### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Introduction to Probability Theory

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Sample Space and Events

Suppose that we are about to perform an experiment whose outcome is not predictable in advance. However, while the outcome of the experiment will not be known in advance, let us suppose that the set of all possible outcomes is known. This set of all possible outcomes of an experiment is known as the sample space of the experiment and is denoted by $S$.
Some examples are the following.

1. If the experiment consists of the flipping of a coin, then
$$S={H, T}$$
where $H$ means that the outcome of the toss is a head and $T$ that it is a tail.
2. If the experiment consists of rolling a die, then the sample space is
$$S={1,2,3,4,5,6}$$
where the outcome $i$ means that $i$ appeared on the die, $i=1,2,3,4,5,6$.
3. If the experiment consists of flipping two coins, then the sample space consists of the following four points:
$$S={(H, H),(H, T),(T, H),(T, T)}$$
The outcome will be $(H, H)$ if both coins come up heads; it will be $(H, T)$ if the first coin comes up heads and the second comes up tails; it will be $(T, H)$ if the

first comes up tails and the second heads; and it will be $(T, T)$ if both coins come up tails.

1. If the experiment consists of rolling two dice, then the sample space consists of the following 36 points:
$$S=\left{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right}$$
where the outcome $(i, j)$ is said to occur if $i$ appears on the first die and $j$ on the second die.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Probabilities Defined on Events

Consider an experiment whose sample space is $S$. For each event $E$ of the sample space $S$, we assume that a number $P(E)$ is defined and satisfies the following three conditions:
(i) $0 \leqslant P(E) \leqslant 1$.
(ii) $P(S)=1$.
(iii) For any sequence of events $E_{1}, E_{2}, \ldots$ that are mutually exclusive, that is, events for which $E_{n} E_{m}=\emptyset$ when $n \neq m$, then
$$P\left(\bigcup_{n=1}^{\infty} E_{n}\right)=\sum_{n=1}^{\infty} P\left(E_{n}\right)$$
We refer to $P(E)$ as the probability of the event $E$.

Example 1.1. In the coin tossing example, if we assume that a head is equally likely to appear as a tail, then we would have
$$P({H})=P({T})=\frac{1}{2}$$
On the other hand, if we had a biased coin and felt that a head was twice as likely to appear as a tail, then we would have
$$P({H})=\frac{2}{3}, \quad P({T})=\frac{1}{3}$$
Example 1.2. In the die tossing example, if we supposed that all six numbers were equally likely to appear, then we would have
$$P({1})=P({2})=P({3})=P({4})=P({5})=P({6})=\frac{1}{6}$$
From (iii) it would follow that the probability of getting an even number would equal
\begin{aligned} P({2,4,6}) &=P({2})+P({4})+P({6}) \ &=\frac{1}{2} \end{aligned}

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Conditional Probabilities

Suppose that we toss two dice and that each of the 36 possible outcomes is equally likely to occur and hence has probability $\frac{1}{36}$. Suppose that we observe that the first die is a four. Then, given this information, what is the probability that the sum of the two dice equals six? To calculate this probability we reason as follows: Given that the initial die is a four, it follows that there can be at most six possible outcomes of our experiment, namely, $(4,1),(4,2),(4,3),(4,4),(4,5)$, and $(4,6)$. Since each of these outcomes originally had the same probability of occurring, they should still have equal probabilities. That is, given that the first die is a four, then the (conditional) probability of each of the outcomes $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$ is $\frac{1}{6}$ while the (conditional) probability of the other 30 points in the sample space is 0 . Hence, the desired probability will be $\frac{1}{6}$.

If we let $E$ and $F$ denote, respectively, the event that the sum of the dice is six and the event that the first die is a four, then the probability just obtained is called the conditional probability that $E$ occurs given that $F$ has occurred and is denoted by
$$P(E \mid F)$$
A general formula for $P(E \mid F)$ that is valid for all events $E$ and $F$ is derived in the same manner as the preceding. Namely, if the event $F$ occurs, then in order for $E$ to occur it is necessary for the actual occurrence to be a point in both $E$ and in $F$, that is, it must be in $E F$. Now, because we know that $F$ has occurred, it follows that $F$ becomes our new sample space and hence the probability that the event $E F$ occurs will equal the probability of $E F$ relative to the probability of $F$. That is,
$$P(E \mid F)=\frac{P(E F)}{P(F)}$$
Note that Eq. (1.5) is only well defined when $P(F)>0$ and hence $P(E \mid F)$ is only defined when $P(F)>0$.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Sample Space and Events

1. 如果实验包括掷硬币，那么
小号=H,吨
在哪里H表示掷骰子的结果是正面，并且吨它是一条尾巴。
2. 如果实验由掷骰子组成，则样本空间为
小号=1,2,3,4,5,6
结果在哪里一世意思是一世出现在模具上，一世=1,2,3,4,5,6.
3. 如果实验是抛两枚硬币，那么样本空间由以下四点组成：
小号=(H,H),(H,吨),(吨,H),(吨,吨)
结果将是(H,H)如果两个硬币都出现正面；这将是(H,吨)如果第一枚硬币正面朝上，第二枚硬币反面；这将是(吨,H)如果

1. 如果实验由掷两个骰子组成，那么样本空间由以下 36 个点组成：
S=\left{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \ ( 2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \ (3,1),(3,2),(3, 3),(3,4),(3,5),(3,6) \ (4,1),(4,2),(4,3),(4,4),(4,5) ,(4,6) \ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \ (6,1),( 6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right}S=\left{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \ ( 2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \ (3,1),(3,2),(3, 3),(3,4),(3,5),(3,6) \ (4,1),(4,2),(4,3),(4,4),(4,5) ,(4,6) \ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \ (6,1),( 6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right}
结果在哪里(一世,j)据说如果发生一世出现在第一个骰子上并且j在第二个模具上。

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Probabilities Defined on Events

(i)0⩽磷(和)⩽1.
(二)磷(小号)=1.
(iii) 对于任何事件序列和1,和2,…是互斥的，也就是说，事件和n和米=∅什么时候n≠米， 然后

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## MATLAB代写

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