CBMS-2024-11

CBMS-2024A-11

题目来源：Question 11 日期：2024-08-01 题目主题：Math-概率论-连续分布与顺序统计量

题目背景

令 $X_i$ $(i=1,\dots ,n;n\ge 2)$ 为 $n$ 个独立的非负实值随机变量，它们具有相同的概率密度函数 $f(x)=e^{-x}$。

1. Compute the mean and variance of $X_i$

For the exponential distribution $f(x)=e^{-x}$, we can calculate the mean and variance as follows:

Mean

$$\begin{array}{rl}E[X_i]& ={\int }_0^{\infty }xf(x)dx\\ & ={\int }_0^{\infty }x{e}^{-x}dx\\ & =[-x{e}^{-x}{]}_0^{\infty }+{\int }_0^{\infty }{e}^{-x}dx\\ & =0+[-e^{-x}{]}_0^{\infty }\\ & =1\end{array}$$

Variance

$$\begin{array}{rl}Var(X_i)& =E[X_i^2]-(E[X_i]{)}^2\\ & ={\int }_0^{\infty }{x}^2{e}^{-x}dx-1^2\\ & =[-x^2{e}^{-x}{]}_0^{\infty }+2{\int }_0^{\infty }x{e}^{-x}dx-1\\ & =0+2-1\\ & =1\end{array}$$

Therefore, $E[X_i]=1$ and $Var(X_i)=1$.

2. Compute probability $\mathbb{P}(X_i\le X_k\mid X_i=x_i)$

Given $X_i=x_i$, the distribution of $X_k$ remains $f(x)=e^{-x}$. Thus:

$$\begin{array}{rl}\mathbb{P}(X_i\le X_k\mid X_i=x_i)& =\mathbb{P}(x_i\le X_k)\\ & =1-\mathbb{P}(X_k<x_i)\\ & =1-{\int }_0^{x_i}{e}^{-x}dx\\ & =1-[-e^{-x}{]}_0^{x_i}\\ & =1-(1-e^{-x_i})\\ & =e^{-x_i}\end{array}$$

3. Compute probability $\mathbb{P}(X_i\le X_k)$

We need to integrate over $x_i$:

$$\begin{array}{rl}\mathbb{P}(X_i\le X_k)& ={\int }_0^{\infty }\mathbb{P}(X_i\le X_k\mid X_i=x)f(x)dx\\ & ={\int }_0^{\infty }{e}^{-x}\cdot e^{-x}dx\\ & ={\int }_0^{\infty }{e}^{-2x}dx\\ & =[-\frac{1}{2}{e}^{-2x}{]}_0^{\infty }\\ & =\frac{1}{2}\end{array}$$

4. Compute probability $\mathbb{P}(x\le X_{\min })$

The probability that $X_{\min }$ is greater than or equal to $x$ is equal to the probability that all $X_i$ are greater than or equal to $x$:

$$\begin{array}{rl}\mathbb{P}(x\le X_{\min })& =\mathbb{P}(\bigcap _{i=1}^n\{x\le X_i\})\\ & =\prod _{i=1}^n\mathbb{P}(x\le X_i)\text{(since Xi are independent)}\\ & =(\mathbb{P}(x\le X_1){)}^n\\ & =(1-\mathbb{P}(X_1<x){)}^n\\ & =(1-{\int }_0^x{e}^{-t}dt{)}^n\\ & =(1-(1-e^{-x}){)}^n\\ & =e^{-nx}\end{array}$$

5. Compute probability $\mathbb{P}(Z_i\le Z_k\mid X_i=x_i)$

Given $X_i=x_i$, we have:

$$\begin{array}{rl}\mathbb{P}(Z_i\le Z_k\mid X_i=x_i)& =\mathbb{P}(\frac{x_i}{{\lambda }_i}\le \frac{X_k}{{\lambda }_k})\\ & =\mathbb{P}(X_k\ge \frac{{\lambda }_k}{{\lambda }_i}{x}_i)\\ & ={\int }_{\frac{{\lambda }_k}{{\lambda }_i}{x}_i}^{\infty }{e}^{-x}dx\\ & =[-e^{-x}{]}_{\frac{{\lambda }_k}{{\lambda }_i}{x}_i}^{\infty }\\ & =e^{-\frac{{\lambda }_k}{{\lambda }_i}{x}_i}\end{array}$$

6. Compute probability $\mathbb{P}\left({\bigcap }_{k=1,\dots ,n;k\ne i}\{Z_i\le Z_k\}\mid X_i=x_i\right)$

This probability is the product of all $\mathbb{P}(Z_i\le Z_k\mid X_i=x_i)$ for $k\ne i$:

$$\begin{array}{rl}\mathbb{P}\left(\bigcap _{k=1,\dots ,n;k\ne i}\{Z_i\le Z_k\}\mid X_i=x_i\right)& =\prod _{k=1,\dots ,n;k\ne i}\mathbb{P}(Z_i\le Z_k\mid X_i=x_i)\\ & =\prod _{k=1,\dots ,n;k\ne i}{e}^{-\frac{{\lambda }_k}{{\lambda }_i}{x}_i}\\ & =\exp \left(-\sum _{k=1,\dots ,n;k\ne i}\frac{{\lambda }_k}{{\lambda }_i}{x}_i\right)\\ & =\exp \left(-\frac{x_i}{{\lambda }_i}\sum _{k=1,\dots ,n;k\ne i}{\lambda }_k\right)\end{array}$$

7. Compute the probability distribution $\mathbb{P}(I=i)$

To calculate $\mathbb{P}(I=i)$, we need to integrate over $x_i$:

$$\begin{array}{rl}\mathbb{P}(I=i)& ={\int }_0^{\infty }\mathbb{P}\left(\bigcap _{k=1,\dots ,n;k\ne i}\{Z_i\le Z_k\}\mid X_i=x_i\right)f(x_i)d{x}_i\\ & ={\int }_0^{\infty }\exp \left(-\frac{x_i}{{\lambda }_i}\sum _{k=1,\dots ,n;k\ne i}{\lambda }_k\right)e^{-x_i}d{x}_i\\ & ={\int }_0^{\infty }\exp \left(-x_i\left(\frac{1}{{\lambda }_i}\sum _{k=1,\dots ,n;k\ne i}{\lambda }_k+1\right)\right)d{x}_i\\ & ={\left[-\frac{1}{\frac{1}{{\lambda }_i}{\sum }_{k=1,\dots ,n;k\ne i}{\lambda }_k+1}\exp \left(-x_i\left(\frac{1}{{\lambda }_i}\sum _{k=1,\dots ,n;k\ne i}{\lambda }_k+1\right)\right)\right]}_0^{\infty }\\ & =\frac{1}{\frac{1}{{\lambda }_i}{\sum }_{k=1,\dots ,n;k\ne i}{\lambda }_k+1}\\ & =\frac{{\lambda }_i}{{\sum }_{k=1}^n{\lambda }_k}\end{array}$$

Therefore, $\mathbb{P}(I=i)=\frac{{\lambda }_i}{{\sum }_{k=1}^n{\lambda }_k}$.

知识点

概率论指数分布条件概率顺序统计量

难点思路

这道题的难点在于处理条件概率和多个随机变量的最小值。特别是在第 6 和第 7 问中，需要仔细处理多个条件的交集概率。

解题技巧和信息

1. 对于指数分布，要熟悉其基本性质，如均值、方差、累积分布函数等。
2. 在处理多个独立随机变量时，要善于利用独立性质简化计算。
3. 在计算条件概率时，要清楚地区分给定条件下的随机变量和非随机变量。
4. 在处理最小值问题时，可以转化为所有变量都大于某个值的概率。
5. 在积分计算中，要善于使用指数函数的性质，如 $\int e^{ax}dx=\frac{1}{a}{e}^{ax}+C$。

重点词汇

• Probability density function: 概率密度函数
• Exponential distribution: 指数分布
• Conditional probability: 条件概率
• Minimum value: 最小值
• Order statistics: 顺序统计量
• Independent random variables: 独立随机变量