is $(13)(48)$ and
$$P(\text { four of a kind })=\frac{(13)(48)}{2,598,960}=\frac{624}{2,598,960}$$

# 统计推断代写

## 统计代写|统计推断代考Statistical Inference代写|Counting

$$\underbrace{\left(1 \times n_2\right)+\left(1 \times n_2\right)+\cdots+\left(1 \times n_2\right)}_{n_1 \text { terms }}=n_1 \times n_2$$

## 统计代写|统计推断代考Statistical Inference代写|Enumerating Outcomes

$$P(A)=\sum_{s_i \in A} P\left(\left{s_i\right}\right)=\sum_{s_i \in A} \frac{1}{N}=\frac{# \text { of elements in } A}{# \text { of elements in } S} .$$

$$P(\text { four aces })=\frac{48}{2,598,960}$$

$$P(\text { four of a kind })=\frac{(13)(48)}{2,598,960}=\frac{624}{2,598,960}$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。