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

Author

zephyr

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 \(p\).

---

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。

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:

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

\[ X_i = \sum_{j=0}^{i} a^{i-j}V_j + a^i \]

(4)

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)

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:

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

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

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

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:

\[ \lim_{i \to \infty} \mathbb{E}(X_i) = \lim_{i \to \infty} (pi + p + 1) = \infty \]

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:

\[ \lim_{i \to \infty} \mathbb{E}(X_i) = \lim_{i \to \infty} \frac{a^{i+1}(p + 1) - a^i - p}{a - 1} = \frac{0 - p}{a - 1} = -\frac{p}{a - 1} \]

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.

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