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 2016年8月実施 問題11

标签：

• Tokyo-University
• Autoregressive-Order-One-with-Bernoulli-Innovations
• Expectation-and-Variance

Author

zephyr, 祭音Myyura

Description

Let $\mathbf{S} = \{V_1, V_2, V_3, \ldots\}$ be a sequence of mutually independent random variables such that each $V_i$ takes the value of 1 with probability $p$, and 0 with probability $(1 - p)$ ($0 < p < 1$). We define a sequence $X_i$ ($i = 0, 1, 2, \ldots$) as follows:

$$X_0 = 1, \ X_i = aX_{i-1} + V_i, \ (i \geq 1)$$

Here, $a$ is a positive real number. Answer the following questions.

(1) Find the expected value $\mathbb{E}(X_1)$ of $X_1$.

(2) Find the variance $\mathrm{Var}(X_1) = \mathbb{E}(X_1^2) - (\mathbb{E}(X_1))^2$ of $X_1$.

(3) Express $X_i$ as a function of $a$ and the elements of $\mathbf{S}$ ($i \geq 1$).

(4) Find $\mathbb{E}(X_i)$ ($i \geq 1$).

(5) Find $x_\infty = \lim_{i \to \infty} \mathbb{E}(X_i)$ as a function of $a$ and $p$.

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设 $\mathbf{S} = \{V_1, V_2, V_3, \ldots\}$ 为一组相互独立的随机变量，使得每个 $V_i$ 以概率 $p$ 取值为 1， 以概率 $(1 - p)$ 取值为 0（$0 < p < 1$）。我们定义一个序列 $X_i$ （$i = 0, 1, 2, \ldots$）如下：

$$X_0 = 1, \ X_i = aX_{i-1} + V_i, \ (i \geq 1)$$

其中，$a$ 是一个正实数。回答以下问题。

(1) 求 $X_1$ 的期望值 $\mathbb{E}(X_1)$。

(2) 求 $X_1$ 的方差 $\mathrm{Var}(X_1) = \mathbb{E}(X_1^2) - (\mathbb{E}(X_1))^2$。

(3) 表示 $X_i$ 作为 $a$ 和 $\mathbf{S}$ 元素的函数（$i \geq 1$）。

(4) 求 $\mathbb{E}(X_i)$（$i \geq 1$）。

(5) 求 $x_\infty = \lim_{i \to \infty} \mathbb{E}(X_i)$，作为 $a$ 和 $p$ 的函数。

题目描述

令 $V_1,V_2,\ldots$ 相互独立，且每个 $V_i$ 都以概率 $p$ 取 $1$、以概率 $1-p$ 取 $0$，其中 $0<p<1$。给定正实数 $a$，定义随机过程

$$X_0=1,\qquad X_i=aX_{i-1}+V_i\quad(i\ge1).$$

回答下列问题：

1. 求 $E[X_1]$。

2. 求 $\operatorname{Var}(X_1)=E[X_1^2]-E[X_1]^2$。

3. 对 $i\ge1$，把 $X_i$ 显式表示为 $a$ 与 $V_1,\ldots,V_i$ 的函数。

4. 求一般的 $E[X_i]$。

5. 按 $a$ 的不同取值讨论并求

   $$x_\infty=\lim_{i\to\infty}E[X_i],$$

   将结果表示为 $a,p$ 的函数；若极限不为有限值，也需据公式判定。

Kai

(1)

Given:

$$X_0 = 1, \ X_1 = aX_0 + V_1$$

Thus,

$$X_1 = a \cdot 1 + V_1 = a + V_1$$

To find $\mathbb{E}(X_1)$:

$$\mathbb{E}(X_1) = \mathbb{E}(a + V_1) = a + \mathbb{E}(V_1)$$

Since $V_1$ takes the value 1 with probability $p$ and 0 with probability $1-p$, we have:

$$\mathbb{E}(V_1) = 1 \cdot p + 0 \cdot (1 - p) = p$$

Therefore,

$$\mathbb{E}(X_1) = a + p$$

(2)

To find $\mathrm{Var}(X_1)$, we first compute $\mathbb{E}(X_1^2)$:

$$X_1 = a + V_1$$

$$X_1^2 = (a + V_1)^2 = a^2 + 2aV_1 + V_1^2$$

Thus,

$$\mathbb{E}(X_1^2) = \mathbb{E}(a^2 + 2aV_1 + V_1^2)$$

$$\mathbb{E}(X_1^2) = a^2 + 2a\mathbb{E}(V_1) + \mathbb{E}(V_1^2)$$

Since $V_1$ is a Bernoulli random variable:

$$\mathbb{E}(V_1^2) = \mathbb{E}(V_1) = p$$

$$\mathbb{E}(X_1^2) = a^2 + 2ap + p$$

The variance of $X_1$ is:

$$\mathrm{Var}(X_1) = \mathbb{E}(X_1^2) - (\mathbb{E}(X_1))^2$$

$$\mathrm{Var}(X_1) = (a^2 + 2ap + p) - (a + p)^2$$

$$\mathrm{Var}(X_1) = a^2 + 2ap + p - (a^2 + 2ap + p^2)$$

$$\mathrm{Var}(X_1) = p - p^2 = p(1 - p)$$

(3)

To find the general form of $X_i$, we solve the recurrence relation:

$$X_i = aX_{i-1} + V_i$$

Starting from $X_0 = 1$, we have:

$$X_1 = aX_0 + V_1 = a + V_1$$

$$X_2 = aX_1 + V_2 = a(a + V_1) + V_2 = a^2 + aV_1 + V_2$$

$$X_3 = aX_2 + V_3 = a(a^2 + aV_1 + V_2) + V_3 = a^3 + a^2V_1 + aV_2 + V_3$$

It can be observed that:

$$X_i = a^i + a^{i-1}V_1 + a^{i-2}V_2 + \dots + aV_{i-1} + V_i$$

$$X_i = a^i+\sum_{j=1}^{i} a^{i-j}V_j$$

(4)

Using linearity of expectation:

$$\mathbb{E}(X_i) = \mathbb{E}\left(a^i+\sum_{j=1}^{i} a^{i-j}V_j\right)$$

$$\mathbb{E}(X_i) = a^i+\sum_{j=1}^{i} a^{i-j}\mathbb{E}(V_j)$$

Since $\mathbb{E}(V_j) = p$ for all $j$:

$$\mathbb{E}(X_i) = a^i+\sum_{j=1}^{i} a^{i-j}p$$

$$\mathbb{E}(X_i) = a^i+p\sum_{j=1}^{i} a^{i-j}$$

For $a\ne1$, the sum is a geometric series:

$$\sum_{j=1}^{i} a^{i-j} = \frac{a^i - 1}{a - 1}$$

$$\mathbb{E}(X_i) = a^i+p\left(\frac{a^i - 1}{a - 1}\right).$$

For $a=1$, $\mathbb{E}(X_i)=1+ip$.

(5)

Let's consider the limit by first simplifying the expression for $\mathbb{E}(X_i)$.

For $a\ne1$:

$$\mathbb{E}(X_i) = a^i+p\left(\frac{a^i - 1}{a - 1}\right)$$

Let's combine the terms by putting them over a common denominator:

$$\mathbb{E}(X_i) = \frac{p(a^i - 1) + a^i(a - 1)}{a - 1}$$

Simplifying the numerator:

$$\mathbb{E}(X_i) = \frac{p a^i - p + a^{i+1} - a^i}{a - 1}$$

$$\mathbb{E}(X_i) = a^i\left(1+\frac{p}{a-1}\right)-\frac{p}{a-1}$$

Now, let's find the limit for different values of $a$.

Case 1: $a > 1$

When $a > 1$, $a^i$ grows exponentially and $1+p/(a-1)>0$. Thus:

$$\lim_{i \to \infty} \mathbb{E}(X_i) =\lim_{i\to\infty}\left[a^i\left(1+\frac{p}{a-1}\right)-\frac{p}{a-1}\right] =\infty.$$

Thus, the limit does not exist in a finite value; it diverges to infinity.

Case 2: $a = 1$

When $a = 1$, we have:

$$\mathbb{E}(X_i) = 1+ip$$

As $i \to \infty$, the expected value becomes:

$$\lim_{i \to \infty} \mathbb{E}(X_i) = \lim_{i \to \infty} (1+ip) = \infty$$

Thus, the limit also does not exist in a finite value; it diverges to infinity.

Case 3: $0 < a < 1$

When $0 < a < 1$, $a^i\to0$. Therefore:

$$\lim_{i \to \infty} \mathbb{E}(X_i) =-\frac{p}{a-1} =\frac{p}{1-a}.$$

This limit exists and is finite.

In summary:

• For $a > 1$, $\lim_{i \to \infty} \mathbb{E}(X_i)$ does not exist as a finite value (diverges to infinity).
• For $a = 1$, $\lim_{i \to \infty} \mathbb{E}(X_i)$ does not exist as a finite value (diverges to infinity).
• For $0 < a < 1$, $\lim_{i \to \infty} \mathbb{E}(X_i) = \frac{p}{1-a}$, which is finite.

Knowledge

随机过程 期望值 几何级数

重点词汇

• Expected value: 期望值
• Variance: 方差
• Geometric series: 几何级数

参考资料

1. Probability and Statistics for Engineering and the Sciences, Chap. 4