Exercise 20A.1 | 练习 20A.1

📚 Exercise 20A.1 | 练习 20A.1

Exercise 20A.1 marks the beginning of a critical topic in IB Mathematics Analysis and Approaches HL: probability generating functions (PGFs). In this exercise, students are typically asked to find the PGF for a simple discrete random variable, such as a binomial or Bernoulli distribution, laying the foundation for more advanced applications. Mastering this skill is essential for efficiently computing probabilities, means, and variances, and for understanding sums of independent random variables.

练习 20A.1 是 IB 数学分析与方法高阶课程中一个关键主题的开端：概率生成函数（PGF）。在这道练习中，学生通常需要求一个简单离散型随机变量的概率生成函数，例如二项分布或伯努利分布，从而为更高级的应用奠定基础。掌握这一技能对于高效计算概率、均值与方差，以及理解独立随机变量之和至关重要。

1. Understanding the Context of Exercise 20A.1 | 理解练习 20A.1 的背景

instance, X could follow a binomial distribution with parameters n = 2 and p = 0.4. This serves as an ideal starting point to grasp the definition and mechanics of a PGF.

在许多标准的 IB 高阶教材中，第 20 章专门讨论概率生成函数。练习 20A.1 通常给出一个简单的场景：给定一个具有已知概率分布的离散随机变量 X，要求推导其概率生成函数 G(t)。例如，X 服从参数 n = 2 和 p = 0.4 的二项分布。这正是掌握 PGF 定义与计算机制的理想起点。

2. Definition of Probability Generating Function | 概率生成函数的定义

non-negative integer values, the probability generating function is defined as G(t) = E(t^X) = Σ P(X = x) t^x, where the sum runs over all possible values of x. This power series encodes all probabilities: the coefficient of t^x is exactly P(X = x). The variable t is a dummy variable, typically considered for |t| ≤ 1 to guarantee convergence.

对于取非负整数值的离散随机变量 X，概率生成函数定义为 G(t) = E(t^X) = Σ P(X = x) t^x，其中求和遍历 x 的所有可能取值。这个幂级数包含了所有概率信息：t^x 的系数恰好就是 P(X = x)。变量 t 是一个哑变量，通常在 |t| ≤ 1 的条件下考虑以保证收敛。

Key properties include: G(1) = 1, the expected value E(X) = G'(1), and the variance Var(X) = G''(1) + G'(1) – [G'(1)]². These relationships make PGFs extremely powerful for moment calculations.

关键性质包括：G(1) = 1，期望值 E(X) = G'(1)，以及方差 Var(X) = G''(1) + G'(1) – [G'(1)]²。这些关系使 PGF 在矩运算中极具威力。

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3. Step-by-Step Solution: Finding the PGF for Binomial Distribution | 逐步求解：求二项分布的 PGF

Consider a typical Exercise 20A.1 prompt: "Find the probability generating function for X ~ B(2, 0.4)." The probability mass function is P(X = x) = C(2, x) (0.4)^x (0.6)^2−x, for x = 0, 1, 2. Compute each term:

设想一道典型的练习 20A.1 题目：“求 X ~ B(2, 0.4) 的概率生成函数。”其概率质量函数为 P(X = x) = C(2, x) (0.4)^x (0.6)^2−x，其中 x = 0, 1, 2。计算各项：

• P(X=0) = C(2,0)·(0.4)⁰·(0.6)² = 0.36
• P(X=1) = 2 · 0.4 · 0.6 = 0.48
• P(X=2) = (0.4)² = 0.16

中文对照：

• P(X=0) = C(2,0)·(0.4)⁰·(0.6)² = 0.36
• P(X=1) = 2 · 0.4 · 0.6 = 0.48
• P(X=2) = (0.4)² = 0.16

Now build the PGF: G(t) = Σ P(X=x) t^x = 0.36 t⁰ + 0.48 t¹ + 0.16 t² = 0.36 + 0.48t + 0.16t².

现在构造 PGF：G(t) = Σ P(X=x) t^x = 0.36 t⁰ + 0.48 t¹ + 0.16 t² = 0.36 + 0.48t + 0.16t²。

4. Expressing the PGF in Simplified Form | 简化概率生成函数的形式

Notice that the expression 0.36 + 0.48t + 0.16t² can be written as (0.6 + 0.4t)². Indeed, for a binomial distribution B(n, p), the PGF is given by (q + pt)ⁿ where q = 1 – orised form is more concise and reveals the connection to the underlying Bernoulli trials.

注意到表达式 0.36 + 0.48t + 0.16t² 可以写成 (0.6 + 0.4t)²。事实上，对于二项分布 B(n, p)，概率生成函数为 (q + pt)ⁿ，其中 q = 1 – p。这里 q = 0.6，p = 0.4，n = 2，因此 G(t) = (0.6 + 0.4t)²。这种因式

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