📚 Using Probability Generating Functions to Find the Distribution of the Sum of Discrete Random Variables | 使用概率生成函数求离散随机变量和的分布
Probability generating functions (PGFs) offer a remarkably elegant method for determining the distribution of the sum of independent discrete random variables. Instead of wrestling with convolutions or tedious arithmetic, you simply multiply their PGFs and identify the resulting form. This article unpacks the theory, walks through classic examples, and highlights exam‑ready techniques tailored to the IB Cambridge Mathematics syllabus.
概率生成函数 (PGF) 为求独立离散随机变量之和的分布提供了一种极为简洁的方法。无需进行复杂的卷积计算或繁琐的算术，只需将它们的 PGF 相乘并识别结果的形式即可。本文解析相关理论，剖析经典例题，并突出适用于 IB 剑桥数学考试大纲的应试技巧。
1. Introduction to Probability Generating Functions (PGFs) | 概率生成函数简介
A probability generating function is defined for a discrete random variable X taking values in {0, 1, 2, …}. It is given by GX(t) = E(tX) = Σx=0∞ tx P(X = x), valid for |t| fficients directly recover the probability mass function: P(X = x) is the coefficient of tx. Furthermore, the PGF uniquely identifies the distribution; if two random variables have the same PGF, they share the same distribution.
概率生成函数定义在取值为 {0, 1, 2, …} 的离散随机变量上。其形式为 GX(t) = E(tX) = Σx=0∞ tx P(X = x)，对 |t| ≤ 1 有效。级数的系数直接还原概率质量函数：P(X = x) 就是 tx 的系数。此外，PGF 唯一确定分布；若两个随机变量的 PGF 相同，则它们的分布相同。
2. The Multiplication Property for Independent Random Variables | 独立随机变量的乘积性质
If X and Y are independent discrete random variables, the PGF of their sum is the product of their individual PGFs: GX+Y(t) = GX(t) × GY(t). Proof: GX+Y(t) = E(tX+Y) = E(tX tY) = E(tX) E(tY) because independence ensures the expectation of the product equals the product of expectations. This simple property transforms a potentially complex convolution problem into a manageable algebraic one.
若 X 与 Y 为独立的离散随机变量，则它们之和的 PGF 等于各自 PGF 的乘积：GX+Y(t) = GX(t) × GY(t)。证明：GX+Y(t) = E(tX+Y) = E(tX tY) = E(tX) E(tY)，因为独立性保证了乘积的期望等于期望的乘积。这一简单性质将困难的卷积问题转化成了易于处理的代数问题。
3. Why Sums? Formulating the PGF of a Sum | 为何聚焦于和？构造和的 PGF
When confronted with S = X + Y, we often need the distribution of S. Without PGFs, we would compute P(S = k) = Σi P(X = i) P(Y = k−i). With PGFs we compute GS(t) = GX(t) GY(t), simplify the expression, and then recognise it as the PGF of a known family. This strategy works for any finite sum of independent variables: GX₁+⋯+Xₙ(t) = Πi=1n GXᵢ(t). The key lies in memorising the standard PGF forms.
当遇到 S = X + Y 时，我们往往需要 S 的分布。若没有 PGF，需计算 P(S = k) = Σi P(X = i) P(Y = k−i)。而有了 PGF，我们只需算出 GS(t) = GX(t) GY(t)，化简表达式，然后识别其属于哪个已知分布族。该策略适用于任意有限个独立变量之和：GX₁+⋯+Xₙ(t) = Πi=1n GXᵢ(t)。关键在于熟记标准 PGF 形式。
4. Example 1: Sum of Independent Poisson Random Variables | 例 1：独立泊松随机变量之和
Let X ~ Po(λ) and Y ~ Po(μ) be independent. The PGF of a Poisson random variable is G(t) = eλ(t−1). Therefore, GX+Y(t) = eλ(t−1) · eμ(t−1) = e(λ+μ)(t−1). This is exactly the PGF of a Poisson random variable with parameter λ+μ. Hence, X + Y ~ Po(λ+μ). The result extends naturally to the sum of n independent Poisson variables: it remains Poisson with parameter equal to the sum of the individual rates.
设 X ~ Po(λ) 与 Y ~ Po(μ) 相互独立。泊松随机变量的 PGF 为 G(t) = eλ(t−1)。于是 GX+Y(t) = eλ(t−1) · eμ(t−1) = e(λ+μ)(t−1)。这正是参数为 λ+μ 的泊松分布 PGF。因此，X + Y ~ Po(λ+μ)。此结果自然推广至 n 个独立泊松变量之和：其和仍为泊松分布，参数为各速率之和。
5. Example 2: Sum of Independent Binomial Random Variables (Same p) | 例 2：独立二项分布之和（相同 p）
Consider X ~ B(n, p) and Y ~ with the same success probability p. The PGF of a binomial distribution is G(t) = (q + pt)n, where q = 1−p. Multiplying gives GX+Y(t) = (q + pt)n (q + pt)m = (q + pt)n+m. This matches the PGF of a binomial distribution with parameters n+m and p. Thus, X + Y ~ B(n+m, p). Caution: the probability p must be identical for both variables. If p differs, the sum is not binomial and the PGF product will not simplify to a standard single‑parameter form.
考虑 X ~ B(n, p) 与 Y ~ B(m, p)，成功概率 p 相同。二项分布的 PGF 为 G(t) = (q + pt)n，其中 q = 1−p。相乘得 GX+Y(t) = (q + pt)n (q + pt)m = (q + pt)n+m。这与参数为 n+m 和 p 的二项分布 PGF 一致。因此，X + Y ~ B(n+m, p)。注意：两个变量的 p 必须相同。若 p 不同，和将不再是二项分布，PGF 乘积也无法简化为标准的单参数形式。
6. Example 3: Sum of Independent Geometric Random Variables | 例 3：独立几何分布之和
ution counts the number of trials up to and including the first success (support {1, 2, 3, … PGF of such a geometric distribution is G(t) = pt / (1 − qt), with q = 1−p. Multiplying, GX+Y(t) = [pt / (1 − qt)]2. This is not another geometric PGF; instead it corresponds to a negative binomial distribution that counts the number of trials until the second success. In general, a negative binomial NB(r, p) has PGF [pt / (1 − qt)]r. Therefore, X + Y ~ NB(2, p). The sum of r independent geometric variables gives NB(r, p).
设 X 与 Y 为独立的几何随机变量，成功概率均为 p，且定义为直到首次成功所需的试验次数（支持集 {1, 2, 3, …}）。此几何分布的 PGF 为 G(t) = pt / (1 − qt)，q = 1−p。相乘得 GX+Y(t) = [pt / (1 − qt)]2。这并非另一个几何分布，而是对应于计数直到第 2 次成功的负二项分布。一般而言，负二项分布 NB(r, p) 的 PGF 为 [pt / (1 − qt)]r。因此，X + Y ~ NB(2, p)。r 个独立几何变量之和服从 NB(r, p)。
7. Using PGFs to Derive the Distribution: Negative Binomial Case | 利用 PGF 推导分布：负二项情形
Starting from the product [pt/(1−qt)]2, we can expand the PGF to confirm the probability mass function. Write G(t) = p2 t2 (1−qt)−2. Using the binomial series (1−u)−k = Σj=0∞ C(k+j−1, j) uj, we set u = qt and k = 2. Then G(t) = p2 t2 Σj=0∞ C(j+1, j) (qt)j = Σj=0∞ (j+1) p2 qj tj+2. Let x = j+2 ≥ 2, then coefficient of tx is (x−1) p2 qx−2 s explicit extraction demonstrates how PGFs encode the entire distribution; however, in exams, recognition of the closed form is usually sufficient.
从乘积 [pt/(1−qt)]2 出发，我们可以将其展开以验证概率质量函数。写 G(t) = p2 t2 (1−qt)−2。利用二项级数 (1−u)−k = Σj=0∞ C(k+j−1, j) uj，设 u = qt 且 k = 2。则 G(t) = p2 t2 Σj=0∞ C(j+1, j) (qt)j = Σj=0∞ (j+1) p2 qj tj+2。令 x = j+2 ≥ 2，则 tx 的系数为 (x−1) p2 qx−2，这正是 P(NB(2,p) = x)。这一显式提取展示了 PGF 如何编码完整分布；然而考试中通常只需识别出闭形即可。
8. Extension: Sum of a Random Number of Random Variables | 扩展：随机个随机变量之和
A more advanced application concerns a sum where the number of terms is itself a random variable. Let N be a non‑negative integer‑valued random variable, independent of an i.i.d. sequence X1, X2, … with common PGF GX(t). Define S = Σi=1N Xi. Then the PGF of S is given by the compound formula GS(t) = GN(GX(t)). For instance, if N ~ Po(λ) and each Xi ~ Po(μ), then S is compound Poisson, and its PGF becomes exp(λ(eμ(t−1) − 1)). While this topic may stretch beyond the core syllabus, exploring it deepens understanding of the versatility of PGFs.
一个更高级的应用涉及项数本身为随机变量的情形。设 N 为非负整数值随机变量，与独立同分布序列 X1, X2, … 独立，它们有共同的 PGF GX(t)。定义 S = Σi=1N Xi。则 S 的 PGF 由复合公式给出：GS(t) = GN(GX(t))。例如，若 N ~ Po(λ) 且每个 Xi ~ Po(μ)，则 S 为复合泊松，其 PGF 变为 exp(λ(eμ(t−1) − 1))。虽然本主题可能超出核心考纲，但探索它能加深对 PGF 多功能性的理解。
9. Common Pitfalls and the Independence Condition | 常见误区与独立性条件
Independence is non‑negotiable. The factorisation GX+Y(t) = GX(t) GY dom variables share the same support convention (e.g., geometric counting from 1). Mistakes often arise when mixing parameterisations, such as using Geo(p) starting at 0 versus starting at 1. Finally, always check the domain |t| ≤ 1 to ensure convergence, especially when differentiating at t = 1 to find moments.
独立性是必须严格遵守的条件。因子分解 GX+Y(t) = GX(t) GY(t) 成立当且仅当 X 与 Y 独立。对不独立的变量应用乘积规则将得到错误的 PGF。此外，需确认随机变量使用相同的支持集约定（例如几何分布从 1 开始计数）。混淆参数化形式时常出错，如使用从 0 开始的 Geo(p) 与从 1 开始的混用。最后，务必检查 |t| ≤ 1 的定义域以确保收敛，特别是在 t = 1 处求导求矩时。
10. Summary and Exam Tips | 总结与考试技巧
Mastering PGF sum problems boils down to a three‑step routine: (1) write the PGFs of the independent variables, (2) multiply them together, (3) simplify and match to a known PGF. Memorise the standard PGF table — Poisson eλ(t−1), binomial (q+pt)n, geometric pt/(1−qt), negative binomial [pt/(1−qt)]r. In IB exams, you may be asked to prove a sum distribution or to derive the PGF of a sum from first principles. Show your working clearly, emphasising where independence is invoked. With practice, PGFs become a swift and reliable tool for tackling distribution of sums.
掌握 PGF 求和的题目可归结为三步流程：（1）写出独立变量的 PGF，（2）将它们相乘，（3）化简并匹配到已知的 PGF 形式。熟记标准 PGF 表——泊松 eλ(t−1)、二项 (q+pt)n、几何 pt/(1−qt)、负二项 [pt/(1−qt)]r。在 IB 考试中，你可能需要证明和的分布，或从基本原理推导和的 PGF。清晰展示步骤，重点说明何处用到了独立性。经过练习，PGF 将成为处理随机变量和分布的一个快捷且可靠的工具。
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