### 数学代写|计量经济学原理代写Principles of Econometrics代考|Calculating a Conditional Probability

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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|计量经济学原理代写Principles of Econometrics代考|Statistical Independence

When selecting a shaded slip from Table P.1, the probability of selecting each possible outcome, $x=1,2,3$, and 4 is $0.25$. In the population of shaded slips the numeric values are equally likely. The probability of randomly selecting $X=2$ from the entire population, from the marginal $p d f$, is $P(X=2)=f_{X}(2)=0.2$. This is different from the conditional probability. Knowing that the slip is shaded tells us something about the probability of obtaining $X=2$. Such random variables are dependent in a statistical sense. Two random variables are statistically independent, or simply independent, if the conditional probability that $X=x$ given that $Y=y$ is the same as the unconditional probability that $X=x$. This means, if $X$ and $Y$ are independent random variables, then
$$P(X=x \mid Y=y)=P(X=x)$$

Equivalently, if $X$ and $Y$ are independent, then the conditional $p d f$ of $X$ given $Y=y$ is the same as the unconditional, or marginal, $p d f$ of $X$ alone.
$$f(x \mid y)=\frac{f(x, y)}{f_{Y}(y)}=f_{X}(x)$$
Solving (P.6) for the joint $p d f$, we can also say that $X$ and $Y$ are statistically independent if their joint $p d f$ factors into the product of their marginal $p d f \mathrm{~s}$
$$P(X=x, Y=y)=f(x, y)=f_{X}(x) f_{Y}(y)=P(X=x) \times P(Y=y)$$
If $(\mathbf{P} .5)$ or $(\mathbf{P} .7)$ is true for each and every pair of values $x$ and $y$, then $X$ and $Y$ are statistically independent. This result extends to more than two random variables. The rule allows us to check the independence of random variables $X$ and $Y$ in Table P.4. If (P.7) is violated for any pair of values, then $X$ and $Y$ are not statistically independent. Consider the pair of values $X=1$ and $Y=1$.
$$P(X=1, Y=1)=f(1,1)=0.1 \neq f_{X}(1) f_{Y}(1)=P(X=1) \times P(Y=1)=0.1 \times 0.4=0.04$$
The joint probability is $0.1$ and the product of the individual probabilities is $0.04$. Since these are not equal, we can conclude that $X$ and $Y$ are not statistically independent.

## 数学代写|计量经济学原理代写Principles of Econometrics代考|A Digression: Summation Notation

Throughout this book we will use a summation sign, denoted by the symbol $\sum$, to shorten algebraic expressions. Suppose the random variable $X$ takes the values $x_{1}, x_{2}, \ldots, x_{15}$. The sum of these values is $x_{1}+x_{2}+\cdots+x_{15}$. Rather than write this sum out each time we will represent it as $\sum_{i=1}^{15} x_{i}$, so that $\sum_{i=1}^{15} x_{i}=x_{1}+x_{2}+\cdots+x_{15}$. If we sum $n$ terms, a general number, then the summation will be $\sum_{i=1}^{n} x_{i}=x_{1}+x_{2}+\cdots+x_{n}$. In this notation

• The symbol $\sum$ is the capital Greek letter sigma and means “the sum of.”
• The letter $i$ is called the index of summation. This letter is arbitrary and may also appear as $t, j$, or $k$.
• The expression $\sum_{i=1}^{n} x_{i}$ is read “the sum of the terms $x_{i}$, from $i$ equals 1 to $n . “$
• The numbers 1 and $n$ are the lower limit and upper limit of summation.
The following rules apply to the summation operation.
Sum 1. The sum of $n$ values $x_{1}, \ldots, x_{n}$ is
$$\sum_{i=1}^{n} x_{i}=x_{1}+x_{2}+\cdots+x_{n}$$
Sum 2. If $a$ is a constant, then
$$\sum_{i=1}^{n} a x_{i}=a \sum_{i=1}^{n} x_{i}$$
Sum 3. If $a$ is a constant, then
$$\sum_{i=1}^{n} a=a+a+\cdots+a=n a$$

Sum 4. If $X$ and $Y$ are two variables, then
$$\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} y_{i}$$
Sum 5. If $X$ and $Y$ are two variables, then
$$\sum_{i=1}^{n}\left(a x_{i}+b y_{i}\right)=a \sum_{i=1}^{n} x_{i}+b \sum_{i=1}^{n} y_{i}$$
Sum 6. The arithmetic mean (average) of $n$ values of $X$ is
$$\bar{x}=\frac{\sum_{i=1}^{n} x_{i}}{n}=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$$
Sum 7. A property of the arithmetic mean (average) is that
$$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)=\sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} \bar{x}=\sum_{i=1}^{n} x_{i}-n \bar{x}=\sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} x_{i}=0$$
Sum 8. We often use an abbreviated form of the summation notation. For example, if $f(x)$ is a function of the values of $X$,
\begin{aligned} \sum_{i=1}^{n} f\left(x_{i}\right) &=f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right) \ &=\sum_{i} f\left(x_{i}\right)(” \text { Sum over all values of the index } i \text { “) }\ &=\sum_{x} f(x)\left(” \text { Sum over all possible values of } X^{\prime \prime}\right) \end{aligned}
Sum 9. Several summation signs can be used in one expression. Suppose the variable $Y$ takes $n$ values and $X$ takes $m$ values, and let $f(x, y)=x+y$. Then the double summation of this function is
$$\sum_{i=1}^{m} \sum_{j=1}^{n} f\left(x_{i}, y_{j}\right)=\sum_{i=1}^{m} \sum_{j=1}^{n}\left(x_{i}+y_{j}\right)$$
To evaluate such expressions work from the innermost sum outward. First set $i=1$ and sum over all values of $j$, and so on. That is,
$$\sum_{i=1}^{m} \sum_{j=1}^{n} f\left(x_{i}, y_{j}\right)=\sum_{i=1}^{m}\left[f\left(x_{i}, y_{1}\right)+f\left(x_{i}, y_{2}\right)+\cdots+f\left(x_{i}, y_{n}\right)\right]$$
The order of summation does not matter, so
$$\sum_{i=1}^{m} \sum_{j=1}^{n} f\left(x_{i}, y_{j}\right)=\sum_{j=1}^{n} \sum_{i=1}^{m} f\left(x_{i}, y_{j}\right)$$

## 数学代写|计量经济学原理代写Principles of Econometrics代考|Calculating an Expected Value

For a discrete random variable the probability that $X$ takes the value $x$ is given by its $p d f f(x)$, $P(X=x)=f(x)$. The expected value in (P.8) can be written equivalently as
\begin{aligned} \mu_{X} &=E(X)=x_{1} f\left(x_{1}\right)+x_{2} f\left(x_{2}\right)+\cdots+x_{n} f\left(x_{n}\right) \ &=\sum_{i=1}^{n} x_{i} f\left(x_{i}\right)=\sum_{x} x f(x) \end{aligned}
Using (P.9), the expected value of $X$, the numeric value on a randomly drawn slip from Table P.1 is
$$\mu_{x}=E(X)=\sum_{x=1}^{4} x f(x)=(1 \times 0.1)+(2 \times 0.2)+(3 \times 0.3)+(4 \times 0.4)=3$$
What does this mean? Draw one “slip” at random from Table P.1, and observe its numerical value $X$. This constitutes an experiment. If we repeat this experiment many times, the values $x=1,2,3$, and 4 will appear $10 \%, 20 \%, 30 \%$, and $40 \%$ of the time, respectively. The arithmetic average of all the numerical values will approach $\mu_{X}=3$, as the number of experiments becomes large. The key point is that the expected value of the random variable is the average value that occurs in many repeated trials of an experiment.

For continuous random variables, the interpretation of the expected value of $X$ is unchanged it is the average value of $X$ if many values are obtained by repeatedly performing the underlying random experiment. ${ }^{4}$

## 数学代写|计量经济学原理代写Principles of Econometrics代考|Statistical Independence

F(X∣是)=F(X,是)F是(是)=FX(X)

## 数学代写|计量经济学原理代写Principles of Econometrics代考|A Digression: Summation Notation

• 符号∑是大写的希腊字母 sigma，意思是“总和”。
• 信一世称为求和指数。这封信是任意的，也可能显示为吨,j， 或者ķ.
• 表达方式∑一世=1nX一世读作“条款的总和X一世， 从一世等于 1 到n.“
• 数字 1 和n是求和的下限和上限。
以下规则适用于求和运算。
总和 1. 总和n价值观X1,…,Xn是
∑一世=1nX一世=X1+X2+⋯+Xn
总和 2. 如果一个是一个常数，那么
∑一世=1n一个X一世=一个∑一世=1nX一世
总和 3. 如果一个是一个常数，那么
∑一世=1n一个=一个+一个+⋯+一个=n一个

∑一世=1n(X一世+是一世)=∑一世=1nX一世+∑一世=1n是一世

∑一世=1n(一个X一世+b是一世)=一个∑一世=1nX一世+b∑一世=1n是一世

X¯=∑一世=1nX一世n=X1+X2+⋯+Xnn

∑一世=1n(X一世−X¯)=∑一世=1nX一世−∑一世=1nX¯=∑一世=1nX一世−nX¯=∑一世=1nX一世−∑一世=1nX一世=0
Sum 8. 我们经常使用求和符号的缩写形式。例如，如果F(X)是值的函数X,

∑一世=1nF(X一世)=F(X1)+F(X2)+⋯+F(Xn) =∑一世F(X一世)(” 对索引的所有值求和 一世 “)  =∑XF(X)(” 对所有可能的值求和 X′′)
Sum 9. 在一个表达式中可以使用多个求和符号。假设变量是需要n价值观和X需要米值，并让F(X,是)=X+是. 那么这个函数的双重求和是

∑一世=1米∑j=1nF(X一世,是j)=∑一世=1米∑j=1n(X一世+是j)

∑一世=1米∑j=1nF(X一世,是j)=∑一世=1米[F(X一世,是1)+F(X一世,是2)+⋯+F(X一世,是n)]

∑一世=1米∑j=1nF(X一世,是j)=∑j=1n∑一世=1米F(X一世,是j)

## 数学代写|计量经济学原理代写Principles of Econometrics代考|Calculating an Expected Value

μX=和(X)=X1F(X1)+X2F(X2)+⋯+XnF(Xn) =∑一世=1nX一世F(X一世)=∑XXF(X)

μX=和(X)=∑X=14XF(X)=(1×0.1)+(2×0.2)+(3×0.3)+(4×0.4)=3

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