Probability Generating Functions of Standard Distributions — A-Level Edexcel Further Statistics 1

概率生成函数的标准分布应用 — A-Level Edexcel 进阶统计 1

Introduction

Probability Generating Functions (PGFs) are powerful tools in probability theory that encode the entire probability distribution of a discrete random variable into a single function. For A-Level Edexcel Further Statistics 1 (FS1), mastering PGFs of standard distributions is essential — it allows you to derive means, variances, and handle sums of independent random variables with remarkable efficiency.

概率生成函数（PGF）是概率论中的强大工具，它将一个离散随机变量的整个概率分布编码为一个单一函数。对于 A-Level Edexcel 进阶统计 1（FS1），掌握标准分布的 PGF 至关重要——它可以让你高效地推导均值、方差，并处理独立随机变量的求和问题。

1. What is a Probability Generating Function?

什么是概率生成函数？

A Probability Generating Function G(t) for a discrete random variable X taking non-negative integer values is defined as:

对于取值为非负整数的离散随机变量 X，其概率生成函数 G(t) 定义为：

G(t) = E(t^X) = Σ P(X = x) · t^x (summed over all possible x)

This is essentially the expected value of t raised to the power of X. The variable t is a dummy variable, and the PGF is typically defined for |t| ≤ 1, though this constraint can be relaxed for many practical calculations.

这本质上是 t 的 X 次幂的期望值。变量 t 是一个虚拟变量，PGF 通常定义在 |t| ≤ 1 的范围内，尽管在许多实际计算中可以放宽这一限制。

The beauty of the PGF lies in how it packages information: knowing G(t) is mathematically equivalent to knowing the entire probability distribution P(X = x). Each probability P(X = x) is simply the coefficient of t^x in the power series expansion of G(t).

PGF 的美妙之处在于它如何封装信息：知道 G(t) 在数学上等同于知道整个概率分布 P(X = x)。每个概率 P(X = x) 就是 G(t) 的幂级数展开式中 t^x 的系数。

2. Key Properties of PGFs

PGF 的关键性质

Property 1: G(1) = 1
Since Σ P(X = x) = 1 for any probability distribution, substituting t = 1 gives:
因为对于任何概率分布都有 Σ P(X = x) = 1，代入 t = 1 得到：
G(1) = Σ P(X = x) · 1^x = Σ P(X = x) = 1

Property 2: Mean from the First Derivative
The expected value (mean) of X is given by the first derivative evaluated at t = 1:
X 的期望值（均值）由在 t = 1 处的一阶导数给出：
E(X) = G'(1)

This is because G'(t) = Σ x · P(X = x) · t^(x-1), and setting t = 1 yields Σ x · P(X = x) = E(X).
这是因为 G'(t) = Σ x · P(X = x) · t^(x-1)，令 t = 1 得到 Σ x · P(X = x) = E(X)。

Property 3: Variance from the Second Derivative
The variance can be obtained using the second derivative:
方差可以通过二阶导数求得：
Var(X) = G”(1) + G'(1) – [G'(1)]²

Derivation: G”(1) = E(X(X-1)) = E(X²) – E(X), so E(X²) = G”(1) + G'(1). Then Var(X) = E(X²) – [E(X)]² = G”(1) + G'(1) – [G'(1)]².
推导：G”(1) = E(X(X-1)) = E(X²) – E(X)，因此 E(X²) = G”(1) + G'(1)，然后 Var(X) = E(X²) – [E(X)]² = G”(1) + G'(1) – [G'(1)]²。

Property 4: Sum of Independent Random Variables
If X and Y are independent discrete random variables with PGFs G_X(t) and G_Y(t), then the PGF of Z = X + Y is:
如果 X 和 Y 是独立的离散随机变量，其 PGF 分别为 G_X(t) 和 G_Y(t)，则 Z = X + Y 的 PGF 为：
G_Z(t) = G_X(t) · G_Y(t)

This is because E(t^(X+Y)) = E(t^X · t^Y) = E(t^X) · E(t^Y) (by independence).
这是因为 E(t^(X+Y)) = E(t^X · t^Y) = E(t^X) · E(t^Y)（由独立性）。

3. PGF of the Binomial Distribution

二项分布的 PGF

X ~ B(n, p)

The probability mass function is P(X = x) = ⁿCₓ · pˣ · (1-p)ⁿ⁻ˣ, where x = 0, 1, 2, …, n.
概率质量函数为 P(X = x) = ⁿCₓ · pˣ · (1-p)ⁿ⁻ˣ，其中 x = 0, 1, 2, …, n。

The PGF is derived as follows:
PGF 推导如下：

G(t) = Σ (from x=0 to n) ⁿCₓ · pˣ · (1-p)ⁿ⁻ˣ · tˣ
= Σ ⁿCₓ · (pt)ˣ · (1-p)ⁿ⁻ˣ
= (1 – p + pt)ⁿ

Let q = 1 – p, we get the elegant form:
令 q = 1 – p，得到简洁形式：

G(t) = (q + pt)ⁿ

This is a beautifully compact expression. Let’s verify using the binomial theorem — the expansion (q + pt)ⁿ = Σ ⁿCₓ · qⁿ⁻ˣ · (pt)ˣ = Σ ⁿCₓ · pˣ · qⁿ⁻ˣ · tˣ, and the coefficient of tˣ is exactly P(X = x). ✓
这是一个非常紧凑的表达式。我们用二项式定理验证——展开式 (q + pt)ⁿ = Σ ⁿCₓ · qⁿ⁻ˣ · (pt)ˣ = Σ ⁿCₓ · pˣ · qⁿ⁻ˣ · tˣ，tˣ 的系数恰好是 P(X = x)。✓

Verify G(1) = 1: (q + p)ⁿ = 1ⁿ = 1 ✓

Finding E(X): G'(t) = n · (q + pt)ⁿ⁻¹ · p. So G'(1) = n · (q + p)ⁿ⁻¹ · p = np ✓
求 E(X)：G'(t) = n · (q + pt)ⁿ⁻¹ · p。所以 G'(1) = n · (q + p)ⁿ⁻¹ · p = np ✓

Finding Var(X): G”(t) = n(n-1) · (q + pt)ⁿ⁻² · p². So G”(1) = n(n-1)p².
Var(X) = G”(1) + G'(1) – [G'(1)]² = n(n-1)p² + np – n²p² = n²p² – np² + np – n²p² = np – np² = np(1-p) = npq ✓
求 Var(X)：G”(t) = n(n-1) · (q + pt)ⁿ⁻² · p²。所以 G”(1) = n(n-1)p²。
Var(X) = G”(1) + G'(1) – [G'(1)]² = n(n-1)p² + np – n²p² = n²p² – np² + np – n²p² = np – np² = np(1-p) = npq ✓

4. PGF of the Poisson Distribution

泊松分布的 PGF

X ~ Po(λ)

The probability mass function is P(X = x) = e^(-λ) · λˣ / x!, where x = 0, 1, 2, …
概率质量函数为 P(X = x) = e^(-λ) · λˣ / x!，其中 x = 0, 1, 2, …

The PGF derivation:
PGF 推导：

G(t) = Σ (from x=0 to ∞) [e^(-λ) · λˣ / x!] · tˣ
= e^(-λ) · Σ (λt)ˣ / x!
= e^(-λ) · e^(λt)

Recall that the Taylor series expansion of e^y is Σ yˣ / x! from x=0 to ∞. So:
回顾 e^y 的泰勒级数展开为 Σ yˣ / x!（x 从 0 到 ∞）。因此：

G(t) = e^(λ(t – 1))

Verify G(1) = 1: e^(λ(1-1)) = e⁰ = 1 ✓

Finding E(X): G'(t) = e^(λ(t-1)) · λ. So G'(1) = e⁰ · λ = λ ✓
求 E(X)：G'(t) = e^(λ(t-1)) · λ。所以 G'(1) = e⁰ · λ = λ ✓

Finding Var(X): G”(t) = e^(λ(t-1)) · λ². So G”(1) = λ².
Var(X) = λ² + λ – λ² = λ ✓
求 Var(X)：G”(t) = e^(λ(t-1)) · λ²。所以 G”(1) = λ²。
Var(X) = λ² + λ – λ² = λ ✓

The Poisson distribution is remarkable in that its mean equals its variance, both being λ. This property is unique among common distributions and serves as a diagnostic check.
泊松分布的一个显著特征是均值等于方差，均为 λ。这一性质在常见分布中是独一无二的，可作为诊断检查。

Key Insight: The PGF of Poisson also reveals an additive property — if X ~ Po(λ₁) and Y ~ Po(λ₂) are independent, then the PGF of X+Y is:
关键洞察：泊松分布的 PGF 也揭示了可加性——如果 X ~ Po(λ₁) 和 Y ~ Po(λ₂) 独立，则 X+Y 的 PGF 为：
G_{X+Y}(t) = e^(λ₁(t-1)) · e^(λ₂(t-1)) = e^((λ₁+λ₂)(t-1))
This is the PGF of a Po(λ₁+λ₂) distribution — a concise proof that the sum of independent Poisson variables is also Poisson!
这是 Po(λ₁+λ₂) 分布的 PGF——简洁地证明了独立泊松变量之和仍为泊松分布！

5. PGF of the Geometric Distribution

几何分布的 PGF

X ~ Geo(p)

The probability mass function is P(X = x) = p · (1-p)^(x-1), where x = 1, 2, 3, … (counting the number of trials until the first success).
概率质量函数为 P(X = x) = p · (1-p)^(x-1)，其中 x = 1, 2, 3, …（计数直到第一次成功的试验次数）。

Let q = 1 – p. The PGF is:
令 q = 1 – p。PGF 为：

G(t) = Σ (from x=1 to ∞) p · q^(x-1) · tˣ
= pt · Σ (from x=1 to ∞) (qt)^(x-1)
= pt · Σ (from k=0 to ∞) (qt)^k
= pt / (1 – qt) [using the geometric series formula, for |qt| < 1]
[使用几何级数公式，当 |qt| < 1 时]

G(t) = pt / (1 – qt)

Alternative form: Some textbooks define geometric distribution counting failures before the first success, giving PGF = p / (1 – qt). For Edexcel FS1, the standard form above is used.
另一种形式：有些教科书定义几何分布为计数第一次成功前的失败次数，得到 PGF = p / (1 – qt)。对于 Edexcel FS1，使用上述标准形式。

Verify G(1) = 1: p / (1 – q) = p / p = 1 ✓

Finding E(X):
G'(t) = p · (1 – qt)^(-1)
Using quotient/product rule: G'(t) = p · [(1-qt)·1 – t·(-q)] / (1-qt)² = p / (1-qt)²
So G'(1) = p / (1-q)² = p / p² = 1/p ✓

Finding Var(X):
G”(t) = 2pq / (1-qt)³
So G”(1) = 2pq / p³ = 2q / p²
Var(X) = G”(1) + G'(1) – [G'(1)]² = 2q/p² + 1/p – 1/p² = (2q + p – 1) / p² = (2(1-p) + p – 1) / p² = (2 – 2p + p – 1) / p² = (1 – p) / p² = q/p² ✓

6. PGF of the Negative Binomial Distribution

负二项分布的 PGF

X ~ NB(r, p)

The negative binomial distribution counts the number of trials needed to achieve r successes, where each trial has success probability p. The PMF is:
负二项分布计数实现 r 次成功所需的试验次数，每次试验成功概率为 p。PMF 为：
P(X = x) = ^(x-1)C_(r-1) · p^r · q^(x-r), for x = r, r+1, r+2, …

The PGF takes a remarkably simple form:
PGF 具有一个非常简单的形式：

G(t) = [pt / (1 – qt)]^r

This makes intuitive sense: the negative binomial is like the sum of r independent Geometric(p) random variables. Since the PGF of a sum of independent variables is the product of their individual PGFs, we get [pt/(1-qt)]^r.
这在直观上很有意义：负二项分布就像是 r 个独立 Geo(p) 随机变量的和。由于独立变量之和的 PGF 是各自 PGF 的乘积，我们得到 [pt/(1-qt)]^r。

E(X) = r/p, Var(X) = rq/p² — consistent with r times the geometric mean and variance.
E(X) = r/p, Var(X) = rq/p² — 与 r 倍的几何分布均值和方差一致。

7. PGF of the Discrete Uniform Distribution

离散均匀分布的 PGF

X ~ Uniform{1, 2, …, n}

P(X = x) = 1/n for x = 1, 2, …, n.

The PGF is:
PGF 为：

G(t) = Σ (from x=1 to n) (1/n) · tˣ = (1/n) · Σ tˣ
= (1/n) · t(1 – tⁿ) / (1 – t) [geometric series sum]
[等比数列求和]

G(t) = t(1 – tⁿ) / [n(1 – t)]

E(X) = (n+1)/2, a classic result easily verified: G'(1) requires careful evaluation using L’Hôpital’s rule or expansion.
E(X) = (n+1)/2，一个经典结果，可轻松验证：G'(1) 需要使用洛必达法则或展开式仔细计算。

8. Working with Sums of Independent Variables

处理独立变量之和

One of the most elegant applications of PGFs is finding the distribution of sums:

PGF 最优雅的应用之一是求和的分布：

Example: If X ~ B(n₁, p) and Y ~ B(n₂, p) are independent (same p), find the distribution of Z = X + Y.
例子：如果 X ~ B(n₁, p) 和 Y ~ B(n₂, p) 独立（相同 p），求 Z = X + Y 的分布。

G_Z(t) = G_X(t) · G_Y(t) = (q + pt)^(n₁) · (q + pt)^(n₂) = (q + pt)^(n₁+n₂)

This is the PGF of B(n₁+n₂, p)! So Z ~ B(n₁+n₂, p). Much simpler than convolution.
这是 B(n₁+n₂, p) 的 PGF！所以 Z ~ B(n₁+n₂, p)。比卷积方法简单得多。

Example: If X ~ Po(λ₁) and Y ~ Po(λ₂) are independent, find the distribution of Z = X + Y.
例子：如果 X ~ Po(λ₁) 和 Y ~ Po(λ₂) 独立，求 Z = X + Y 的分布。

G_Z(t) = e^(λ₁(t-1)) · e^(λ₂(t-1)) = e^((λ₁+λ₂)(t-1))

This is the PGF of Po(λ₁+λ₂)! So Z ~ Po(λ₁+λ₂).
这是 Po(λ₁+λ₂) 的 PGF！所以 Z ~ Po(λ₁+λ₂)。

This additive property (closure under convolution) holds for Binomial (same p), Poisson, Negative Binomial (same p), and Normal distributions — PGFs provide an elegant proof for the discrete cases.
这种可加性（在卷积下封闭）对二项分布（相同 p）、泊松分布、负二项分布（相同 p）和正态分布都成立——PGF 为离散情况提供了优雅的证明。

9. Summary Table of Standard PGFs

标准分布 PGF 汇总表

Distribution 分布	Notation 记号	PGF G(t)	E(X)	Var(X)
Binomial 二项	B(n, p)	(q + pt)ⁿ	np	npq
Poisson 泊松	Po(λ)	e^(λ(t-1))	λ	λ
Geometric 几何	Geo(p)	pt/(1-qt)	1/p	q/p²
Neg. Binomial 负二项	NB(r, p)	[pt/(1-qt)]^r	r/p	rq/p²
Uniform 均匀	U(1,n)	t(1-tⁿ)/[n(1-t)]	(n+1)/2	(n²-1)/12

10. Exam Tips for Edexcel FS1

Edexcel FS1 考试技巧

Always verify G(1) = 1 before proceeding — it’s a quick sanity check worth one mark in many questions.
在进行下一步前始终验证 G(1) = 1——这是一个快速的合理性检查，在许多题目中值一分。
Memorise the four standard PGFs (Binomial, Poisson, Geometric, Neg. Binomial). The Uniform PGF is less common but derivable.
记住四个标准 PGF（二项、泊松、几何、负二项）。均匀分布的 PGF 不太常见但可推导。
For sums of independent variables, the product rule G_{X+Y}(t) = G_X(t)·G_Y(t) is your most powerful weapon — always check for independence first.
对于独立变量之和，乘积法则 G_{X+Y}(t) = G_X(t)·G_Y(t) 是你最强大的武器——始终首先检查独立性。
When finding E(X) from G'(1), you may need the product rule, chain rule, or quotient rule. Practice differentiating each standard form until it becomes automatic.
当从 G'(1) 求 E(X) 时，你可能需要乘积法则、链式法则或商法则。练习对每种标准形式进行微分，直到熟练自如。
The second derivative formula Var(X) = G”(1) + G'(1) – [G'(1)]² is given in the formula booklet, but knowing it saves time.
二阶导数公式 Var(X) = G”(1) + G'(1) – [G'(1)]² 在公式手册中提供，但记住它可以节省时间。
For “given that” questions: If you’re told G(t) and asked to find a probability distribution, expand G(t) as a power series in t — the coefficient of tˣ is P(X = x).
对于”已知”类问题：如果给出 G(t) 并要求求概率分布，将 G(t) 展开为 t 的幂级数——tˣ 的系数即为 P(X = x)。

Conclusion

结论

Probability Generating Functions transform the task of working with probability distributions from summation to differentiation — a significant simplification. The standard distributions each have a characteristic PGF form that encodes their essential properties. By mastering these standard forms and the key properties (G(1)=1, G'(1)=E(X), product rule for sums), you gain a powerful toolkit for tackling the most challenging FS1 questions. Practice deriving each PGF from first principles — not just memorising them — and you’ll develop the fluency needed to handle any problem Edexcel can throw at you.

概率生成函数将与概率分布相关的工作从求和转化为微分——这是一个显著的简化。每个标准分布都有其特征性的 PGF 形式，编码了其基本性质。通过掌握这些标准形式和关键性质（G(1)=1，G'(1)=E(X)，求和的乘积规则），你就能获得应对最具挑战性的 FS1 问题的强大工具箱。练习从基本原理推导每个 PGF——而不仅仅是记忆它们——你将培养出应对 Edexcel 可能出的任何问题所需的熟练度。

This article covers the complete PGF syllabus content for Edexcel A-Level Further Mathematics, Statistics 1. For more resources and past paper practice, visit our study hub.

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