IS Math-2017-03

题目来源：Problem 3 日期：2024-07-28 题目主题：Math-Probability Theory-Exponential Distribution

Solution

Question 1

To compute the average of the exponential random variable $T$, we use the definition of the expected value:

$$\mathbb{E}[T]={\int }_0^{\infty }tf(t)\mathrm{d}t$$

Given the probability density function (PDF):

$$f(t)=\{\begin{array}{ll}\lambda e^{-\lambda t}& (t\ge 0)\\ 0& (t<0)\end{array}$$

The expected value calculation is:

$$\mathbb{E}[T]={\int }_0^{\infty }t\lambda e^{-\lambda t}\mathrm{d}t$$

We use integration by parts:

Let $u=t$ and $\mathrm{d}v=\lambda e^{-\lambda t}\mathrm{d}t$.

Then $\mathrm{d}u=\mathrm{d}t$ and $v=-e^{-\lambda t}$.

Thus,

$$\mathbb{E}[T]={\left[-t{e}^{-\lambda t}\right]}_0^{\infty }+{\int }_0^{\infty }{e}^{-\lambda t}\mathrm{d}t$$

The boundary term evaluates to zero:

$${\left[-t{e}^{-\lambda t}\right]}_0^{\infty }=0$$

The integral term evaluates to:

$${\int }_0^{\infty }{e}^{-\lambda t}\mathrm{d}t=\frac{1}{\lambda }$$

So, the expected value is:

$$\mathbb{E}[T]=\frac{1}{\lambda }$$

Now, to derive the cumulative distribution function (CDF) $F(t)=P(T\le t)$:

$$F(t)={\int }_0^{t}f(u)\mathrm{d}u={\int }_0^t\lambda e^{-\lambda u}\mathrm{d}u$$

Evaluating this integral, we get:

$$F(t)={\left[-e^{-\lambda u}\right]}_0^t=1-e^{-\lambda t}$$

So, the CDF is:

$$F(t)=\{\begin{array}{ll}1-e^{-\lambda t}& (t\ge 0)\\ 0& (t<0)\end{array}$$

Question 2

To show that the exponential distribution is memoryless, we need to show that:

$$P(T>s+t\mid T>s)=P(T>t)$$

By definition of conditional probability:

$$P(T>s+t\mid T>s)=\frac{P((T>s+t)\cap (T>s))}{P(T>s)}=\frac{P(T>s+t)}{P(T>s)}$$

Since $T>s+t$ implies $T>s$, we have:

$$P(T>s+t)=e^{-\lambda (s+t)}$$

and

$$P(T>s)=e^{-\lambda s}$$

Thus,

$$P(T>s+t\mid T>s)=\frac{e^{-\lambda (s+t)}}{e^{-\lambda s}}=e^{-\lambda t}=P(T>t)$$

This confirms the memoryless property of the exponential distribution.

Question 3

Let $T_i$ be the time required for solution of the $i$-th student, where $T_i\sim \text{Exp}({\lambda }_0)$. The time required for solution of the student who finishes earliest is:

$$T_{\min }=\min (T_1,T_2,\dots ,T_n)$$

The CDF of $T_{\min }$ is:

$$F_{T_{\min }}(t)=P(T_{\min }\le t)=1-P(T_{\min }>t)$$

Since the $T_i$ are independent:

$$P(T_{\min }>t)=P(T_1>t)P(T_2>t)\cdots P(T_n>t)={\left(e^{-{\lambda }_{0}t}\right)}^n=e^{-n{\lambda }_{0}t}$$

Thus,

$$F_{T_{\min }}(t)=1-e^{-n{\lambda }_{0}t}$$

The PDF of $T_{\min }$ is:

$$f_{T_{\min }}(t)=\frac{\mathrm{d}}{\mathrm{d}t}{F}_{T_{\min }}(t)=n{\lambda }_0{e}^{-n{\lambda }_{0}t}$$

The expected value is:

$$\mathbb{E}[T_{\min }]=\frac{1}{n{\lambda }_0}$$

Question 4

Let $T_A\sim \text{Exp}({\lambda }_A)$ and $T_B\sim \text{Exp}({\lambda }_B)$.

We want to find $P(T_A<T_B)$:

$$P(T_A<T_B)={\int }_0^{\infty }P(T_B>t)f_{T_A}(t)\mathrm{d}t$$

Since $T_B>t$ follows $P(T_B>t)=e^{-{\lambda }_{B}t}$ and $f_{T_A}(t)={\lambda }_A{e}^{-{\lambda }_{A}t}$, we have:

$$P(T_A<T_B)={\int }_0^{\infty }{e}^{-{\lambda }_{B}t}{\lambda }_A{e}^{-{\lambda }_{A}t}\mathrm{d}t={\lambda }_A{\int }_0^{\infty }{e}^{-({\lambda }_A+{\lambda }_B)t}\mathrm{d}t$$

This integral evaluates to:

$$P(T_A<T_B)={\lambda }_A{\left[\frac{1}{{\lambda }_A+{\lambda }_B}\right]}_0^{\infty }=\frac{{\lambda }_A}{{\lambda }_A+{\lambda }_B}$$

Question 5

Given that Hideo’s solving time $T_H\sim \text{Exp}(10\lambda )$ and the other ten students’ solving times $T_i\sim \text{Exp}(\lambda )$, we need to find the probability that Hideo finishes solving the problem first and fourth.

Probability that Hideo finishes first

To find the probability that Hideo finishes first, we use the fact that the exponential distribution is memoryless and the given rates of the distributions. The probability that Hideo finishes before any of the other ten students is:

$$P(T_H<T_i\text{ for all }i)=\prod _{i=1}^{10}P(T_H<T_i)$$

Since $T_H$ and $T_i$ are independent exponential random variables with rates $10\lambda$ and $\lambda$ respectively, we can use the following result:

$$P(T_H<T_i)=\frac{10\lambda }{10\lambda +\lambda }=\frac{10\lambda }{11\lambda }=\frac{10}{11}$$

Thus, the probability that Hideo finishes before all other ten students is:

$$P(T_H<T_1,T_H<T_2,\dots ,T_H<T_{10})={\left(\frac{10}{11}\right)}^{10}$$

Calculating this, we get:

$$P(\text{Hideo finishes first})={\left(\frac{10}{11}\right)}^{10}\approx 0.3487$$

Probability that Hideo finishes fourth

To find the probability that Hideo finishes fourth, we need to consider that exactly three students finish before Hideo and the remaining seven finish after him. The solving times of these students are independently distributed as exponential random variables.

The probability that Hideo finishes fourth can be computed by considering the order statistics of the exponential distribution. We use the multinomial coefficient and the corresponding probabilities for Hideo to be in the fourth position:

The probability that exactly three students finish before Hideo and seven students finish after Hideo is given by:

$$P(\text{Hideo finishes fourth})=\left(\genfrac{}{}{0ex}{}{10}{3}\right){\left(\frac{1}{11}\right)}^3{\left(\frac{10}{11}\right)}^7$$

Calculating the binomial coefficient:

$$\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$

Thus, the probability becomes:

$$P(\text{Hideo finishes fourth})=120{\left(\frac{1}{11}\right)}^3{\left(\frac{10}{11}\right)}^7\approx 0.04626$$

知识点

概率论指数分布条件概率期望值顺序统计量

难点思路

计算一个特定学生在多个独立变量中排名的问题需要利用顺序统计量的知识，具体计算较为复杂，需要结合组合数学和概率分布的性质。

解题技巧和信息

• 利用指数分布的无记忆性质简化条件概率问题。
• 通过期望值计算和顺序统计量的组合性质来分析特定学生的排名问题。
• 理解并应用组合数学的基本原理来计算概率。

重点词汇

• Exponential distribution: 指数分布
• Memoryless property: 无记忆性质
• Expected value: 期望值
• Order statistics: 顺序统计量
• Probability density function (PDF): 概率密度函数

参考资料

1. “Introduction to Probability Models” by Sheldon Ross, Chap. 5
2. “Probability and Statistics” by Morris H. DeGroot and Mark J. Schervish, Chap. 4
3. “A First Course in Probability” by Sheldon Ross, Chap. 5