### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|E 2204

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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|The Discrete Case

If $X$ is a discrete random variable having a probability mass function $p(x)$, then the expected value of $X$ is defined by
$$E[X]=\sum_{x: p(x)>0} x p(x)$$
In other words, the expected value of $X$ is a weighted average of the possible values that $X$ can take on, each value being weighted by the probability that $X$ assumes that value. For example, if the probability mass function of $X$ is given by
$$p(1)=\frac{1}{2}=p(2)$$
then
$$E[X]=1\left(\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=\frac{3}{2}$$
is just an ordinary average of the two possible values 1 and 2 that $X$ can assume. On the other hand, if
$$p(1)=\frac{1}{3}, \quad p(2)=\frac{2}{3}$$
then
$$E[X]=1\left(\frac{1}{3}\right)+2\left(\frac{2}{3}\right)=\frac{5}{3}$$
is a weighted average of the two possible values 1 and 2 where the value 2 is given twice as much weight as the value 1 since $p(2)=2 p(1)$.
Example 2.15. Find $E[X]$ where $X$ is the outcome when we roll a fair die.
Solution: $\quad$ Since $p(1)=p(2)=p(3)=p(4)=p(5)=p(6)=\frac{1}{6}$, we obtain
$$E[X]=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right)=\frac{7}{2}$$
Example 2.16 (Expectation of a Bernoulli Random Variable). Calculate $E[X]$ when $X$ is a Bernoulli random variable with parameter $p$.
Solution: Since $p(0)=1-p, p(1)=p$, we have
$$E[X]=0(1-p)+1(p)=p$$
Thus, the expected number of successes in a single trial is just the probability that the trial will be a success.

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

We may also define the expected value of a continuous random variable. This is done as follows. If $X$ is a continuous random variable having a probability density function $f(x)$, then the expected value of $X$ is defined by
$$E[X]=\int_{-\infty}^{\infty} x f(x) d x$$
Example 2.20 (Expectation of a Uniform Random Variable). Calculate the expectation of a random variable uniformly distributed over $(\alpha, \beta)$.

Solution: From Eq. (2.8) we have
\begin{aligned} E[X] &=\int_{\alpha}^{\beta} \frac{x}{\beta-\alpha} d x \ &=\frac{\beta^{2}-\alpha^{2}}{2(\beta-\alpha)} \ &=\frac{\beta+\alpha}{2} \end{aligned}
In other words, the expected value of a random variable uniformly distributed over the interval $(\alpha, \beta)$ is just the midpoint of the interval.

Example 2.21 (Expectation of an Exponential Random Variable). Let $X$ be exponentially distributed with parameter $\lambda$. Calculate $E[X]$.
Solution:
$$E[X]=\int_{0}^{\infty} x \lambda e^{-\lambda x} d x$$
Integrating by parts $\left(d v=\lambda e^{-\lambda x} d x, u=x\right)$ yields
\begin{aligned} E[X] &=-\left.x e^{-\lambda x}\right|{0} ^{\infty}+\int{0}^{\infty} e^{-\lambda x} d x \ &=0-\left.\frac{e^{-\lambda x}}{\lambda}\right|_{0} ^{\infty} \ &=\frac{1}{\lambda} \end{aligned}

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Expectation of a Function of a Random Variable

Suppose now that we are given a random variable $X$ and its probability distribution (that is, its probability mass function in the discrete case or its probability density function in the continuous case). Suppose also that we are interested in calculating not the expected value of $X$, but the expected value of some function of $X$, say, $g(X)$. How do we go about doing this? One way is as follows. Since $g(X)$ is itself a random variable, it must have a probability distribution, which should be computable from a knowledge of the distribution of $X$. Once we have obtained the distribution of $g(X)$, we can then compute $E[g(X)]$ by the definition of the expectation.
Example 2.23. Suppose $X$ has the following probability mass function:
$$p(0)=0.2, \quad p(1)=0.5, \quad p(2)=0.3$$
Calculate $E\left[X^{2}\right]$
Solution: Letting $Y=X^{2}$, we have that $Y$ is a random variable that can take on one of the values $0^{2}, 1^{2}, 2^{2}$ with respective probabilities
\begin{aligned} &p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \ &p_{Y}(1)=P\left{Y=1^{2}\right}=0.5 \ &p_{Y}(4)=P\left{Y=2^{2}\right}=0.3 \end{aligned}
Hence,
$$E\left[X^{2}\right]=E[Y]=0(0.2)+1(0.5)+4(0.3)=1.7$$
Note that
$$1.7=E\left[X^{2}\right] \neq(E[X])^{2}=1.21$$

p(1)=12=p(2)

p(1)=13,p(2)=23

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

\begin{aligned} E[X] &=-\left.xe^{-\lambda x}\right| {0} ^{\infty}+\int {0}^{\infty} e^{-\lambda x} dx \ &=0-\left.\frac{e^{-\lambda x}}{\ lambda}\right|_{0} ^{\infty} \ &=\frac{1}{\lambda} \end{aligned}

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Expectation of a Function of a Random Variable

p(0)=0.2,p(1)=0.5,p(2)=0.3

\begin{对齐} &p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \ &p_{Y}(1)=P\left{Y=1^{2}\右}=0.5 \ &p_{Y}(4)=P\left{Y=2^{2}\right}=0.3 \end{对齐}\begin{对齐} &p_{Y}(0)=P\left{Y=0^{2}\right}=0.2 \ &p_{Y}(1)=P\left{Y=1^{2}\右}=0.5 \ &p_{Y}(4)=P\left{Y=2^{2}\right}=0.3 \end{对齐}

1.7=和[X2]≠(和[X])2=1.21

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