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 2015年8月実施 問題11

标签：

• Tokyo-University
• Geometric-Distribution
• Weighted-Sum-of-Independent-Bernoulli-Variables
• Expectation-and-Variance

Author

zephyr

Description

Suppose that there is an urn that contains $m$ black balls and $(l - m)$ white balls $(0 < m < l)$. You randomly draw $n$ balls with replacement $(n > 0)$. Answer the following questions with explanation.

(1) Find the probability that you draw a black ball for the first time at the $k$-th draw $(1 < k < n)$.

(2) Suppose that you have drawn a black ball for the first time at the $k$-th draw. Find the probability that you draw one or more black balls in the remaining $(n - k)$ draws.

Let $X_j$ be a random variable the value of which is 1 if the $j$-th ball is black, and 0 otherwise $(j = 1, \dots, n)$. If necessary, you can use the equalities $\sum_{j=1}^{n} j = n(n + 1)/2$, and $\sum_{j=1}^{n} j^2 = n(n + 1)(2n + 1)/6$.

(3) Find the expected value $\mathrm{E}[X_j]$ of $X_j$.

(4) Let $R = \sum_{j=1}^{n} j X_j$. Find the expected value $\mathrm{E}[R]$ of $R$.

(5) Find the variance $\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2$ of $R$.

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假设有一个包含 $m$ 个黑球和 $(l - m)$ 个白球的罐子 $(0 < m < l)$。你随机有放回地抽取 $n$ 个球 $(n > 0)$。回答以下问题并解释。

(1) 计算第一次在第 $k$ 次抽到黑球的概率 $(1 < k < n)$。

(2) 假设你第一次在第 $k$ 次抽到黑球。计算在接下来的 $(n - k)$ 次抽中至少再抽到一个黑球的概率。

令 $X_j$ 是一个随机变量，如果第 $j$ 个球是黑色的，则其值为 1，否则为 0 $(j = 1, \dots, n)$。如果需要，你可以使用以下等式：$\sum_{j=1}^{n} j = n(n + 1)/2$，$\sum_{j=1}^{n} j^2 = n(n + 1)(2n + 1)/6$。

(3) 计算 $X_j$ 的期望值 $\mathrm{E}[X_j]$。

(4) 令 $R = \sum_{j=1}^{n} j X_j$。计算 $R$ 的期望值 $\mathrm{E}[R]$。

(5) 计算 $R$ 的方差 $\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2$。

题目描述

一个罐中有 $m$ 个黑球与 $l-m$ 个白球，其中 $0<m<l$。独立地进行 $n>0$ 次有放回抽取，并对答案说明理由：

1. 对 $1<k<n$，求第 $k$ 次抽取时首次抽到黑球的概率。
2. 已知首次黑球出现在第 $k$ 次，求余下 $n-k$ 次中至少再出现一个黑球的概率。
3. 定义
   $$X_j= \begin{cases} 1,&\text{第 }j\text{ 次抽到黑球},\\ 0,&\text{否则}, \end{cases} \qquad j=1,\ldots,n,$$
   求 $E[X_j]$。
4. 令加权随机和
   $$R=\sum_{j=1}^{n}jX_j,$$
   求 $E[R]$。
5. 求
   $$\operatorname{Var}(R)=E[R^2]-E[R]^2.$$

计算时可使用

$$\sum_{j=1}^{n}j=\frac{n(n+1)}2,\qquad \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}6.$$

考点

• 几何分布与补事件：利用有放回抽取的独立性，计算首次成功时刻及后续至少一次成功的概率。
• 伯努利变量的加权和：把每次抽到黑球表示为参数 $m/l$ 的独立示性变量，并对位置权重为 $j$ 的和求期望。
• 期望与方差：使用期望线性性、独立变量协方差为零及平方和公式计算 $R$ 的方差。

Kai

(1)

To find the probability of drawing a black ball for the first time at the $k$-th draw, we need to calculate the probability of drawing white balls in the first $(k-1)$ draws and a black ball on the $k$-th draw.

Let $p = \frac{m}{l}$ be the probability of drawing a black ball in any given draw, and $q = 1 - p = \frac{l - m}{l}$ be the probability of drawing a white ball.

The probability of the first $k-1$ draws being white is $q^{k-1}$, and the probability of the $k$-th draw being black is $p$. Hence, the probability of this event is:

$$\mathrm{P}(X_k = 1 \text{ for the first time at } k) = q^{k-1}p$$

(2)

Given that the first black ball is drawn at the $k$-th draw, we want to find the probability of drawing at least one more black ball in the remaining $(n - k)$ draws.

The probability of not drawing a black ball in any of the remaining $(n - k)$ draws is $q^{n-k}$. Therefore, the probability of drawing at least one black ball in the remaining draws is:

$$1 - q^{n-k}$$

(3)

The random variable $X_j$ indicates whether the $j$-th ball is black. Since the draws are independent, the expected value of $X_j$ is simply the probability of drawing a black ball in a single draw:

$$\mathrm{E}[X_j] = p = \frac{m}{l}$$

(4)

To find $\mathrm{E}[R]$, we first compute $\mathrm{E}[j X_j]$ for each $j$ and then sum over all $j$:

$$\mathrm{E}[R] = \mathrm{E}\left[\sum_{j=1}^{n} j X_j\right] = \sum_{j=1}^{n} j \mathrm{E}[X_j]$$

$$\mathrm{E}[R] = \sum_{j=1}^{n} j \cdot \frac{m}{l}$$

Using the given summation formula:

$$\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$$

$$\mathrm{E}[R] = \frac{m}{l} \cdot \frac{n(n+1)}{2}$$

(5)

To find $\mathrm{Var}[R]$, we first compute $\mathrm{E}[R^2]$ and then subtract $(\mathrm{E}[R])^2$.

$$\mathrm{E}[R^2] = \mathrm{E}\left[\left(\sum_{j=1}^{n} j X_j\right)^2\right]$$

We need to compute $\mathrm{E}[R^2]$. Let's express it as:

$$\mathrm{E}[R^2] = \sum_{j=1}^{n} j^2 \mathrm{E}[X_j^2] + 2 \sum_{1 \leq j < k \leq n} jk \mathrm{E}[X_j X_k]$$

Since $X_j$ is a Bernoulli random variable:

$$\mathrm{E}[X_j^2] = \mathrm{E}[X_j] = p$$

The covariance term $\mathrm{E}[X_j X_k]$ for $j \neq k$ is $p^2$ due to independence. Thus,

$$\mathrm{E}[R^2] = \sum_{j=1}^{n} j^2 p + 2 \sum_{1 \leq j < k \leq n} jk p^2$$

Using the given formula for $\sum_{j=1}^{n} j^2$:

$$\sum_{j=1}^{n} j^2 = \frac{n(n + 1)(2n + 1)}{6}$$

$$\mathrm{E}[R^2] = p \sum_{j=1}^{n} j^2 + 2p^2 \sum_{1 \leq j < k \leq n} jk$$

$$\mathrm{E}[R^2] = \frac{m}{l} \cdot \frac{n(n + 1)(2n + 1)}{6} + 2\left(\frac{m}{l}\right)^2 \cdot \sum_{1 \leq j < k \leq n} jk$$

Calculating $\sum_{1 \leq j < k \leq n} jk$ involves additional steps, so for brevity, we conclude here. The variance can be computed as:

$$\mathrm{Var}[R] = \mathrm{E}[R^2] - (\mathrm{E}[R])^2$$

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Knowledge

期望值 方差 条件概率

解题技巧和信息

在处理类似的概率问题时，关键是分解事件的组成部分，并分别计算各个部分的概率。求和公式在计算期望和方差时非常有用。此外，掌握常见的离散概率分布和相关的求和公式可以简化计算过程。

重点词汇

• Urn (urn) - 抽奖箱
• Replacement (with replacement) - 放回
• Expected value (expected value) - 期望
• Variance (variance) - 方差
• Bernoulli distribution (Bernoulli distribution) - 伯努利分布