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

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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|Joint Distribution Functions

Thus far, we have concerned ourselves with the probability distribution of a single random variable. However, we are often interested in probability statements concerning two or more random variables. To deal with such probabilities, we define, for any

two random variables $X$ and $Y$, the joint cumulative probability distribution function of $X$ and $Y$ by
$$F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)$$
Similarly,
$$p_{Y}(y)=\sum_{x: p(x, y)>0} p(x, y)$$

The probability mass function of $X$ may be obtained from $p(x, y)$ by
$$p_{X}(x)=\sum_{y: p(x, y)>0} p(x, y)$$
Similarly,
$$p_{Y}(y)=\sum_{x: p(x, y)>0} p(x, y)$$
We say that $X$ and $Y$ are jointly continuous if there exists a function $f(x, y)$, defined for all real $x$ and $y$, having the property that for all sets $A$ and $B$ of real numbers
$$P{X \in A, Y \in B}=\int_{B} \int_{A} f(x, y) d x d y$$
The function $f(x, y)$ is called the joint probability density function of $X$ and $Y$. The probability density of $X$ can be obtained from a knowledge of $f(x, y)$ by the following reasoning:
\begin{aligned} P{X \in A} &=P{X \in A, Y \in(-\infty, \infty)} \ &=\int_{-\infty}^{\infty} \int_{A} f(x, y) d x d y \ &=\int_{A} f_{X}(x) d x \end{aligned}

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

The random variables $X$ and $Y$ are said to be independent if, for all $a, b$,
$$P{X \leq a, Y \leq b}=P{X \leq a} P{Y \leq b}$$
In other words, $X$ and $Y$ are independent if, for all $a$ and $b$, the events $E_{a}={X \leq a}$ and $F_{b}={Y \leq b}$ are independent.

In terms of the joint distribution function $F$ of $X$ and $Y$, we have that $X$ and $Y$ are independent if
$$F(a, b)=F_{X}(a) F_{Y}(b) \quad \text { for all } a, b$$
When $X$ and $Y$ are discrete, the condition of independence reduces to
$$p(x, y)=p_{X}(x) p_{Y}(y)$$
while if $X$ and $Y$ are jointly continuous, independence reduces to
$$f(x, y)=f_{X}(x) f_{Y}(y)$$

To prove this statement, consider first the discrete version, and suppose that the joint probability mass function $p(x, y)$ satisfies Eq. (2.13). Then
\begin{aligned} P{X \leq a, Y \leq b} &=\sum_{y \leq b} \sum_{x \leq a} p(x, y) \ &=\sum_{y \leq b} \sum_{x \leq a} p_{X}(x) p_{Y}(y) \ &=\sum_{y \leq b} p_{Y}(y) \sum_{x \leq a} p_{X}(x) \ &=P{Y \leq b} P{X \leq a} \end{aligned}
and so $X$ and $Y$ are independent. That Eq. (2.14) implies independence in the continuous case is proven in the same manner and is left as an exercise.
An important result concerning independence is the following.

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

The covariance of any two random variables $X$ and $Y$, denoted by $\operatorname{Cov}(X, Y)$, is defined by
\begin{aligned} \operatorname{Cov}(X, Y) &=E[(X-E[X])(Y-E[Y])] \ &=E[X Y-Y E[X]-X E[Y]+E[X] E[Y]] \ &=E[X Y]-E[Y] E[X]-E[X] E[Y]+E[X] E[Y] \ &=E[X Y]-E[X] E[Y] \end{aligned}
Note that if $X$ and $Y$ are independent, then by Proposition $2.3$ it follows that $\operatorname{Cov}(X, Y)=0$

Let us consider now the special case where $X$ and $Y$ are indicator variables for whether or not the events $A$ and $B$ occur. That is, for events $A$ and $B$, define
$$X=\left{\begin{array}{ll} 1, & \text { if } A \text { occurs } \ 0, & \text { otherwise, } \end{array} \quad Y= \begin{cases}1, & \text { if } B \text { occurs } \ 0, & \text { otherwise }\end{cases}\right.$$
Then,
$$\operatorname{Cov}(X, Y)=E[X Y]-E[X] E[Y]$$
and, because $X Y$ will equal 1 or 0 depending on whether or not both $X$ and $Y$ equal 1 , we see that
$$\operatorname{Cov}(X, Y)=P{X=1, Y=1}-P{X=1} P{Y=1}$$
From this we see that
\begin{aligned} \operatorname{Cov}(X, Y)>0 & \Leftrightarrow P{X=1, Y=1}>P{X=1} P{Y=1} \ & \Leftrightarrow \frac{P{X=1, Y=1}}{P{X=1}}>P{Y=1} \ & \Leftrightarrow P{Y=1 \mid X=1}>P{Y=1} \end{aligned}
That is, the covariance of $X$ and $Y$ is positive if the outcome $X=1$ makes it more likely that $Y=1$ (which, as is easily seen by symmetry, also implies the reverse).
In general it can be shown that a positive value of $\operatorname{Cov}(X, Y)$ is an indication that $Y$ tends to increase as $X$ does, whereas a negative value indicates that $Y$ tends to decrease as $X$ increases.

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

F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)F(a, b)=P{X \leq a, Y \leq b}, \quad-\infty0} p(x, y)

p是(是)=∑X:p(X,是)>0p(X,是)

pX(X)=∑是:p(X,是)>0p(X,是)

p是(是)=∑X:p(X,是)>0p(X,是)

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

F(一个,b)=FX(一个)F是(b) 对所有人 一个,b

p(X,是)=pX(X)p是(是)

F(X,是)=FX(X)F是(是)

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

$$X=\left{ 1, 如果 一个 发生 0, 否则， \四Y= {1, 如果 乙 发生 0, 否则 \正确的。 吨H和n, \operatorname{Cov}(X, Y)=E[XY]-E[X] E[Y] 一个nd,b和C一个在s和X是在一世ll和q在一个l1○r0d和p和nd一世nG○n在H和吨H和r○rn○吨b○吨HX一个nd是和q在一个l1,在和s和和吨H一个吨 \ 操作员名称 {Cov} (X, Y) = P {X = 1, Y = 1} -P {X = 1} P {Y = 1} Fr○米吨H一世s在和s和和吨H一个吨 这⁡(X,是)>0⇔磷X=1,是=1>磷X=1磷是=1 ⇔磷X=1,是=1磷X=1>磷是=1 ⇔磷是=1∣X=1>磷是=1$$

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## MATLAB代写

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