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

标签：

• Tokyo-University
• Order-Statistics
• Uniform-Order-Statistics

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.