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|Moments and Moment Generating Functions

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## 统计代写|统计推断代考Statistical Inference代写|Moments and Moment Generating Functions

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|Differentiating Under an Integral Sign

In the previous section we encountered an instance in which we desired to interchange the order of integration and differentiation. This situation is encountered frequently in theoretical statistics. The purpose of this section is to characterize conditions under which this operation is legitimate. We will also discuss interchanging the order of differentiation and summation.

Many of these conditions can be established using standard theorems from calculus and detailed proofs can be found in most calculus textbooks. Thus, detailed proofs will not be presented here.
We first want to establish the method of calculating
$$\frac{d}{d \theta} \int_{a(\theta)}^{b(\theta)} f(x, \theta) d x$$
where $-\infty<a(\theta), b(\theta)<\infty$ for all $\theta$. The rule for differentiating (2.4.1) is called Leibnitz’s Rule and is an application of the Fundamental Theorem of Calculus and the chain rule.

Theorem 2.4.1 (Leibnitz’s Rule) If $f(x, \theta), a(\theta)$, and $b(\theta)$ are differentiable with respect to $\theta$, then
$$\frac{d}{d \theta} \int_{a(\theta)}^{b(\theta)} f(x, \theta) d x=f(b(\theta), \theta) \frac{d}{d \theta} b(\theta)-f(a(\theta), \theta) \frac{d}{d \theta} a(\theta)+\int_{a(\theta)}^{b(\theta)} \frac{\partial}{\partial \theta} f(x, \theta) d x .$$
Notice that if $a(\theta)$ and $b(\theta)$ are constant, we have a special case of Leibnitz’s Rule:
$$\frac{d}{d \theta} \int_a^b f(x, \theta) d x=\int_a^b \frac{\partial}{\partial \theta} f(x, \theta) d x$$

# 统计推断代写

## 统计代写|统计推断代考Statistical Inference代写|Moments and Moment Generating Functions

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

## 统计代写|统计推断代考Statistical Inference代写|Differentiating Under an Integral Sign

$$\frac{d}{d \theta} \int_{a(\theta)}^{b(\theta)} f(x, \theta) d x$$

$$\frac{d}{d \theta} \int_{a(\theta)}^{b(\theta)} f(x, \theta) d x=f(b(\theta), \theta) \frac{d}{d \theta} b(\theta)-f(a(\theta), \theta) \frac{d}{d \theta} a(\theta)+\int_{a(\theta)}^{b(\theta)} \frac{\partial}{\partial \theta} f(x, \theta) d x .$$

$$\frac{d}{d \theta} \int_a^b f(x, \theta) d x=\int_a^b \frac{\partial}{\partial \theta} f(x, \theta) d x$$

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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。