CBMS-2017-11

题目来源：[[文字版题库/CBMS/2017#Problem 11|2017#Problem 11]] 日期：2024-07-29 题目主题：CS-概率论-随机过程

解题思路

这个问题涉及一个由互相独立的随机变量 $V_i$ 组成的序列和一个递归关系定义的序列 $X_i$。问题要求我们计算 $X_1$ 的期望值和方差，表示 $X_i$ 为 $a$ 和序列 $\mathbf{S}$ 的函数，并求出 $X_i$ 的期望值和在 $i\to \infty$ 时的极限。

Solution

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

Given:

$$X_0=1$$ $$X_1=a{X}_0+V_1$$

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

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

$$X_1=a+V_1$$ $$X_1^2=(a+V_1{)}^2=a^2+2a{V}_1+V_1^2$$

Thus,

$$\mathbb{E}(X_1^2)=\mathbb{E}(a^2+2a{V}_1+V_1^2)$$ $$\mathbb{E}(X_1^2)=a^2+2a\mathbb{E}(V_1)+\mathbb{E}(V_1^2)$$

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

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

$$X_i=a{X}_{i-1}+V_i$$

Starting from $X_0=1$, we have:

$$X_1=a{X}_0+V_1=a+V_1$$ $$X_2=a{X}_1+V_2=a(a+V_1)+V_2=a^2+a{V}_1+V_2$$ $$X_3=a{X}_2+V_3=a(a^2+a{V}_1+V_2)+V_3=a^3+a^2{V}_1+a{V}_2+V_3$$

It can be observed that:

$$X_i=a^i+a^{i-1}{V}_1+a^{i-2}{V}_2+\cdots +a{V}_{i-1}+V_i$$ $$X_i=\sum _{j=0}^i{a}^{i-j}{V}_j+a^i$$

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

Using linearity of expectation:

$$\mathbb{E}(X_i)=\mathbb{E}\left(\sum _{j=0}^i{a}^{i-j}{V}_j+a^i\right)$$ $$\mathbb{E}(X_i)=\sum _{j=0}^i{a}^{i-j}\mathbb{E}(V_j)+a^i$$

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

$$\mathbb{E}(X_i)=\sum _{j=0}^i{a}^{i-j}p+a^i$$ $$\mathbb{E}(X_i)=p\sum _{j=0}^i{a}^{i-j}+a^i$$

The sum is a geometric series:

$$\sum _{j=0}^i{a}^{i-j}=\frac{a^{i+1}-1}{a-1}$$ $$\mathbb{E}(X_i)=p\left(\frac{a^{i+1}-1}{a-1}\right)+a^i$$

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

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

Given:

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

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

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

Simplifying the numerator:

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

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

Case 1: $a>1$

When $a>1$, as $i\to \infty$, $a^{i+1}$ and $a^i$ grow exponentially, and the dominant term will be $a^{i+1}$. Thus, the limit is:

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

Since $a^{i+1}$ grows much faster than $a^i$ and $p$, we have:

$$\underset{i\to \infty }{\lim }\mathbb{E}(X_i)=\frac{{\lim }_{i\to \infty }{a}^{i+1}(p+1)}{a-1}=\infty$$

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

Case 2: $a=1$

When $a=1$, we have:

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

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

$$\underset{i\to \infty }{\lim }\mathbb{E}(X_i)=\underset{i\to \infty }{\lim }(pi+p+1)=\infty$$

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

Case 3: $0<a<1$

When $0<a<1$, as $i\to \infty$, both $a^{i+1}$ and $a^i$ approach 0. The terms involving $a^{i+1}$ and $a^i$ become negligible, and the limit can be simplified as:

$$\underset{i\to \infty }{\lim }\mathbb{E}(X_i)=\underset{i\to \infty }{\lim }\frac{a^{i+1}(p+1)-a^i-p}{a-1}=\frac{0-p}{a-1}=-\frac{p}{a-1}$$

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 $0<a<1$, ${\lim }_{i\to \infty }\mathbb{E}(X_i)=-\frac{p}{a-1}$, which is finite.

知识点

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