東京大学 新領域創成科学研究科 メディカル情報生命専攻 2023年8月実施 問題11

Author

zephyr

Description

Let \(X_i\) \((i = 1, \dots, n; n \geq 2)\) be \(n\) independent nonnegative real-valued random variables with the same probability density function \(f(x) = e^{-x}\). Answer the following questions with mathematical derivation.

(1) Compute the mean and variance of \(X_i\).

(2) Suppose the value of \(X_i\) is given as \(x_i\) for some \(i\). Compute probability \(\mathbb{P}(X_i \leq X_k \mid X_i = x_i)\) that \(X_i\) is less than or equal to \(X_k\) for a given index \(k\) with \(k \neq i\).

(3) Compute probability \(\mathbb{P}(X_i \leq X_k)\) by multiplying the probability of (2) by \(f(x_i)\) and integrating it with respect to \(x_i\).

(4) Let \(X_{\min} = \min_{k=1,\dots,n} X_k\) be the minimum value of set \(\{ X_i \mid i = 1, \dots, n \}\). Compute the probability \(\mathbb{P}(x \leq X_{\min})\) that \(X_{\min}\) is greater than or equal to \(x\) for a given positive real number \(x\).

For \(n\) given positive real numbers \(\lambda_i\) \((i = 1, \dots, n)\), define random variables \(Z_i\) \((i = 1, \dots, n)\) as \(Z_i = \frac{X_i}{\lambda_i}\).

(5) Suppose the value of \(X_i\) is given as \(x_i\) for some \(i\). Compute probability \(\mathbb{P}(Z_i \leq Z_k \mid X_i = x_i)\) that \(Z_i\) is less than or equal to \(Z_k\) for a given index \(k\) with \(k \neq i\).

(6) Suppose the value of \(X_i\) is given as \(x_i\) for some \(i\). Compute probability \(\mathbb{P}\left(\bigcap_{k=1,\dots,n; k \neq i} \{ Z_i \leq Z_k \} \mid X_i = x_i\right)\) that \(Z_i\) is less than or equal to \(Z_k\) for all the indices \(k\) with \(k \neq i\).

(7) Let \(Z_I\) be a minimum element in set \(\{ Z_i \mid i = 1, \dots, n \}\). Answer the probability distribution \(\mathbb{P}(I = i)\) \((i = 1, \dots, n)\) of index \(I = \mathop{\arg\min}\limits_{k=1,\dots,n} Z_k\).

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设 \(X_i\) \((i = 1, \dots, n; n \geq 2)\) 是 \(n\) 个独立的非负实值随机变量，其概率密度函数为 \(f(x) = e^{-x}\)。请用数学推导回答以下问题。

(1) 计算 \(X_i\) 的均值和方差。

(2) 假设 \(X_i\) 的值为 \(x_i\)，计算 \(\mathbb{P}(X_i \leq X_k \mid X_i = x_i)\) 的概率，即 \(X_i\) 小于或等于 \(X_k\) 的概率，其中 \(k \neq i\)。

(3) 计算 \(\mathbb{P}(X_i \leq X_k)\) 的概率，通过 (2) 的概率乘以 \(f(x_i)\) 并对 \(x_i\) 积分。

(4) 令 \(X_{\min} = \min_{k=1,\dots,n} X_k\) 为集合 \(\{ X_i \mid i = 1, \dots, n \}\) 的最小值。计算 \(\mathbb{P}(x \leq X_{\min})\) 的概率，即 \(X_{\min}\) 大于或等于给定正实数 \(x\) 的概率。

对于 \(n\) 个给定的正实数 \(\lambda_i\) \((i = 1, \dots, n)\)，定义随机变量 \(Z_i\) \((i = 1, \dots, n)\) 为 \(Z_i = \frac{X_i}{\lambda_i}\)。

(5) 假设 \(X_i\) 的值为 \(x_i\)，计算 \(\mathbb{P}(Z_i \leq Z_k \mid X_i = x_i)\) 的概率，即 \(Z_i\) 小于或等于 \(Z_k\) 的概率，其中 \(k \neq i\)。

(6) 假设 \(X_i\) 的值为 \(x_i\)，计算 \(\mathbb{P}\left(\bigcap_{k=1,\dots,n; k \neq i} \{ Z_i \leq Z_k \} \mid X_i = x_i\right)\) 的概率，即 \(Z_i\) 小于或等于 \(Z_k\) 的概率，对于所有 \(k \neq i\)。

(7) 令 \(Z_I\) 为集合 \(\{ Z_i \mid i = 1, \dots, n \}\) 中的最小元素。回答索引 \(I = \mathop{\arg\min}\limits_{k=1,\dots,n} Z_k\) 的概率分布 \(\mathbb{P}(I = i)\) \((i = 1, \dots, n)\)。

Kai

Written by zephyr

题目背景

令 \(X_i\) \((i = 1, \dots, n; n \geq 2)\) 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

\[ \begin{align*} E[X_i] &= \int_0^\infty x f(x) dx \\ &= \int_0^\infty x e^{-x} dx \\ &= [-xe^{-x}]_0^\infty + \int_0^\infty e^{-x} dx \\ &= 0 + [-e^{-x}]_0^\infty \\ &= 1 \end{align*} \]

Variance

\[ \begin{align*} Var(X_i) &= E[X_i^2] - (E[X_i])^2 \\ &= \int_0^\infty x^2 e^{-x} dx - 1^2 \\ &= [-x^2e^{-x}]_0^\infty + 2\int_0^\infty xe^{-x} dx - 1 \\ &= 0 + 2 - 1 \\ &= 1 \end{align*} \]

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

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

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

\[ \begin{align*} \mathbb{P}(X_i \leq X_k \mid X_i = x_i) &= \mathbb{P}(x_i \leq X_k) \\ &= 1 - \mathbb{P}(X_k < x_i) \\ &= 1 - \int_0^{x_i} e^{-x} dx \\ &= 1 - [-e^{-x}]_0^{x_i} \\ &= 1 - (1 - e^{-x_i}) \\ &= e^{-x_i} \end{align*} \]

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

We need to integrate over \(x_i\):

\[ \begin{align*} \mathbb{P}(X_i \leq X_k) &= \int_0^\infty \mathbb{P}(X_i \leq X_k \mid X_i = x) f(x) dx \\ &= \int_0^\infty e^{-x} \cdot e^{-x} dx \\ &= \int_0^\infty e^{-2x} dx \\ &= [-\frac{1}{2}e^{-2x}]_0^\infty \\ &= \frac{1}{2} \end{align*} \]

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

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 \(x\):

\[ \begin{align*} \mathbb{P}(x \leq X_{\min}) &= \mathbb{P}(\bigcap_{i=1}^n \{x \leq X_i\}) \\ &= \prod_{i=1}^n \mathbb{P}(x \leq X_i) \quad \text{(since $X_i$ are independent)} \\ &= (\mathbb{P}(x \leq X_1))^n \\ &= (1 - \mathbb{P}(X_1 < x))^n \\ &= (1 - \int_0^x e^{-t} dt)^n \\ &= (1 - (1 - e^{-x}))^n \\ &= e^{-nx} \end{align*} \]

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

Given \(X_i = x_i\), we have:

\[ \begin{align*} \mathbb{P}(Z_i \leq Z_k \mid X_i = x_i) &= \mathbb{P}(\frac{x_i}{\lambda_i} \leq \frac{X_k}{\lambda_k}) \\ &= \mathbb{P}(X_k \geq \frac{\lambda_k}{\lambda_i}x_i) \\ &= \int_{\frac{\lambda_k}{\lambda_i}x_i}^\infty e^{-x} dx \\ &= [-e^{-x}]_{\frac{\lambda_k}{\lambda_i}x_i}^\infty \\ &= e^{-\frac{\lambda_k}{\lambda_i}x_i} \end{align*} \]

6. Compute probability \(\mathbb{P}\left(\bigcap_{k=1,\dots,n; k \neq i} \{ Z_i \leq Z_k \} \mid X_i = x_i\right)\)

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

\[ \begin{align*} \mathbb{P}\left(\bigcap_{k=1,\dots,n; k \neq i} \{ Z_i \leq Z_k \} \mid X_i = x_i\right) &= \prod_{k=1,\dots,n; k \neq i} \mathbb{P}(Z_i \leq Z_k \mid X_i = x_i) \\ &= \prod_{k=1,\dots,n; k \neq i} e^{-\frac{\lambda_k}{\lambda_i}x_i} \\ &= \exp\left(-\sum_{k=1,\dots,n; k \neq i} \frac{\lambda_k}{\lambda_i}x_i\right) \\ &= \exp\left(-\frac{x_i}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k\right) \end{align*} \]

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

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

\[ \begin{align*} \mathbb{P}(I = i) &= \int_0^\infty \mathbb{P}\left(\bigcap_{k=1,\dots,n; k \neq i} \{ Z_i \leq Z_k \} \mid X_i = x_i\right) f(x_i) dx_i \\ &= \int_0^\infty \exp\left(-\frac{x_i}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k\right) e^{-x_i} dx_i \\ &= \int_0^\infty \exp\left(-x_i\left(\frac{1}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k + 1\right)\right) dx_i \\ &= \left[-\frac{1}{\frac{1}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k + 1} \exp\left(-x_i\left(\frac{1}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k + 1\right)\right)\right]_0^\infty \\ &= \frac{1}{\frac{1}{\lambda_i}\sum_{k=1,\dots,n; k \neq i} \lambda_k + 1} \\ &= \frac{\lambda_i}{\sum_{k=1}^n \lambda_k} \end{align*} \]

Therefore, \(\mathbb{P}(I = i) = \frac{\lambda_i}{\sum_{k=1}^n \lambda_k}\).

Knowledge

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