Probability Generating Functions of Standard Distributions | 标准分布的概率生成函数

📚 Probability Generating Functions of Standard Distributions | 标准分布的概率生成函数

For Edexcel A-Level Further Mathematics, probability generating functions (PGFs) are a compact way to encode the entire probability distribution of a non-negative integer-valued random variable. Recognising the PGFs of standard distributions allows you to find means, variances, probabilities, and distributions of sums efficiently.

在 Edexcel A-Level 进阶数学中，概率生成函数（PGF）是编码非负整数随机变量整个概率分布的一种简洁方式。识别标准分布的概率生成函数，可以帮助你高效地求均值、方差、概率以及随机变量之和的分布。

1. What is a probability generating function? | 什么是概率生成函数

For a discrete random variable X taking values in {0, 1, 2, …}, the probability generating function is defined by G_X(t) = E(t^X) = Σ P(X = x) t^x, where the sum is taken over all possible values of X.

对于取值在 {0, 1, 2, …} 的离散随机变量 X，概率生成函数定义为 G_X(t) = E(t^X) = Σ P(X = x) t^x，其中求和取遍 X 的所有可能取值。

The PGF is a power series whose coefficients are exactly the probabilities p_x = P(X = x), so it contains all information about the distribution. It is usually considered for values of t where the series converges, often |t| ≤ 1.

概率生成函数是一个幂级数，其系数恰好是概率 p_x = P(X = x)，因此它包含了分布的全部信息。通常考虑使级数收敛的 t 值，常见范围为 |t| ≤ 1。

G_X(t) = p₀ + p₁t + p₂t² + p₃t³ + … = Σ P(X = x) tˣ

---

2. Core properties of the PGF | 概率生成函数的核心性质

Since the probabilities sum to 1, a PGF must satisfy G_X(1) = 1. This is a quick sanity check for any PGF you derive.

因为概率之和为 1，所以 PGF 必须满足 G_X(1) = 1。这是检验你所推导的 PGF 是否正确的一个快速方法。

Differentiating the PGF and substituting t = 1 gives moments. The first derivative gives the mean, and the second derivative gives the second factorial moment.

对 PGF 求导并代入 t = 1 可以得到矩。一阶导数给出均值，二阶导数给出二阶阶乘矩。

G′_X(1) = E(X) and G″_X(1) = E[X(X − 1)]

Therefore, the variance can be calculated using Var(X) = G″_X(1) + G′_X(1) − [G′_X(1)]².

因此，方差可以用公式 Var(X) = G″_X(1) + G′_X(1) − [G′_X(1)]² 计算。

More generally, the kth derivative at t = 0 is related to the probability P(X = k) by P(X = k) = G_X^(k)(0) / k!, where G_X^(k) denotes the kth derivative.

更一般地，t = 0 处的 k 阶导数与概率 P(X = k) 相关，公式为 P(X = k) = G_X^(k)(0) / k!，其中 G_X^(k) 表示 k 阶导数。

---

3. Bernoulli distribution | 伯努利分布

If X follows a Bernoulli distribution with parameter p, then X can only take the values 0 and 1, with P(X = 1) = p and P(X = 0) = q = 1 − p.

如果 X 服从参数为 p 的伯努利分布，那么 X 只能取值 0 和 1，其中 P(X = 1) = p，P(X = 0) = q = 1 − p。

The PGF is simply the linear expression G_X(t) = q + pt. This is the building block for the binomial PGF.

其 PGF 就是线性表达式 G_X(t) = q + pt。这是二项分布 PGF 的基础。

X ~ B(1, p): G_X(t) = q + pt, E(X) = p, Var(X) = pq

For example, if X ~ B(1, 0.3), then G_X(t) = 0.7 + 0.3t, confirming that the coefficient of t is P(X = 1).

例如，若 X ~ B(1, 0.3)，则 G_X(t) = 0.7 + 0.3t，这验证了 t 的系数就是 P(X = 1)。

---

4. Binomial distribution | 二项分布

If X ~ B(n, p), then X counts the number of successes in n independent Bernoulli trials, each with success probability p. Its probability mass function is P(X = x) = C(n, x) p^x q^(n − x) for x = 0, 1, …, n.

如果 X ~ B(n, p)，则 X 表示在 n 次独立伯努利试验中成功的次数，每次成功概率为 p。其概率质量函数为 P(X = x) = C(n, x) p^x q^(n − x)，其中 x = 0, 1, …, n。

Since a binomial random variable is the sum of n independent Bernoulli random variables, its PGF is the nth power of the Bernoulli PGF:

由于二项随机变量是 n 个独立伯努利随机变量之和，其 PGF 是伯努利 PGF 的 n 次方：

X ~ B(n, p): G_X(t) = (q + pt)ⁿ

Alternatively, this follows directly from the binomial expansion of (q + pt)ⁿ. Differentiating and setting t = 1 gives E(X) = np and Var(X) = npq.

这也可以直接由 (q + pt)ⁿ 的二项展开得到。求导并令 t = 1 可得 E(X) = np 和 Var(X) = npq。

---

5. Poisson distribution | 泊松分布

If X ~ Po(λ), then P(X = x) = e^(−λ) λ^x / x! for x = 0, 1, 2, … . The PGF is derived by summing the exponential series:

如果 X ~ Po(λ)，则 P(X = x) = e^(−λ) λ^x / x!，其中 x = 0, 1, 2, …。其 PGF 通过指数级数求和得到：

G_X(t) = E(t^X) = e^(−λ) Σ (λt)^x / x! = e^(λ(t − 1))

Thus the Poisson PGF has the distinctive form e^(λ(t − 1)). Differentiating gives G′_X(t) = λe^(λ(t − 1)) and G″_X(t) = λ²e^(λ(t − 1)), so E(X) = λ and Var(X) = λ.

因此泊松分布的 PGF 具有独特形式 e^(λ(t − 1))。求导得到 G′_X(t) = λe^(λ(t − 1)) 和 G″_X(t) = λ²e^(λ(t − 1))，所以 E(X) = λ，Var(X) = λ。

This is why the Poisson PGF is often immediately recognised in exam questions when an exponential of a linear function appears.

这就是为什么在考试中，一旦出现线性函数的指数形式，通常可以立即识别为泊松分布的 PGF。

---

6. Geometric distribution | 几何分布

The geometric distribution has two common versions. The version used in most Edexcel questions counts the number of trials up to and including the first success: P(X = x) = q^(x − 1) p for x = 1, 2, 3, … .

几何分布有两种常见形式。大多数 Edexcel 题目使用的形式是：记录直到并包括第一次成功所需的试验次数，即 P(X = x) = q^(x − 1) p，其中 x = 1, 2, 3, …。

For this version, the PGF is G_X(t) = pt / (1 − qt), valid for |qt| < 1. It is obtained by summing a geometric series:

对于这种形式，PGF 为 G_X(t) = pt / (1 − qt)，在 |qt| < 1 时成立。它通过几何级数求和得到：

G_X(t) = pt + pqt² + pq²t³ + … = pt / (1 − qt)

Differentiating gives E(X) = 1/p and Var(X) = q / p². If instead Y counts failures before the first success, so Y ∈ {0, 1, 2, …}, then G_Y(t) = p / (1 − qt). Always state which geometric convention you are using.

求导可得 E(X) = 1/p 和 Var(X) = q / p²。如果改用 Y 表示第一次成功前的失败次数，则 Y ∈ {0, 1, 2, …}，且 G_Y(t) = p / (1 − qt)。解题时务必说明你使用的几何分布形式。

---

7. Negative binomial distribution | 负二项分布

The negative binomial distribution is a natural generalisation of the geometric distribution. If X counts the number of trials needed to obtain the rth success, then X ~ NB(r, p), with P(X = x) = C(x − 1, r − 1) p^r q^(x − r) for x = r, r + 1, … .

负二项分布是几何分布的自然推广。如果 X 表示获得第 r 次成功所需的试验次数，则 X ~ NB(r, p)，且 P(X = x) = C(x − 1, r − 1) p^r q^(x − r)，其中 x = r, r + 1, …。

Because X can be expressed as the sum of r independent geometric random variables, its PGF is the rth power of the geometric PGF:

由于 X 可以表示为 r 个独立几何随机变量之和，其 PGF 是几何分布 PGF 的 r 次方：

X ~ NB(r, p): G_X(t) = [pt / (1 − qt)]^r

From this, E(X) = r/p and Var(X) = rq / p². If the alternative version counts failures before the rth success, the PGF is [p / (1 − qt)]^r.

由此可得 E(X) = r/p 和 Var(X) = rq / p²。如果采用另一种形式，即记录第 r 次成功前的失败次数，则其 PGF 为 [p / (1 − qt)]^r。

---

8. Discrete uniform distribution | 离散均匀分布

If X is uniformly distributed on {1, 2, …, n}, then P(X = x) = 1/n for each x in that set. Its PGF is a finite geometric series:

如果 X 在 {1, 2, …, n} 上服从均匀分布，则对于该集合中的每个 x，P(X = x) = 1/n。其 PGF 是一个有限几何级数：

G_X(t) = (t + t² + … + tⁿ) / n = t(1 − tⁿ) / [n(1 − t)] for t ≠ 1

At t = 1, the value is G_X(1) = 1. For this distribution, E(X) = (n + 1)/2 and Var(X) = (n² − 1)/12.

当 t = 1 时，G_X(1) = 1。对于该分布，E(X) = (n + 1)/2，Var(X) = (n² − 1)/12。

If the uniform support is {0, 1, …, n − 1}, then the PGF becomes G_X(t) = (1 − tⁿ) / [n(1 − t)]. This small difference is important when identifying distributions from a given PGF.

如果均匀分布的支持集是 {0, 1, …, n − 1}，则 PGF 变为 G_X(t) = (1 − tⁿ) / [n(1 − t)]。在从给定 PGF 识别分布时，这一细微差别很重要。

---

9. Summary table and recognition | 汇总表与识别

The following table summarises the standard PGFs you should memorise for Edexcel Further Maths. Recognising these forms quickly is often the key to solving exam problems.

下表总结了 Edexcel 进阶数学中需要记住的标准 PGF。快速识别这些形式通常是解决考试题目的关键。

Distribution	PGF G_X(t)	Mean	Variance
Bernoulli B(1, p)	q + pt	p	pq
Binomial B(n, p)	(q + pt)ⁿ	np	npq
Poisson Po(λ)	e^(λ(t − 1))	λ	λ
Geometric Geo(p) on {1, 2, …}	pt / (1 − qt)	1/p	q / p²
Negative binomial NB(r, p) on {r, r + 1, …}	[pt / (1 − qt)]^r	r/p	rq / p²
Discrete uniform on {1, …, n}	t(1 − tⁿ) / [n(1 − t)]	(n + 1)/2	(n² − 1)/12

When a PGF is given, compare its structure with these standard forms before attempting long differentiation. For example, 0.2 + 0.8t is Bernoulli, while (0.4 + 0.6t)⁵ is binomial with n = 5 and p = 0.6.

当给定一个 PGF 时，先将其结构与这些标准形式进行比较，再进行冗长的求导。例如，0.2 + 0.8t 是伯努利分布，而 (0.4 + 0.6t)⁵ 是 n = 5、p = 0.6 的二项分布。

---

10. Using PGFs to find moments | 使用 PGF 求矩

Differentiating a standard PGF is usually easier than summing the original probability mass function. For example, if X ~ Po(3), then G_X(t) = e^(3(t − 1)).

对标准 PGF 求导通常比直接对原始概率质量函数求和更容易。例如，如果 X ~ Po(3)，则 G_X(t) = e^(3(t − 1))。

Then G′

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）