📚 Probability Generating Functions: Core Concepts | 概率生成函数的核心概念
A Probability Generating Function (PGF) is a powerful mathematical tool used to encapsulate the entire probability distribution of a discrete random variable in a single algebraic function. It transforms problems involving sums, expectations, and variance into more tractable algebraic manipulations.
概率生成函数（PGF）是一种强大的数学工具，它将离散随机变量的整个概率分布封装在一个代数函数中。它将涉及求和、期望和方差的问题转化为更易处理的代数运算。
For the Edexcel A-Level Further Mathematics syllabus, the PGF is defined for a discrete random variable X taking values in the non-negative integers. It is a compact representation that allows us to derive moments, find distributions of sums of independent variables, and identify probability mass functions efficiently.
在爱德思（Edexcel）A-Level进阶数学大纲中，PGF定义为取值于非负整数的离散随机变量 X 的函数。它是一种紧凑的表示方法，使我们能够高效地推导矩、求独立变量之和的分布以及识别概率质量函数。
This article systematically walks through the definition, its key properties, and worked examples aligned with the examination style of Edexcel.
本文系统地讲解概率生成函数的定义、关键性质以及与爱德思考试风格一致的例题。
1. Definition of the PGF | 概率生成函数的定义
The probability generating function of a discrete random variable X is defined as the expected value of t raised to the power X:
离散随机变量 X 的概率生成函数定义为 t 的 X 次幂的期望值：
Gₓ(t) = E(tˣ) = Σ p(x) tˣ
where the summation is over all possible values x of X, and p(x) = P(X = x).
其中求和覆盖 X 的所有可能取值 x，p(x) = P(X = x)。
The notation Gₓ(t) is standard. The variable t is a dummy variable and has no probabilistic meaning on its own; it is a placeholder used to encode probabilities as coefficients of powers of t.
记号 Gₓ(t) 是标准用法。变量 t 是一个哑变量，本身没有概率含义；它只是一个占位符，用于将概率编码为 t 的幂的系数。
For example, if P(X = 0) = 0.2, P(X = 1) = 0.3, and P(X = 2) = 0.5, then:
例如，若 P(X = 0) = 0.2，P(X = 1) = 0.3，P(X = 2) = 0.5，则：
Gₓ(t) = 0.2 + 0.3t + 0.5t²
Notice that the probability of a particular value x appears as the coefficient of tˣ. This one-to-one correspondence is the fundamental working principle of the PGF.
注意特定值 x 的概率作为 tˣ 的系数出现。这种一一对应关系是概率生成函数的基本工作原理。
2. Convergence and Domain | 收敛性与定义域
Since Gₓ(t) = E(tˣ) = Σ p(x) tˣ is an infinite series when X has infinitely many possible values, we need to consider convergence.
由于当 X 有无限多个可能取值时，Gₓ(t) = E(tˣ) = Σ p(x) tˣ 是一个无穷级数，我们需要考虑其收敛性。
The series always converges (absolutely) for |t| < 1, because |tˣ| ≤ |t| and Σ p(x) = 1. In practice, we usually consider t in a small neighbourhood of t = 1, where Gₓ(1) = Σ p(x) = 1.
级数在 |t| < 1 时总是绝对收敛，因为 |tˣ| ≤ |t| 且 Σ p(x) = 1。在实际应用中，我们通常考虑 t = 1 的一个小邻域，其中 Gₓ(1) = Σ p(x) = 1。
It is important to note that Gₓ(1) = 1 always, as it simply sums all probabilities. This serves as a quick check when constructing or verifying a PGF.
重要的一点是 Gₓ(1) = 1 始终成立，因为它只是对所有概率求和。这在构造或验证PGF时可以作为快速检验手段。
A function can be a valid PGF only if it can be expressed as a power series with non-negative coefficients that sum to 1 when evaluated at t = 1. This is called the regularity condition.
一个函数只有在能够表示为非负系数的幂级数，并且在 t = 1 处求和为 1 时，才能作为有效的PGF。这称为正则性条件。
3. Recovering the Probability Mass Function | 从PGF恢复概率质量函数
One of the most important properties of the PGF is that the probability distribution can be fully recovered from it. Since p(x) is the coefficient of tˣ, we can retrieve individual probabilities by differentiation.
PGF最重要的性质之一在于，概率分布可以从中完全恢复。由于 p(x) 是 tˣ 的系数，我们可以通过微分来提取单个概率。
Observe that differentiating Gₓ(t) with respect to t each time "brings down" the exponent and multiplies the coefficient accordingly. After differentiating k times and setting t = 0, only the constant term remains.
观察到每对 Gₓ(t) 关于 t 求一次导数，就会将指数"拉下来"并与对应系数相乘。在求导 k 次并将 t = 0 代入后，只剩常数项。
The key formula for recovering probabilities is:
恢复概率的关键公式为：
p(x) = P(X = x) = Gₓ⁽ˣ⁾(0) / x!
Thus, the probability of exactly k occurrences is the k-th derivative of Gₓ evaluated at t = 0, divided by k factorial.
因此，恰好 k 次发生的概率等于 Gₓ 在 t = 0 处的 k 阶导数除以 k 的阶乘。
Example: If Gₓ(t) = (0.3 + 0.7t)¹⁰, then P(X = 3) = coefficient of t³, which can be found via the binomial expansion formula.
例：若 Gₓ(t) = (0.3 + 0.7t)¹⁰，则 P(X = 3) = t³ 的系数，可通过二项展开公式求得。
This property is especially useful when the PGF is a compact function such as a binomial or a geometric series, where direct differentiation is straightforward.
这一性质在PGF为紧凑函数（如二项式或几何级数）时尤为有用，因为此时直接微分非常简便。
4. Moments: Mean and Variance | 矩：均值与方差
The PGF can be used to derive the mean and variance of a distribution without needing the full probability mass function. This is accomplished through evaluating derivatives of Gₓ at t = 1.
PGF可用于在不需要完整概率质量函数的情况下导出分布的均值和方差。这是通过在 t = 1 处求 Gₓ 的导数来实现的。
First, consider the first derivative. Since Gₓ′(t) = Σ x p(x) tˣ⁻¹, evaluating at t = 1 gives:
首先，考虑一阶导数。由于 Gₓ′(t) = Σ x p(x) tˣ⁻¹，在 t = 1 处取值得到：
E(X) = Gₓ′(1)
The mean is simply the first derivative evaluated at t = 1.
均值就是 t = 1 处的一阶导数值。
For the second moment, we use the second derivative. Note that Gₓ″(t) = Σ x(x−1) p(x) tˣ⁻², hence:
对于二阶矩，我们使用二阶导数。注意 Gₓ″(t) = Σ x(x−1) p(x) tˣ⁻²，因此：
E[X(X−1)] = Gₓ″(1)
Using the identity X² = X(X−1) + X, the second moment is given by:
利用恒等式 X² = X(X−1) + X，二阶矩为：
E(X²) = Gₓ″(1) + Gₓ′(1)
Finally, Var(X) = E(X²) − [E(X)]² = Gₓ″(1) + Gₓ′(1) − [Gₓ′(1)]².
最后，Var(X) = E(X²) − [E(X)]² = Gₓ″(1) + Gₓ′(1) − [Gₓ′(1)]²。
5. The Factorial Moments | 阶乘矩
In the context of PGFs, factorial moments often arise more naturally than ordinary moments. The r-th factorial moment is defined as E[X(X−1)...(X−r+1)].
在PGF的语境中，阶乘矩往往比普通矩更为自然。第 r 阶阶乘矩定义为 E[X(X−1)...(X−r+1)]。
The relationship between the PGF and factorial moments is remarkably clean. Indeed, evaluating the r-th derivative of Gₓ at t = 1 yields:
PGF与阶乘矩之间的关系非常简洁。实际上，Gₓ 在 t = 1 处的 r 阶导数值为：
Gₓ⁽ʳ⁾(1) = E[X(X−1)(X−2)…(X−r+1)]
Hence, the first factorial moment equals the mean, and the second factorial moment E[X(X−1)] can be combined with the first to obtain the variance.
因此，一阶阶乘矩等于均值，二阶阶乘矩 E[X(X−1)] 可与其结合以求得方差。
For example, for a geometric distribution with PGF G(t) = p/(1−qt), we have G′(t) = pq/(1−qt)², so E(X) = pq/p² = q/p.
例如，对于PGF为 G(t) = p/(1−qt) 的几何分布，有 G′(t) = pq/(1−qt)²，所以 E(X) = pq/p² = q/p。
This approach avoids the summation of infinite series and proves to be considerably faster in examinations.
这种方法避免了无穷级数的求和，在考试中被证明效率显著更高。
6. Sums of Independent Random Variables | 独立随机变量之和
One of the most remarkable applications of the PGF lies in determining the distribution of the sum of independent random variables. If X and Y are independent, then the PGF of X + Y is the product of their individual PGFs.
PGF最卓越的应用之一在于确定独立随机变量之和的分布。若 X 与 Y 独立，则 X + Y 的PGF等于它们各自PGF的乘积。
The proof follows directly from the definition and the property of independence:
证明直接来自定义和独立性的性质：
GX+Y(t) = E(tX+Y) = E(tˣ tʸ) = E(tˣ)E(tʸ) = Gₓ(t)GY(t)
This result extends to the sum of n independent random variables: the PGF of the sum is equal to the product of the n individual PGFs.
这一结果可推广到 n 个独立随机变量之和：和的PGF等于 n 个单独PGF的乘积。
Example: Let X ~ Poisson(2) and Y ~ Poisson(3) be independent. The PGF of X is e2(t−1) and that of Y is e3(t−1). Therefore, the PGF of X+Y is e5(t−1), which is the PGF of a Poisson(5) distribution.
例：设 X ~ 泊松(2) 和 Y ~ 泊松(3) 相互独立。X 的PGF为 e2(t−1)，Y 的PGF为 e3(t−1)。因此，X+Y 的PGF为 e5(t−1)，即泊松(5)分布的PGF。
This technique provides an elegant solution to many problems that would otherwise require lengthy convolution sums.
这一技术为许多问题提供了优雅的解答，否则这些问题将需要冗长的卷积求和。
7. PGFs of Standard Distributions | 标准分布的PGF
It is essential to be familiar with the PGFs of the standard discrete distributions that appear frequently in the Edexcel specification.
熟悉爱德思大纲中高频出现的标准离散分布的PGF是至关重要的。
| Distribution | PGF Gₓ(t) | Mean E(X) | Variance |
| Bernoulli(p) | 1 − p + pt | p | p(1−p) |
| Binomial(n, p) | (1 − p + pt)ⁿ | np | np(1−p) |
| Geometric(p) | p / (1 − qt) | q/p | q/p² |
| Poisson(λ) | eλ(t−1) | λ | λ |
Where q = 1 − p. Notice that for the geometric distribution, we use the form P(X = x) = qˣ⁻¹p for x = 1, 2, 3, … in the Edexcel syllabus.
其中 q = 1 − p。注意在爱德思大纲中，几何分布采用 P(X = x) = qˣ⁻¹p，x = 1, 2, 3, … 的形式。
These PGFs are not merely memorised — candidates should be able to derive them from first principles using the definitions of the underlying distributions.
这些PGF不应只是死记硬背——考生应能够直接从底层分布的定义出发，从基本原理推导它们。
8. Using PGFs to Combine Independent Variables | 利用PGF合并独立变量
A classic examination question asks you to find the distribution of the sum of independent variables, each with the same distribution, and to use the PGF approach to justify your answer.
一个经典的考试问题是要求求若干个同分布独立变量之和的分布，并利用PGF方法论证答案。
Let X₁, X₂, …, Xₙ be independent and identically distributed with PGF Gₓ(t). Then the sum Sₙ has PGF:
设 X₁, X₂, …, Xₙ 独立同分布，其PGF为 Gₓ(t)。则和 Sₙ 的PGF为：
GSₙ(t) = [Gₓ(t)]ⁿ
For instance, if each Xᵢ ~ Bernoulli(p), then Sₙ ~ Binomial(n, p), since the PGF of the sum is (1 − p + pt)ⁿ.
例如，若每个 Xᵢ ~ 伯努利(p)，则 Sₙ ~ 二项(n, p)，因为和的PGF为 (1 − p + pt)ⁿ。
Similarly, the sum of independent Poisson variables is Poisson whose parameter is the sum of the individual parameters. This is known as the additive (or reproductive) property of the Poisson distribution.
类似地，独立泊松变量之和仍是泊松分布，其参数为各个参数之和。这称为泊松分布的可加性（或再生性）。
These results are particularly significant as they allow you to state a distribution result with confidence, supported by the PGF derivation.
这些结果尤为重要，因为它们允许你在PGF推导的支持下，自信地陈述分布结论。
9. Worked Example: Applying PGFs | 例题：PGF的应用
Problem: The discrete random variable X has a PGF given by Gₓ(t) = (0.2 + 0.8t)⁴.
题目：离散随机变量 X 的PGF为 Gₓ(t) = (0.2 + 0.8t)⁴。
(a) Find P(X = 2).
(a) 求 P(X = 2)。
Solution: The coefficient of t² is obtained from the binomial expansion:
解：t² 的系数来自二项展开：
P(X = 2) = C(4,2)(0.2)²(0.8)² = 6 × 0.04 × 0.64 = 0.1536
(b) Find the mean and variance of X.
(b) 求 X 的均值和方差。
Solution: Differentiate to get Gₓ′(t) = 4(0.2 + 0.8t)³ × 0.8 and Gₓ″(t) = 12(0.2 + 0.8t)² × 0.64.
解：求导得 Gₓ′(t) = 4(0.2 + 0.8t)³ × 0.8，Gₓ″(t) = 12(0.2 + 0.8t)² × 0.64。
Therefore E(X) = Gₓ′(1) = 4 × 1 × 0.8 = 3.2.
因此 E(X) = Gₓ′(1) = 4 × 1 × 0.8 = 3.2。
And E[X(X−1)] = Gₓ″(1) = 12 × 1 × 0.64 = 7.68.
且 E[X(X−1)] = Gₓ″(1) = 12 × 1 × 0.64 = 7.68。
Thus Var(X) = 7.68 + 3.2 − (3.2)² = 7.68 + 3.2 − 10.24 = 0.64.
因此 Var(X) = 7.68 + 3.2 − (3.2)² = 7.68 + 3.2 − 10.24 = 0.64。
This aligns with the known result that for X ~ Binomial(4, 0.8), E(X) = 3.2 and Var(X) = 4 × 0.8 × 0.2 = 0.64.
这与已知结果一致：对于 X ~ 二项(4, 0.8)，E(X) = 3.2，Var(X) = 4 × 0.8 × 0.2 = 0.64。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Several recurring errors appear in examination responses involving PGFs. Awareness of these pitfalls is key to securing full marks.
在涉及PGF的考试作答中，存在若干反复出现的错误。对这些陷阱保持警觉是获得满分的关键。
First, many candidates forget to divide by k! when recovering probabilities. Remember that p(x) = Gₓ⁽ˣ⁾(0) / x! — the factorial is essential unless x = 0 or x = 1.
第一，许多考生在恢复概率时忘记除以 k!。记住 p(x) = Gₓ⁽ˣ⁾(0) / x! ——除非 x = 0 或 x = 1，阶乘是不可或缺的。
Second, when computing variance, candidates sometimes use E[X(X−1)] directly as if it were E(X²). The correct formula is E(X²) = Gₓ″(1) + Gₓ′(1).
第二，计算方差时，考生有时将 E[X(X−1)] 直接当作 E(X²)。正确公式为 E(X²) = Gₓ″(1) + Gₓ′(1)。
Third, when finding the PGF of a sum, it is vital to check independence. Without independence, the PGF of the sum is not equal to the product of the individual PGFs.
第三，求和的PGF时，必须验证独立性。没有独立性，和的PGF就不等于各自PGF的乘积。
It is also advisable to use Gₓ(1) = 1 as a check before proceeding with calculations — if the sum of coefficients does not equal 1, the PGF is invalid.
同时建议在继续计算之前，使用 Gₓ(1) = 1 进行检验——若系数之和不等于1，则该PGF无效。
Examiners reward clear algebraic layout. When answering, state the differentiation steps explicitly and substitute t = 1 rather than computing the entire derivative expression.
考官重视清晰的代数步骤。作答时，应明确写出求导步骤，并代入 t = 1，而非计算完整的导数表达式。
11. Summary and Key Takeaways | 总结与要点回顾
The probability generating function is a central topic in Edexcel A-Level Further Mathematics. It provides a unified framework for understanding discrete distributions, their moments, and the distributions of sums.
概率生成函数是爱德思A-Level进阶数学的核心主题。它为理解离散分布、其矩以及和的分布提供了一个统一框架。
The essential facts can be summarised as follows:
基本要点概述如下：
Gₓ(t) = E(tˣ) = Σ p(x) tˣ, with p(x) recovered as the coefficient of tˣ.
Gₓ(t) = E(tˣ) = Σ p(x) tˣ，p(x) 作为 tˣ 的系数被提取。
E(X) = Gₓ′(1) and E(X²) = Gₓ″(1) + Gₓ′(1).
E(X) = Gₓ′(1)，E(X²) = Gₓ″(1) + Gₓ′(1)。
For independent X and Y, GX+Y(t) = Gₓ(t)GY(t).
对于独立的 X 和 Y，GX+Y(t) = Gₓ(t)GY(t)。
PGFs of standard distributions (binomial, geometric, Poisson) must be known or derivable.
标准分布（二项、几何、泊松）的PGF必须熟记或能推导。
Gₓ(1) = 1 serves as a validity check.
Gₓ(1) = 1 作为有效性检验。
Mastery of the PGF not only prepares candidates for direct questions on the topic but also provides a tool for solving more complex problems across the entire statistics syllabus.
掌握PGF不仅帮助考生应对直接考查该主题的题目，还提供了一种工具，用于解决整个统计大纲中的更复杂问题。
12. Practice Questions | 练习题目
To consolidate understanding, attempt the following questions before looking at worked solutions.
为巩固理解，请先尝试以下问题，再看解答。
Problem 1: Find the mean and variance of a random variable with PGF G(t) = e3(t−1).
题目1：求PGF为 G(t) = e3(t−1) 的随机变量的均值与方差。
Problem 2: If X and Y are independent, with X ~ Poisson(4) and Y ~ Poisson(6), state the distribution of X + Y and write down its PGF.
题目2：若 X 与 Y 独立，且 X ~ 泊松(4)，Y ~ 泊松(6)，指出 X+Y 的分布并写出其PGF。
Problem 3: The discrete random variable X has PGF G(t) = (2 − t)⁻¹. Find P(X = 3) and Var(X).
题目3：离散随机变量 X 的PGF为 G(t) = (2 − t)⁻¹。求 P(X = 3) 和 Var(X)。
Solutions:
解答：
Solution 1: This is Poisson(3), so E(X) = 3, Var(X) = 3.
解答1：这是泊松(3)分布，故 E(X) = 3，Var(X) = 3。
Solution 2: X+Y ~ Poisson(10), G(t) = e10(t−1).
解答2：X+Y ~ 泊松(10)，G(t) = e10(t−1)。
Solution 3: Expand G(t) = (2 − t)⁻¹ = (1/2)(1 − t/2)⁻¹ = ½[1 + t/2 + (t/2)² + (t/2)³ + …]. Thus P(X = 3) = ½ × (1/2)³ = 1/16 = 0.0625. For mean: E(X) = G′(1), with G′(t) = (2 − t)⁻², so E(X) = 1. For the variance: G″(t) = 2(2 − t)⁻³, so E[X(X−1)] = 2, giving Var(X) = 2 + 1 − 1 = 2.
解答3：展开 G(t) = (2 − t)⁻¹ = (1/2)(1 − t/2)⁻¹ = ½[1 + t/2 + (t/2)² + (t/2)³ + …]。因此 P(X = 3) = ½ × (1/2)³ = 1/16 = 0.0625。求均值：E(X) = G′(1)，其中 G′(t) = (2 − t)⁻²，故 E(X) = 1。求方差：G″(t) = 2(2 − t)⁻³，所以 E[X(X−1)] = 2，得 Var(X) = 2 + 1 − 1 = 2。
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