### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|STAT3004

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• Statistical Inference 统计推断
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• 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|Random Variables

It frequently occurs that in performing an experiment we are mainly interested in some functions of the outcome as opposed to the outcome itself. For instance, in tossing dice we are often interested in the sum of the two dice and are not really concerned about the actual outcome. That is, we may be interested in knowing that the sum is seven and not be concerned over whether the actual outcome was $(1,6)$ or $(2,5)$ or $(3,4)$ or $(4,3)$ or $(5,2)$ or $(6,1)$. These quantities of interest, or more formally, these real-valued functions defined on the sample space, are known as random variables.

Since the value of a random variable is determined by the outcome of the experiment, we may assign probabilities to the possible values of the random variable.

Example 2.1. Letting $X$ denote the random variable that is defined as the sum of two fair dice; then
\begin{aligned} &P{X=2}=P{(1,1)}=\frac{1}{36}, \ &P{X=3}=P{(1,2),(2,1)}=\frac{2}{36} \ &P{X=4}=P{(1,3),(2,2),(3,1)}=\frac{3}{36} \ &P{X=5}=P{(1,4),(2,3),(3,2),(4,1)}=\frac{4}{36} \ &P{X=6}=P{(1,5),(2,4),(3,3),(4,2),(5,1)}=\frac{5}{36} \ &P{X=7}=P{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}=\frac{6}{36} \ &P{X=8}=P{(2,6),(3,5),(4,4),(5,3),(6,2)}=\frac{5}{36} \ &P{X=9}=P{(3,6),(4,5),(5,4),(6,3)}=\frac{4}{36} \ &P{X=10}=P{(4,6),(5,5),(6,4)}=\frac{3}{36}, \ &P{X=11}=P{(5,6),(6,5)}=\frac{2}{36}, \ &P{X=12}=P{(6,6)}=\frac{1}{36} \end{aligned}
In other words, the random variable $X$ can take on any integral value between two and twelve, and the probability that it takes on each value is given by Eq. (2.1). Since $X$ must take on one of the values two through twelve, we must have
$$1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}$$
which may be checked from Eq. (2.1).

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

As was previously mentioned, a random variable that can take on at most a countable number of possible values is said to be discrete. For a discrete random variable $X$, we define the probability mass function $p(a)$ of $X$ by
$$p(a)=P{X=a}$$
The probability mass function $p(a)$ is positive for at most a countable number of values of $a$. That is, if $X$ must assume one of the values $x_{1}, x_{2}, \ldots$, then
$$\begin{array}{ll} p\left(x_{i}\right)>0, & i=1,2, \ldots \ p(x)=0, & \text { all other values of } x \end{array}$$
Since $X$ must take on one of the values $x_{i}$, we have
$$\sum_{i=1}^{\infty} p\left(x_{i}\right)=1$$
The cumulative distribution function $F$ can be expressed in terms of $p(a)$ by
$$F(a)=\sum_{\text {all }} p\left(x_{i} \leq a\right.$$
For instance, suppose $X$ has a probability mass function given by
$$p(1)=\frac{1}{2}, \quad p(2)=\frac{1}{3}, \quad p(3)=\frac{1}{6}$$
then, the cumulative distribution function $F$ of $X$ is given by
$$F(a)= \begin{cases}0, & a<1 \ \frac{1}{2}, & 1 \leq a<2 \ \frac{5}{6}, & 2 \leq a<3 \ 1, & 3 \leq a\end{cases}$$
This is graphically presented in Fig. 2.1.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Binomial Random Variable

Suppose that $n$ independent trials, each of which results in a “success” with probability $p$ and in a “failure” with probability $1-p$, are to be performed. If $X$ represents the number of successes that occur in the $n$ trials, then $X$ is said to be a binomial random variable with parameters $(n, p)$.

The probability mass function of a binomial random variable having parameters $(n, p)$ is given by
$$p(i)=\left(\begin{array}{c} n \ i \end{array}\right) p^{i}(1-p)^{n-i}, \quad i=0,1, \ldots, n$$
where
$$\left(\begin{array}{l} n \ i \end{array}\right)=\frac{n !}{(n-i) ! i !}$$
equals the number of different groups of $i$ objects that can be chosen from a set of $n$ objects. The validity of Eq. (2.3) may be verified by first noting that the probability of any particular sequence of the $n$ outcomes containing $i$ successes and $n-i$ failures is, by the assumed independence of trials, $p^{i}(1-p)^{n-i}$. Eq. (2.3) then follows since there are $\left(\begin{array}{l}n \ i\end{array}\right)$ different sequences of the $n$ outcomes leading to $i$ successes and $n-i$ failures. For instance, if $n=3, i=2$, then there are $\left(\begin{array}{l}3 \ 2\end{array}\right)=3$ ways in which the three trials can result in two successes. Namely, any one of the three outcomes $(s, s, f),(s, f, s),(f, s, s)$, where the outcome $(s, s, f)$ means that the first two trials are successes and the third a failure. Since each of the three outcomes $(s, s, f),(s, f, s),(f, s, s)$ has a probability $p^{2}(1-p)$ of occurring the desired probability is thus $\left(\begin{array}{l}3 \ 2\end{array}\right) p^{2}(1-p)$.

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

1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}1=P\left{\bigcup_{n=2}^{12}{X=n}\right}=\sum_{n=2}^{12} P{X=n}

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

p(一个)=磷X=一个

p(X一世)>0,一世=1,2,… p(X)=0, 所有其他值 X

∑一世=1∞p(X一世)=1

F(一个)=∑全部 p(X一世≤一个

p(1)=12,p(2)=13,p(3)=16

F(一个)={0,一个<1 12,1≤一个<2 56,2≤一个<3 1,3≤一个

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Binomial Random Variable

p(一世)=(n 一世)p一世(1−p)n−一世,一世=0,1,…,n

(n 一世)=n!(n−一世)!一世!

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