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 2017年8月実施 問題11

Author

zephyr

Description

Assume that the distributions of real-valued mutually independent random variables \(\mathbf{X_1}, \ldots, \mathbf{X_n}\) are identical and denoted as \(F(x)\).

\[ F(x) = P(X_i \leq x), i = 1, \ldots, n \]

Denote by \(\mathbf{X_{(1)}}, \ldots, \mathbf{X_{(n)}}\) the random variables obtained by arranging \(\mathbf{X_1}, \ldots, \mathbf{X_n}\) in ascending order. Answer the following questions.

(1) Find the distribution function of \(\mathbf{X_{(1)}}\).

(2) Find the distribution function of \(\mathbf{X_{(n)}}\).

(3) Find the distribution function of \(\mathbf{X_{(k)}}\) for any \(k\).

(4) Find the expectation of \(\mathbf{X_{(1)}}\) when \(F(x)\) is the uniform distribution over \([0,1]\).

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假设实值相互独立随机变量 \(\mathbf{X_1}, \ldots, \mathbf{X_n}\) 的分布是相同的，并记为 \(F(x)\)。

\[ F(x) = P(X_i \leq x), i = 1, \ldots, n \]

记 \(\mathbf{X_{(1)}}, \ldots, \mathbf{X_{(n)}}\) 为将 \(\mathbf{X_1}, \ldots, \mathbf{X_n}\) 按升序排列后得到的随机变量。回答以下问题。

(1) 找到 \(\mathbf{X_{(1)}}\) 的分布函数。

(2) 找到 \(\mathbf{X_{(n)}}\) 的分布函数。

(3) 找到任意 \(k\) 的 \(\mathbf{X_{(k)}}\) 的分布函数。

(4) 当 \(F(x)\) 为 \([0,1]\) 上的均匀分布时，找到 \(\mathbf{X_{(1)}}\) 的期望。

Kai

(1)

To find the distribution function of the smallest order statistic \(\mathbf{X_{(1)}}\), we consider:

\[ F_{\mathbf{X_{(1)}}}(x) = P(\mathbf{X_{(1)}} \leq x). \]

Since \(\mathbf{X_{(1)}}\) is the smallest of the \(\mathbf{X_1}, \ldots, \mathbf{X_n}\), \(\mathbf{X_{(1)}} \gt x\) means that all \(\mathbf{X_i} \gt x\). Thus:

\[ F_{\mathbf{X_{(1)}}}(x) = 1 - P(\mathbf{X_{(1)}} > x). \]

We know that \(\mathbf{X_{(1)}} > x\) if and only if all \(\mathbf{X_i} > x\), so:

\[ P(\mathbf{X_{(1)}} > x) = P(\mathbf{X_1} > x, \mathbf{X_2} > x, \ldots, \mathbf{X_n} > x) = \left( P(\mathbf{X_1} > x) \right)^n = (1 - F(x))^n. \]

Thus, the distribution function of \(\mathbf{X_{(1)}}\) is:

\[ F_{\mathbf{X_{(1)}}}(x) = 1 - (1 - F(x))^n. \]

(2)

To find the distribution function of the largest order statistic \(\mathbf{X_{(n)}}\), we consider:

\[ F_{\mathbf{X_{(n)}}}(x) = P(\mathbf{X_{(n)}} \leq x). \]

Since \(\mathbf{X_{(n)}}\) is the largest of the \(\mathbf{X_1}, \ldots, \mathbf{X_n}\), \(\mathbf{X_{(n)}} \leq x\) means that at least one \(\mathbf{X_i} \leq x\). Thus:

\[ F_{\mathbf{X_{(n)}}}(x) = P(\mathbf{X_1} \leq x, \mathbf{X_2} \leq x, \ldots, \mathbf{X_n} \leq x) = \left( P(\mathbf{X_1} \leq x) \right)^n = (F(x))^n. \]

(3)

To find the distribution function of the \(k\)-th order statistic \(\mathbf{X_{(k)}}\), we need to determine the probability \(F_{\mathbf{X_{(k)}}}(x) = P(\mathbf{X_{(k)}} \leq x)\). This represents the probability that the \(k\)-th smallest value among \(\mathbf{X_1}, \ldots, \mathbf{X_n}\) is less than or equal to \(x\).

Step 1: Basic Concepts and Binomial Probability

Since \(\mathbf{X_i}\) are independent and identically distributed, the probability that any particular \(\mathbf{X_i}\) is less than or equal to \(x\) is \(F(x)\). Similarly, the probability that \(\mathbf{X_i}\) is greater than \(x\) is \(1 - F(x)\).

Step 2: Using Binomial Distribution

We can think of this as a binomial distribution problem. We need to consider the event that at least \(k\) out of \(n\) \(\mathbf{X_i}\) values are less than or equal to \(x\). Mathematically, this can be expressed as:

\[ F_{\mathbf{X_{(k)}}}(x) = P(\mathbf{X_{(k)}} \leq x) = \sum_{j=k}^{n} \binom{n}{j} (F(x))^j (1 - F(x))^{n-j}. \]

Here, \(\binom{n}{j}\) is the binomial coefficient, representing the number of ways to choose \(j\) successes (values \(\leq x\)) out of \(n\) trials.

(4)

If \(F(x)\) is the uniform distribution over \([0,1]\), then \(F(x) = x\) for \(x \in [0,1]\). Therefore:

\[ F_{\mathbf{X_{(1)}}}(x) = 1 - (1 - x)^n. \]

The expectation of \(\mathbf{X_{(1)}}\) is given by:

\[ \mathbb{E}[\mathbf{X_{(1)}}] = \int_{0}^{1} x f_{\mathbf{X_{(1)}}}(x) \, \mathrm{d}x, \]

where \(f_{\mathbf{X_{(1)}}}(x)\) is the derivative of \(F_{\mathbf{X_{(1)}}}(x)\):

\[ f_{\mathbf{X_{(1)}}}(x) = \frac{\mathrm{d}}{\mathrm{dx}} \left[ 1 - (1 - x)^n \right] = n (1 - x)^{n-1}. \]

Therefore:

\[ \mathbb{E}[\mathbf{X_{(1)}}] = \int_{0}^{1} x \cdot n (1 - x)^{n-1} \, \mathrm{d}x. \]

This is a Beta distribution integral:

\[ \mathbb{E}[\mathbf{X_{(1)}}] = n \int_{0}^{1} x (1 - x)^{n-1} \, \mathrm{d}x. \]

Using the Beta function property, we get:

\[ \int_{0}^{1} x (1 - x)^{n-1} \, \mathrm{d}x = \frac{1}{n+1}. \]

Thus:

\[ \mathbb{E}[\mathbf{X_{(1)}}] = \frac{n}{n+1}. \]

Knowledge

顺序统计量 概率分布函数 期望值 Beta分布

难点思路

第 (3) 小问关于任意 \(k\) 阶顺序统计量的分布函数需要理解 Binomial 分布的性质并进行累加，这是一个较难点。

解题技巧和信息

对于顺序统计量，了解如何通过分布函数 \(F(x)\) 来表示最小和最大顺序统计量的分布函数非常重要。对于均匀分布的情况，可以利用 Beta 分布性质简化期望值计算。

重点词汇

• order statistic 顺序统计量
• distribution function 分布函数
• expectation 期望值
• uniform distribution 均匀分布

参考资料

1. "Probability and Statistics" by Morris H. DeGroot and Mark J. Schervish, Chapter 5.
2. "A First Course in Probability" by Sheldon Ross, Chapter 8.