IS Math-2022-03

题目来源：Problem 3 日期：2024-06-27 题目主题：Math-概率-期望值与概率密度函数

具体题目

Consider a region $R$ defined by $0<x<1$ and $0<y<1$ in the $xy$-plane. We randomly select a point on $R$ and refer to the selected point as $A$. We assume that $A$ is uniformly distributed on $R$. Let $AB$ be a perpendicular line from $A$ to the $y$-axis and $AC$ be a perpendicular line from $A$ to the $x$-axis as shown in the figure. We call rectangle $OCAB$ as “the rectangle of $A$”, where $O$ denotes the origin. Let $S$ be a random variable representing the area of the rectangle of $A$. Answer the following questions.

1. Calculate the expectation value of $S$.
2. Calculate the probability that $S\le r$ holds, where $0<r<1$.
3. Calculate the probability density function of $S$.

Again consider the region $R$. Let $n$ be a positive integer. We select $n$ points on $R$ and refer to the selected points as $A_1,A_2,\dots ,A_n$. We assume that each of the points is uniformly distributed on $R$, and $A_i$ and $A_j$ for $i\ne j$ are selected independently. Answer the following question.

1. Let $S_i$ be a random variable representing the area of the rectangle of $A_i$. Let $Z$ be a random variable which is the minimum of $S_1,S_2,\dots ,S_n$. Calculate the probability density function of $Z$.

正确解答

1. Calculate the expectation value of $S$

Given a point $A$ with coordinates $(X,Y)$ uniformly distributed in the region $R$, the area of the rectangle $OCAB$ is $S=XY$. The expectation value of $S$ is given by

$$E[S]=E[XY]$$

Since $X$ and $Y$ are independent and uniformly distributed on $[0,1]$,

$$E[X]={\int }_0^{1}xdx=\frac{1}{2},E[Y]={\int }_0^{1}ydy=\frac{1}{2}$$

Thus,

$$E[S]=E[XY]=E[X]E[Y]=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$$

2. Calculate the probability that $S\le r$ holds, where $0<r<1$

The cumulative distribution function (CDF) of $S$ is defined as

$$F_S(r)=P(S\le r)$$

Since $S=XY$, we have

$$F_S(r)=P(XY\le r)$$

To solve the integral, we divide the integration region into pieces:

Integration Setup

We need to integrate over the region where $XY\le r$.

We can break this region into two parts:

• For $0<y\le r$, $0<x\le 1$
• For $r<y\le 1$, $0<x\le \frac{r}{y}$

First Integration Area ($0<y\le r$)

Here, $x$ ranges from 0 to 1 for $0<y\le r$.

$${\int }_0^r{\int }_0^{1}dxdy={\int }_0^{r}1dy=r$$

Second Integration Area ($r<y\le 1$)

Here, $x$ ranges from 0 to $\frac{r}{y}$ for $r<y\le 1$.

$${\int }_r^1{\int }_0^{\frac{r}{y}}dxdy={\int }_r^1\frac{r}{y}dy$$

To solve the second integral:

$${\int }_r^1\frac{r}{y}dy=r{\int }_r^1\frac{1}{y}dy=r{\left[\ln y\right]}_r^1=r(\ln 1-\ln r)=r(0-\ln r)=-r\ln r$$

Combine Both Areas

Adding the results from both regions:

$$P(S\le r)={\int }_0^r{\int }_0^{1}dxdy+{\int }_r^1{\int }_0^{\frac{r}{y}}dxdy=r+(-r\ln r)=r-r\ln r$$

3. Calculate the probability density function of $S$

==The probability density function (PDF) of $S$ is the derivative of the CDF:==

^bfcb0c

$$f_S(r)=\frac{d}{dr}{F}_S(r)=\frac{d}{dr}\left(r-r\ln r\right)$$

Using the product rule,

$$f_S(r)=\ln \left(\frac{1}{r}\right)+r\left(-\frac{1}{r}\right)=\ln \left(\frac{1}{r}\right)-1=-\ln r-1$$

So the PDF is

$$f_S(r)=\{\begin{array}{ll}-\ln r& \text{if }0<r<1\\ 0& \text{otherwise}\end{array}$$

4. Calculate the probability density function of $Z$

Let $S_i$ be the area of the rectangle corresponding to $A_i$. The random variable $Z$ is the minimum of $S_1,S_2,\dots ,S_n$. The CDF of $Z$ is

$$F_Z(z)=P(Z\le z)=1-P(Z>z)=1-P(S_1>z,S_2>z,\dots ,S_n>z)$$

Since the $S_i$ are independent,

$$P(Z>z)=P(S_1>z)P(S_2>z)\dots P(S_n>z)={\left(1-F_S(z)\right)}^n$$

Thus,

$$F_Z(z)=1-{\left(1-F_S(z)\right)}^n$$

To find the PDF, we differentiate the CDF:

$$f_Z(z)=\frac{d}{dz}{F}_Z(z)=\frac{d}{dz}\left[1-{\left(1-F_S(z)\right)}^n\right]$$

Using the chain rule,

$$f_Z(z)=n{\left(1-F_S(z)\right)}^{n-1}{f}_S(z)$$

Substituting the expressions for $F_S(z)$ and $f_S(z)$,

$$f_Z(z)=n{\left(1-(z-z\ln z)\right)}^{n-1}\left(-\ln z\right)$$

Thus, the PDF of $Z$ is

$$f_Z(z)=\{\begin{array}{ll}n{\left(1-z+z\ln z\right)}^{n-1}\left(-\ln z\right)& \text{if }0<z<1\\ 0& \text{otherwise}\end{array}$$

知识点

概率论概率密度函数累积分布函数二元积分

解题技巧和信息

在解决这类问题时，理解如何定义随机变量并确定它们的分布是至关重要的。对于期望值的计算，可以利用随机变量的独立性和均匀分布特性。对于累积分布函数（CDF）和概率密度函数（PDF）的计算，重要的是掌握积分技巧和微分技巧。对于多变量情况下，理解最小值的分布需要综合使用 CDF 和 PDF 的性质。

重点词汇

• Expectation value 期望值
• Probability density function (PDF) 概率密度函数
• Cumulative distribution function (CDF) 累积分布函数
• Independent 独立
• Uniform distribution 均匀分布

参考资料

1. “Introduction to Probability” by Dimitri P. Bertsekas and John N. Tsitsiklis, Chapter 3
2. “Probability and Statistics” by Morris H. DeGroot and Mark J. Schervish, Chapter 4