数学代写|概率论代写Probability theory代考|STAT7203

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数学代写|概率论代写Probability theory代考|The distribution of a function of a random variable

Sometimes we know the distribution of a random variable, yet we would like to determine the distribution of a function of a random variable. If $X$ is a random variable, then, in order to determine the distribution of $Y=g(X)$, we first should recognize whether $Y$ is discrete or continuous by determining the possible values of random variable $Y$. If $Y=g(X)$ is a discrete random variable, then, in order to determine its distribution, we should obtain its probability function. On the other hand, if $Y=g(X)$ is a continuous random variable, then, in order to obtain its distribution, we should obtain its density function. To do so, firstly, we obtain the cumulative distribution function and then differentiate it. This results in the density function. For better comprehension of this issue, consider the following examples.

If $X$ is a continuous random variable with the density function of
$$f_X(x)=\frac{1}{10} ; 0<x<10,$$
then determine the probability function of random variable $Y=[X]$ (the sign [] denotes the integer part).

Solution. The random variable $Y=[X]$ can take on the values of 1 to 9 . Thus, $Y$ is a discrete random variable and in order to determine its distribution, we should find its probability function. Hence, we have: \begin{aligned} & P(Y=0)=P(0 \leq X<1)=\int_0^1 \frac{1}{10} \mathrm{dx}=\frac{1}{10} \ & P(Y=1)=P(1 \leq X<2)=\int_1^2 \frac{1}{10} \mathrm{dx}=\frac{1}{10} \ & \vdots \ & P(Y=9)=P(9 \leq X<10)=\int_9^{10} \frac{1}{10} \mathrm{dx}=\frac{1}{10} \end{aligned}
Therefore, as it is seen, the random variable $\mathrm{Y}$ adopts discrete values 0 to 9 with probability $\frac{1}{10}$.

数学代写|概率论代写Probability theory代考|Conditioning on continuous space

In Chapter 3 , we showed that, according to the law of total probability, if the should condition on possible parts of the space to calculate the probability of occurrence of that event. Now, if the probability of occurrence of an event depends on the result of a random variable, we should condition on its possible values. Consider the following example: The probability that a shooter’s shot hits the target is $\frac{1}{X^2}$ in which $X$ is the distance of the target from the shooter which is determined randomly and can take on $1,2,3,4$, and 5 with probability $\frac{1}{5}$. What is the probability that the shooter’s shot hits the target?

Solution. Since the probability that the shot hits the target depends on the distance from the target, we should condition on the distance from the target. Therefore, if event $\mathrm{E}$ denotes that the shot hits the target, we have:
$$P(E)=\sum_{x=1}^5 P(E \mid X=x) P(X=x)=\frac{1}{1} \times \frac{1}{5}+\frac{1}{4} \times \frac{1}{5}+\frac{1}{9} \times \frac{1}{5}+\frac{1}{16} \times \frac{1}{5}+\frac{1}{25} \times \frac{1}{5}=\frac{5269}{18000}$$
Now, if $X$ follows a continuous distribution taking on values in the interval $(1,5)$ with equal density $f_X(x)=\frac{1}{5}$, then the probability that the shot hits the target again depends on $X$ and we should condition on its possible values. However, herein, $X$ is a continuous random variable. In such cases, the general approach for conditioning is the same as before. We divide the possible values of the random variable $X$ into very small parts. Then, in contrast with discrete case that we denote the probability of each point by $P(X=x)$, for the continuous case, we denote the probability of each small interval of length $d x$ about point $x$ by $f_X(x) d x$. Finally, we can calculate the probability of occurrence of an event like $E$ for each small interval of the random variable $X$ and add them up as follows:
$$\left.P(E)=\int_1^5 P(E \mid X=x) f_X(x) d x=\int_1^5 \frac{1}{x^2} \times \frac{1}{5-1} d x=\frac{-1}{4 x}\right]_1^5=0.2$$

概率论代考

数学代写|概率论代写Probability theory代考|The distribution of a function of a random variable

$$f_X(x)=\frac{1}{10} ; 0<x<10,$$

$$P(Y=0)=P(0 \leq X<1)=\int_0^1 \frac{1}{10} \mathrm{~d} x=\frac{1}{10}$$

数学代写|概率论代写Probability theory代考|Conditioning on continuous space

$$P(E)=\sum_{x=1}^5 P(E \mid X=x) P(X=x)=\frac{1}{1} \times \frac{1}{5}$$

$$P(E)=\int_1^5 P(E \mid X=x) f_X(x) d x=\int_1^5 \frac{1}{x^2}$$

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MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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