### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|MAP 4102

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• 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

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

Suppose that independent trials, each having probability $p$ of being a success, are performed until a success occurs. If we let $X$ be the number of trials required until the first success, then $X$ is said to be a geometric random variable with parameter $p$. Its probability mass function is given by
$$p(n)=P{X=n}=(1-p)^{n-1} p, \quad n=1,2, \ldots$$

Eq. (2.4) follows since in order for $X$ to equal $n$ it is necessary and sufficient that the first $n-1$ trials be failures and the $n$th trial a success. Eq. (2.4) follows since the outcomes of the successive trials are assumed to be independent.
To check that $p(n)$ is a probability mass function, we note that
$$\sum_{n=1}^{\infty} p(n)=p \sum_{n=1}^{\infty}(1-p)^{n-1}=1$$

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

A random variable $X$, taking on one of the values $0,1,2, \ldots$, is said to be a Poisson random variable with parameter $\lambda$, if for some $\lambda>0$,
$$p(i)=P{X=i}=e^{-\lambda} \frac{\lambda^{i}}{i !}, \quad i=0,1, \ldots$$
Eq. (2.5) defines a probability mass function since
$$\sum_{i=0}^{\infty} p(i)=e^{-\lambda} \sum_{i=0}^{\infty} \frac{\lambda^{i}}{i !}=e^{-\lambda} e^{\lambda}=1$$
The Poisson random variable has a wide range of applications in a diverse number of areas, as will be seen in Chapter $5 .$

An important property of the Poisson random variable is that it may be used to approximate a binomial random variable when the binomial parameter $n$ is large and $p$ is small. To see this, suppose that $X$ is a binomial random variable with parameters $(n, p)$, and let $\lambda=n p$. Then
\begin{aligned} P{X=i} &=\frac{n !}{(n-i) ! i !} p^{i}(1-p)^{n-i} \ &=\frac{n !}{(n-i) ! i !}\left(\frac{\lambda}{n}\right)^{i}\left(1-\frac{\lambda}{n}\right)^{n-i} \ &=\frac{n(n-1) \cdots(n-i+1)}{n^{i}} \frac{\lambda^{i}}{i !} \frac{(1-\lambda / n)^{n}}{(1-\lambda / n)^{i}} \end{aligned}
Now, for $n$ large and $p$ small
$$\left(1-\frac{\lambda}{n}\right)^{n} \approx e^{-\lambda}, \quad \frac{n(n-1) \cdots(n-i+1)}{n^{i}} \approx 1, \quad\left(1-\frac{\lambda}{n}\right)^{i} \approx 1$$
Hence, for $n$ large and $p$ small,
$$P{X=i} \approx e^{-\lambda} \frac{\lambda^{i}}{i !}$$

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

In this section, we shall concern ourselves with random variables whose set of possible values is uncountable. Let $X$ be such a random variable. We say that $X$ is a continuous random variable if there exists a nonnegative function $f(x)$, defined for all real $x \in$ $(-\infty, \infty)$, having the property that for any set $B$ of real numbers
$$P{X \in B}=\int_{B} f(x) d x$$
The function $f(x)$ is called the probability density function of the random variable $X$. In words, Eq. (2.6) states that the probability that $X$ will be in $B$ may be obtained by integrating the probability density function over the set $B$. Since $X$ must assume some value, $f(x)$ must satisfy
$$1=P{X \in(-\infty, \infty)}=\int_{-\infty}^{\infty} f(x) d x$$

All probability statements about $X$ can be answered in terms of $f(x)$. For instance, letting $B=[a, b]$, we obtain from Eq. (2.6) that
$$P{a \leq X \leq b}=\int_{a}^{b} f(x) d x$$
If we let $a=b$ in the preceding, then
$$P{X=a}=\int_{a}^{a} f(x) d x=0$$
In words, this equation states that the probability that a continuous random variable will assume any particular value is zero.

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

p(n)=磷X=n=(1−p)n−1p,n=1,2,…

∑n=1∞p(n)=p∑n=1∞(1−p)n−1=1

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

p(一世)=磷X=一世=和−λλ一世一世!,一世=0,1,…

∑一世=0∞p(一世)=和−λ∑一世=0∞λ一世一世!=和−λ和λ=1

(1−λn)n≈和−λ,n(n−1)⋯(n−一世+1)n一世≈1,(1−λn)一世≈1

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

1=磷X∈(−∞,∞)=∫−∞∞F(X)dX

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## MATLAB代写

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