### 统计代写|统计推断作业代写statistical inference代考|Kolmogorov’s Three Rules

statistics-lab™ 为您的留学生涯保驾护航 在代写 统计推断statistical inference方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写统计推断statistical inference方面经验极为丰富，各种代写 统计推断statistical inference相关的作业也就用不着说。

• 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

## 统计代写|统计推断作业代写statistical inference代考|Kolmogorov’s Three Rules

Given a random experiment (say rolling a die) a probability measure is a population quantity that summarizes the randomness. The brilliant discovery of the father of probability, the Russian mathematician Kolmogorov, was that to satisfy our intuition about how probability should behave, only three rules were needed.
Consider an experiment with a random outcome. Probability takes a possible outcome from an experiment and:

1. assigns it a number between 0 and 1
2. requires that the probability that something occurs is 1
3. required that the probability of the union of any two sets of outcomes that have nothing in common (mutually exclusive) is the sum of their respective probabilities.
From these simple rules all of the familiar rules of probability can be developed. This all might seem a little odd at first and so we’ll build up our intuition with some simple examples based on coin flipping and die rolling.
I would like to reiterate the important definition that we wrote out: mutually exclusive. Two events are mutually exclusive if they cannot both simultaneously occur. For example, we cannot simultaneously get a 1 and a 2 on a die. Rule 3 says that since the event of getting a 1 and 2 on a die are mutually exclusive, the probability of getting at least one (the union) is the sum of their probabilities. So if we know that the probability of getting a 1 is $1 / 6$ and the probability of getting a 2 is $1 / 6$, then the probability of getting a 1 or a 2 is $2 / 6$, the sum of the two probabilities since they are mutually exclusive.

## 统计代写|统计推断作业代写statistical inference代考|Consequences of The Three Rules

Let’s cover some consequences of our three simple rules. Take, for example, the probability that something occurs is 1 minus the probability of the opposite occurring. Let $A$ be the event that we get a 1 or a 2 on a rolled die. Then $A^{c}$ is the opposite, getting a $3,4,5$ or 6 . Since $A$ and $A^{c}$ cannot both simultaneously occur, they are mutually exclusive. So the probability that either $A$ or $A^{c}$ is $P(A)+P\left(A^{c}\right)$. Notice, that the probability that either occurs is the probability of getting a $1,2,3,4,5$ or 6 , or in other words, the probability that something occurs, which is 1 by rule number 2 . So we have that $1=P(A)+P\left(A^{c}\right)$ or that $P(A)=1-P\left(A^{c}\right)$

We won’t go through this tedious exercise (since Kolmogorov already did it for us). Instead here’s a list of some of the consequences of Kolmogorov’s rules that are often useful.
The probability that nothing occurs is 0
The probability that something occurs is 1
The probability of something is 1 minus the probability that the opposite occurs
The probability of at least one of two (or more) things that can not simultaneously occur (mutually exclusive) is the sum of their respective probabilities
For any two events the probability that at least one occurs is the sum of their probabilities minus their intersection.
This last rules states that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ shows what is the issue with adding probabilities that are not mutually exclusive. If we do this, we’ve added the probability that both occur in twice! (Watch the video where I draw a Venn diagram to illustrate this).

## 统计代写|统计推断作业代写statistical inference代考|Example of Implementing Probability Calculus

The National Sleep Foundation (www.sleepfoundation.org) reports that around $3 \%$ of the American population has sleep apnea. They also report that around $10 \%$ of the North American and European population has restless leg syndrome. Does this imply that $13 \%$ of people will have at least one sleep problems of these sorts? In other words, can we simply add these two probabilities?
Answer: No, the events can simultaneously occur and so are not mutually exclusive. To elaborate let:
$A_{2}={$ Person has RLS $}$
Then
$P\left(A_{1} \cup A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)-P\left(A_{1} \cap A_{2}\right)$
$=0.13$ – Probability of having both
Given the scenario, it’s likely that some fraction of the population has both. This example serves as a reminder don’t add probabilities unless the events are mutually exclusive. We’ll have a similar rule for multiplying probabilities and independence.

## 统计代写|统计推断作业代写statistical inference代考|Kolmogorov’s Three Rules

1. 为其分配一个介于 0 和 1 之间的数字
2. 要求某事发生的概率为 1
3. 要求任何两组没有共同点（互斥）的结果的并集概率是它们各自概率的总和。
从这些简单的规则可以发展出所有熟悉的概率规则。乍一看，这一切似乎有点奇怪，因此我们将通过一些基于掷硬币和掷骰子的简单示例来建立我们的直觉。
我想重申我们写出的重要定义：互斥。如果两个事件不能同时发生，则它们是互斥的。例如，我们不能同时在骰子上得到 1 和 2。规则 3 说，由于骰子上得到 1 和 2 的事件是互斥的，因此得到至少一个（并集）的概率是它们的概率之和。所以如果我们知道得到 1 的概率是1/6得到 2 的概率是1/6, 那么得到 1 或 2 的概率为2/6, 两个概率之和，因为它们是互斥的。

=0.13– 两者兼有

## 广义线性模型代考

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## MATLAB代写

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