IS CS-2018S2-06

题目来源：Problem 6 日期：2024-08-10 题目主题：Math-Probability-Exponential Distribution

解题思路

我们需要处理一个以指数分布为基础的概率题目。这类问题通常会涉及到累积分布函数（CDF）、均值、方差、最大值等的计算。关键是要熟练掌握指数分布的性质，并能将这些性质运用于计算。

Solution

1. Cumulative Distribution Function (CDF) and Median of $T$

Given the probability density function (PDF) of $T$:

$$f_T(t)=\lambda e^{-\lambda t},t\ge 0,$$

The cumulative distribution function (CDF) $F_T(t)$ is defined as:

$$F_T(t)={\int }_0^t{f}_T(x)\mathrm{d}x.$$

Let’s compute $F_T(t)$:

$$F_T(t)={\int }_0^t\lambda e^{-\lambda x}\mathrm{d}x$$

To solve this integral:

$$F_T(t)={\left[-e^{-\lambda x}\right]}_0^t=-e^{-\lambda t}+e^0=1-e^{-\lambda t}$$

So, the cumulative distribution function is:

$$F_T(t)=1-e^{-\lambda t},t\ge 0.$$

Median of $T$

The median $m_T$ of $T$ is the value of $t$ such that $F_T(t)=0.5$:

$$1-e^{-\lambda m_T}=0.5$$

Solving for $m_T$:

$$e^{-\lambda m_T}=0.5$$

$-\lambda m_T=\ln (0.5)$

$m_T=\frac{\ln (2)}{\lambda }$

Thus, the median of $T$ is:

$$m_T=\frac{\ln (2)}{\lambda }$$

2. Expected Value and Variance of ${\mu }_T$

We measured the decomposition times $T_i(i=1,\dots ,n)$ of $n$ RNA molecules, where each $T_i$ follows the PDF $f_T(t)$ independently and identically.

The sample mean ${\mu }_T$ is defined as:

$${\mu }_T=\frac{{\sum }_{i=1}^n{T}_i}{n}.$$

Since each $T_i$ is identically and independently distributed:

• The expected value $\mathrm{E}[T_i]$ is:

$$\mathrm{E}[T_i]=\frac{1}{\lambda }$$

• The variance $\mathrm{Var}(T_i)$ is:

$$\mathrm{Var}(T_i)=\frac{1}{{\lambda }^2}$$

Thus, the expected value of ${\mu }_T$ is:

$$\mathrm{E}[{\mu }_T]=\mathrm{E}\left[\frac{{\sum }_{i=1}^n{T}_i}{n}\right]=\frac{1}{n}\sum _{i=1}^n\mathrm{E}[T_i]=\frac{1}{\lambda }$$

The variance of ${\mu }_T$ is:

$$\mathrm{Var}({\mu }_T)=\mathrm{Var}\left(\frac{{\sum }_{i=1}^n{T}_i}{n}\right)=\frac{1}{n^2}\sum _{i=1}^n\mathrm{Var}(T_i)=\frac{1}{n}\cdot \frac{1}{{\lambda }^2}=\frac{1}{n{\lambda }^2}$$

3. Probability of $T_{\text{max}}>t$

Let $T_{\text{max}}=\max \{T_1,\dots ,T_n\}$, which is the maximum of the measured times $T_i$.

The probability that $T_{\text{max}}>t$ can be expressed as:

$$\text{Prob}(T_{\text{max}}>t)=1-\text{Prob}(T_{\text{max}}\le t)$$

Since the $T_i$s are independent:

$$\text{Prob}(T_{\text{max}}\le t)=\text{Prob}(T_1\le t)\times \text{Prob}(T_2\le t)\times \cdots \times \text{Prob}(T_n\le t)$$ $$\text{Prob}(T_{\text{max}}\le t)=(F_T(t){)}^n$$

Thus, the probability that $T_{\text{max}}>t$ is:

$$\text{Prob}(T_{\text{max}}>t)=1-(F_T(t){)}^n=1-{\left(1-e^{-\lambda t}\right)}^n$$

4. PDF and Expected Value of $T_{\text{max}}$

The probability density function $f_{T_{\text{max}}}(t)$ of $T_{\text{max}}$ can be derived from the CDF:

$$f_{T_{\text{max}}}(t)=\frac{\mathrm{d}}{\mathrm{d}t}\text{Prob}(T_{\text{max}}\le t)=\frac{\mathrm{d}}{\mathrm{d}t}\left(1-(F_T(t){)}^n\right)$$

Differentiating:

$$f_{T_{\text{max}}}(t)=n(F_T(t){)}^{n-1}{f}_T(t)=n{\left(1-e^{-\lambda t}\right)}^{n-1}\lambda e^{-\lambda t}$$

To calculate the expected value $\mathrm{E}[T_{\text{max}}]$, one could use integration, but the exact value is often complex to derive directly and might require numerical methods. Generally, the expectation $\mathrm{E}[T_{\text{max}}]$ depends on both $n$ and $\lambda$.

知识点

累积分布函数期望值方差 概率论

解题技巧和信息

对于指数分布问题，熟练掌握其 PDF、CDF、均值和方差的计算是至关重要的。此外，对于多个独立同分布的随机变量，最大值的分布通常可以通过 CDF 来推导。

重点词汇

probability density function (PDF) 概率密度函数

cumulative distribution function (CDF) 累积分布函数

expected value 期望值

variance 方差

maximum 最大值

参考资料

1. Ross, S. M. (2019). Introduction to Probability Models. Academic Press.
2. Papoulis, A., & Pillai, S. U. (2002). Probability, Random Variables, and Stochastic Processes. McGraw-Hill.