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 2014年8月実施 問題11

标签：

• 东京大学
• 概率生成函数
• 随机项数和的概率生成函数
• 几何分布

Author

zephyr, 祭音Myyura

Description

For an arbitrary random variable $X$ that takes values in non-negative integers, we define the probability generating function $\varphi_X(s)$ of $X$ as $\varphi_X(s) = E\{s^X\} = \sum_{k=0}^{\infty} s^k \Pr\{X = k\}$. Here, $E\{A\}$ denotes the expected value of $A$, $\Pr\{X = k\}$ denotes the probability that $X$ assumes value $k$, and $s$ represents a real number.

(1) Show the following equalities (a), (b).

• (a) $\varphi_X(1) = 1$
• (b) $\frac{d \varphi_X}{ds}(1) = E\{X\}$

In the following, $N$ and $U_i$ $(i = 1, 2, \ldots)$ are mutually independent, identically distributed random variables that take values in non-negative integers.

(2) For a positive integer $n$, we define a random variable $Y_n = Y_n(U_1, U_2, \ldots) = \sum_{i=1}^n U_i$. Show $\varphi_{Y_n}(s) = \varphi_N(s)^n$.

(3) We define a random variable $W = W(N, Y_1, Y_2, \ldots) = \sum_{n=1}^{\infty} Y_n I(N = n)$. Here,

$$I(N = n) = \begin{cases} 1 & \text{if } N = n \\ 0 & \text{if } N \neq n \end{cases}$$

Show $\varphi_W(s) = \varphi_N(\varphi_N(s))$.

(Hint: $\Pr\{W = k\} = \sum_{n=1}^{\infty} \Pr\{Y_n = k\} \Pr\{N = n\}$ for a positive integer $k$)

(4) For $\Pr\{N = k\} = q^k (1 - q)$, $(0 < q < 1)$, calculate $E\{N\}$ and $E\{W\}$.

---

对于一个取非负整数值的任意随机变量 $X$，我们定义 $X$ 的概率生成函数 $\varphi_X(s)$ 为 $\varphi_X(s) = E\{s^X\} = \sum_{k=0}^{\infty} s^k \Pr\{X = k\}$。其中，$E\{A\}$ 表示 $A$ 的期望值，$\Pr\{X = k\}$ 表示 $X$ 取值为 $k$ 的概率，$s$ 表示一个实数。

(1) 证明以下等式 (a), (b)。

• (a) $\varphi_X(1) = 1$
• (b) $\frac{d \varphi_X}{ds}(1) = E\{X\}$

在下列情形中，$N$ 和 $U_i$ $(i = 1, 2, \ldots)$ 是相互独立的同分布随机变量，取非负整数值。

(2) 对于一个正整数 $n$，我们定义一个随机变量 $Y_n = Y_n(U_1, U_2, \ldots) = \sum_{i=1}^n U_i$。证明 $\varphi_{Y_n}(s) = \varphi_N(s)^n$。

(3) 我们定义一个随机变量 $W = W(N, Y_1, Y_2, \ldots) = \sum_{n=1}^{\infty} Y_n I(N = n)$。 其中，

$$I(N = n) = \begin{cases} 1 & \text{如果 } N = n \\ 0 & \text{如果 } N \neq n \end{cases}$$

证明 $\varphi_W(s) = \varphi_N(\varphi_N(s))$。

（提示：$\Pr\{W = k\} = \sum_{n=1}^{\infty} \Pr\{Y_n = k\} \Pr\{N = n\}$ 对于一个正整数 $k$）

(4) 对于 $\Pr\{N = k\} = q^k (1 - q)$, $(0 < q < 1)$，计算 $E\{N\}$ 和 $E\{W\}$。

题目描述

对任意取非负整数值的随机变量 $X$，定义其概率生成函数

$$\varphi_X(s)=E\{s^X\}=\sum_{k=0}^{\infty}s^k\Pr\{X=k\},$$

其中 $s$ 为实数。回答下列问题：

1. 证明 $\varphi_X(1)=1$ 以及

   $$\left.\frac{d\varphi_X}{ds}\right|_{s=1}=E\{X\}.$$

2. 以下设 $N,U_1,U_2,\ldots$ 为相互独立、同分布且取非负整数值的随机变量。对正整数 $n$ 定义

   $$Y_n=\sum_{i=1}^{n}U_i,$$

   证明 $\varphi_{Y_n}(s)=\varphi_N(s)^n$。

3. 定义

   $$W=\sum_{n=1}^{\infty}Y_n I(N=n),\qquad I(N=n)= \begin{cases} 1,&N=n,\\ 0,&N\ne n, \end{cases}$$

   证明 $\varphi_W(s)=\varphi_N(\varphi_N(s))$。可使用提示：对正整数 $k$，

   $$\Pr\{W=k\}=\sum_{n=1}^{\infty}\Pr\{Y_n=k\}\Pr\{N=n\}.$$

4. 当 $\Pr\{N=k\}=q^k(1-q)$、$0<q<1$ 时，计算 $E\{N\}$ 与 $E\{W\}$。

Kai

We work on $0\le s\le1$, where the generating functions are finite. The derivative at $s=1$ is understood from the left; it may be $+\infty$ if $E[X]=+\infty$. For $0<s<1$, termwise differentiation is valid, and monotone convergence gives the limit as $s\uparrow1$.

(1)

(a)

$$\begin{aligned} \varphi_X(1) &= E\{1^X\} = \sum_{k=0}^{\infty} 1^k \Pr\{X = k\} \\ &= \sum_{k=0}^{\infty} \Pr\{X = k\} \\ &= 1 \end{aligned}$$

The last step follows from the fact that the sum of probabilities over all possible outcomes is 1.

(b)

$$\begin{aligned} \frac{d \varphi_X}{ds}(s) &= \frac{d}{ds} E\{s^X\} = \frac{d}{ds} \sum_{k=0}^{\infty} s^k \Pr\{X = k\} \\ &= \sum_{k=1}^{\infty} k s^{k-1} \Pr\{X = k\} \end{aligned}$$

Evaluating at $s = 1$:

$$\begin{aligned} \frac{d \varphi_X}{ds}(1) &= \sum_{k=1}^{\infty} k \Pr\{X = k\} \\ &= E\{X\} \end{aligned}$$

(2)

We need to show $\varphi_{Y_n}(s) = \varphi_N(s)^n$ where $Y_n = \sum_{i=1}^n U_i$.

$$\begin{aligned} \varphi_{Y_n}(s) &= E\{s^{Y_n}\} = E\{s^{\sum_{i=1}^n U_i}\} \\ &= E\{s^{U_1} \cdot s^{U_2} \cdot ... \cdot s^{U_n}\} \\ &= E\{s^{U_1}\} \cdot E\{s^{U_2}\} \cdot ... \cdot E\{s^{U_n}\} \quad \text{(due to i.i.d)} \\ &= (E\{s^{N}\})^n \quad \text{(due to i.i.d)} \\ &= \varphi_N(s)^n \end{aligned}$$

(3)

We need to show $\varphi_W(s) = \varphi_N(\varphi_N(s))$ where $W = \sum_{n=1}^{\infty} Y_n I(N = n)$.

Using the hint and the definition of probability generating function:

$$\begin{aligned} \varphi_W(s) &= E\{s^W\}=\sum_{k=0}^{\infty}s^k\Pr\{W=k\} \\ &=\Pr\{N=0\}+\sum_{k=0}^{\infty}s^k \sum_{n=1}^{\infty}\Pr\{Y_n=k\}\Pr\{N=n\} \\ &=\Pr\{N=0\}+\sum_{n=1}^{\infty}\Pr\{N=n\} \sum_{k=0}^{\infty}s^k\Pr\{Y_n=k\} \\ &=\Pr\{N=0\}+\sum_{n=1}^{\infty}\Pr\{N=n\}\varphi_{Y_n}(s) \\ &= \sum_{n=0}^{\infty} \Pr\{N = n\} \varphi_N(s)^n \quad \text{(from result of part 2)} \\ &= \varphi_N(\varphi_N(s)) \end{aligned}$$

(4)

Given $\Pr\{N = k\} = q^k (1 - q)$, $(0 < q < 1)$, we need to calculate $E\{N\}$ and $E\{W\}$.

First, let's calculate $E\{N\}$:

$$\begin{aligned} E\{N\} &= \sum_{k=0}^{\infty} k \Pr\{N = k\} = \sum_{k=0}^{\infty} k q^k (1 - q) \\ &= (1 - q) \sum_{k=0}^{\infty} k q^k = (1 - q) \frac{q}{(1-q)^2} = \frac{q}{1-q} \end{aligned}$$

Now, for $E\{W\}$, we can use the result from part 1(b) and part 3:

$$\begin{aligned} E\{W\} &= \frac{d \varphi_W}{ds}(1) = \frac{d}{ds} \varphi_N(\varphi_N(s)) |_{s=1} \\ &= \varphi_N'(\varphi_N(1)) \cdot \varphi_N'(1) \\ &= E\{N\}^2 \quad \text{(using part 1(b))} \\ &= (\frac{q}{1-q})^2 \end{aligned}$$

Therefore, $E\{W\} = (\frac{q}{1-q})^2$.

Knowledge

概率论 概率生成函数 条件期望 全期望公式 复合分布

难点思路

1. 理解概率生成函数的定义和基本性质
2. 利用独立性推导和的概率生成函数
3. 使用条件期望和全期望公式推导复合随机变型的概率生成函数
4. 应用概率生成函数的性质计算具体分布的期望

解题技巧和信息

1. 概率生成函数的基本性质：
   • $\varphi_X(1) = 1$
   • $\frac{d \varphi_X}{ds}(1) = E\{X\}$
   • $\varphi_{X+Y}(s) = \varphi_X(s) \cdot \varphi_Y(s)$ (对于独立的 $X$ 和 $Y$)
2. 几何分布的概率生成函数：如果 $X \sim Geo(p)$，则 $\varphi_X(s) = \frac{p}{1-(1-p)s}$
3. 利用全期望公式：$E\{Y\} = E\{E\{Y|X\}\}$

重点词汇

• Probability generating function: 概率生成函数
• Independent and identically distributed (i.i.d.): 独立同分布
• Compound distribution: 复合分布
• Conditional expectation: 条件期望
• Law of total expectation: 全期望公式
• Geometric distribution: 几何分布

Reference

• 東京大学 CBMS 公式過去問