📚 Probability Generating Functions of Common Distributions | 常见分布的概率生成函数
The probability generating function (PGF) is a powerful tool in statistics, especially for discrete random variables. It encodes the entire probability mass function into a single power series, which makes calculations of moments and sums of independent variables much simpler.
概率生成函数（PGF）是统计学中一个强有力的工具，尤其适用于离散型随机变量。它将整个概率质量函数浓缩为一个幂级数，从而使得矩的计算以及独立变量之和的处理大为简化。
1. Definition of the Probability Generating Function | 概率生成函数的定义
Let \(X\) be a discrete random variable taking non-negative integer values 0, 1, 2, … The probability generating function of \(X\) is defined as
设 \(X\) 是一个取非负整数值 0, 1, 2, … 的离散型随机变量。\(X\) 的概率生成函数定义为
GX(t) = E(tX) = Σ p(x) tx
where \(p(x) = P(X = x)\) and the sum is taken over all possible values of \(x\). The series converges for \(|t| \le 1\).
其中 \(p(x) = P(X = x)\)，求和遍及 \(x\) 的所有可能取值。该级数在 \(|t| \le 1\) 时收敛。
For example, if \(X\) has probability mass function \(p(0)=0.2\), \(p(1)=0.5\), \(p(2)=0.3\), then \(G_X(t)=0.2+0.5t+0.3t^2\).
例如，若 \(X\) 的概率质量函数为 \(p(0)=0.2\), \(p(1)=0.5\), \(p(2)=0.3\)，则 \(G_X(t)=0.2+0.5t+0.3t^2\)。
2. Key Properties of the PGF | 概率生成函数的基本性质
The PGF has several essential properties that make it useful in solving problems.
概率生成函数具有若干重要性质，使其在解题中非常有用。
Property 1: \(G_X(1) = 1\), because the total probability is 1.
性质1： \(G_X(1) = 1\)，因为总概率为 1。
Property 2: The \(r\)-th derivative at \(t=0\) gives the factorial moment:
性质2：在 \(t=0\) 处的 \(r\) 阶导数给出阶乘矩：
GX(r)(0) = r! P(X = r)
This directly recovers the probability mass function from the PGF.
这可以直接从概率生成函数恢复概率质量函数。
Property 3: The mean and variance can be found using derivatives at \(t=1\):
性质3：均值与方差可通过在 \(t=1\) 处的导数求得：
E(X) = G’X(1), Var(X) = G”X(1) + G’X(1) − [G’X(1)]2
These properties allow quick computation of moments without summing infinite series.
这些性质允许我们无需计算无穷级数即可快速求矩。
3. Uniqueness and Inversion | 唯一性与逆变换
The PGF uniquely determines the probability distribution. If two random variables have the same PGF, they have the same probability mass function.
概率生成函数唯一地确定概率分布。若两个随机变量具有相同的概率生成函数，则它们具有相同的概率质量函数。
To recover probabilities from a known PGF, we can use the relation
要从已知的概率生成函数恢复概率，我们可使用关系式
P(X = k) = GX(k)(0) / k!
This is particularly useful when the PGF is a polynomial or a simple rational function.
当概率生成函数是多项式或简单有理函数时，这一关系尤为有用。
For instance, if \(G(t)=0.2+0.5t+0.3t^2\), then \(G”(0)=0.6\), so \(P(X=2)=0.6/2=0.3\).
例如，若 \(G(t)=0.2+0.5t+0.3t^2\)，则 \(G”(0)=0.6\)，因此 \(P(X=2)=0.6/2=0.3\)。
4. Binomial Distribution | 二项分布
The binomial distribution \(X \sim \text{Binomial}(n,p)\) models the number of successes in \(n\) independent Bernoulli trials, each with success probability \(p\). Its probability mass function is
二项分布 \(X \sim \text{Binomial}(n,p)\) 描述在 \(n\) 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) px (1−p)n−x, x = 0,1,…,n
Using the definition of the PGF,
利用概率生成函数的定义，
GX(t) = Σ C(n,x)(pt)x(1−p)n−x = (1 − p + pt)n
This compact form is easy to differentiate. For example, \(G'(t)=n p (1-p+pt)^{n-1}\), so \(E(X) = G'(1)=np\).
这一紧凑形式很容易求导。例如，\(G'(t)=n p (1-p+pt)^{n-1}\)，所以 \(E(X) = G'(1)=np\)。
Similarly, \(G”(t)=n(n-1)p^2(1-p+pt)^{n-2}\), giving \(E[X(X-1)] = n(n-1)p^2\). Hence \(\text{Var}(X)=np(1-p)\).
类似地，\(G”(t)=n(n-1)p^2(1-p+pt)^{n-2}\)，得到 \(E[X(X-1)] = n(n-1)p^2\)。因此 \(\text{Var}(X)=np(1-p)\)。
5. Poisson Distribution | 泊松分布
Poisson distribution \(X \sim \text{Poisson}(\lambda)\) is often used to model the number of rare events occurring in a fixed interval. Its probability mass function is
泊松分布 \(X \sim \text{Poisson}(\lambda)\) 常用于固定区间内稀有事件发生的次数。其概率质量函数为
P(X = x) = e−λ λx / x!, x = 0,1,2,…
The PGF is obtained by
其概率生成函数为
GX(t) = e−λ Σ (λt)x/x! = eλ(t−1)
Using the PGF, \(G'(t)=\lambda e^{\lambda(t-1)}\), so \(E(X)=\lambda\). Also \(G”(t)=\lambda^2 e^{\lambda(t-1)}\), giving \(\text{Var}(X)=\lambda\).
利用概率生成函数，\(G'(t)=\lambda e^{\lambda(t-1)}\)，因此 \(E(X)=\lambda\)。又 \(G”(t)=\lambda^2 e^{\lambda(t-1)}\)，给出 \(\text{Var}(X)=\lambda\)。
The Poisson PGF is particularly elegant and often used in deriving the distribution of sums of independent Poisson variables.
泊松分布的概率生成函数特别简洁，常用于推导独立泊松变量之和的分布。
6. Geometric Distribution | 几何分布
The geometric distribution models the number of trials until the first success. We adopt the convention \(X\) = number of trials, with \(P(X=x)=p(1-p)^{x-1}\) for \(x=1,2,3,…\).
几何分布描述首次成功所需的试验次数。我们采用 \(X\) 表示试验次数，且 \(P(X=x)=p(1-p)^{x-1}\)，其中 \(x=1,2,3,…\)。
Its PGF is
其概率生成函数为
GX(t) = Σ p qx−1 tx = pt / (1 − qt), where q = 1−p
To find the mean, differentiate:
为求均值，先求导：
G'(t) = p / (1−qt)2, so E(X) = G'(1) = 1/p
The second derivative gives \(G”(t)=2pq/(1-qt)^3\). Hence \(\text{Var}(X) = q/p^2\).
二阶导数为 \(G”(t)=2pq/(1-qt)^3\)。因此 \(\text{Var}(X) = q/p^2\)。
Note: Some texts define the geometric distribution as the number of failures before the first success, giving \(P(X=x)=p q^x\) for \(x=0,1,…\). In that case the PGF becomes \(p/(1-qt)\). Always read the question carefully.
注意：有些教材将几何分布定义为首次成功前的失败次数，即 \(P(X=x)=p q^x\)，其中 \(x=0,1,…\)。此时概率生成函数变为 \(p/(1-qt)\)。做题时务必仔细审题。
7. Negative Binomial Distribution | 负二项分布
The negative binomial distribution models the number of trials needed to obtain \(r\) successes. With convention \(X\) = number of trials, its probability mass function is
负二项分布描述获得 \(r\) 次成功所需的试验次数。若 \(X\) 表示试验次数，其概率质量函数为
P(X=x) = C(x−1, r−1) pr qx−r, x = r, r+1, …
The corresponding PGF is
相应的概率生成函数为
GX(t) = ( pt / (1 − qt) )r
This result is intuitive: a negative binomial variable is the sum of \(r\) independent geometric variables, each with PGF \(\frac{pt}{1-qt}\).
这一结果很直观：负二项变量是 \(r\) 个独立几何变量之和，每个几何变量的概率生成函数均为 \(\frac{pt}{1-qt}\)。
From the PGF we get \(E(X)=r/p\) and \(\text{Var}(X)=r q/p^2\).
由该概率生成函数可得到 \(E(X)=r/p\) 与 \(\text{Var}(X)=r q/p^2\)。
Some specifications define negative binomial as the number of failures before \(r\) successes; then the PGF is \((p/(1-qt))^r\). Always match the definition used in your exam board.
有些考试大纲将负二项分布定义为在 \(r\) 次成功之前的失败次数，此时概率生成函数为 \((p/(1-qt))^r\)。务必与你的考试局所采用的定义保持一致。
8. Uniform Discrete Distribution | 离散均匀分布
For a discrete uniform distribution over \(1,2,…,n\), each value has probability \(1/n\). Its PGF is
对于取值 \(1,2,…,n\) 的离散均匀分布，每个取值的概率均为 \(1/n\)。其概率生成函数为
GX(t) = (t + t2 + … + tn) / n = t(1−tn) / [n(1−t)]
Using this PGF, the mean is
利用该概率生成函数，均值为
E(X) = (n+1)/2
and the variance is
方差为
Var(X) = (n2−1)/12
These results are useful in many combinatorial probability problems.
这些结果在许多组合概率问题中非常有用。
9. Sums of Independent Random Variables | 独立随机变量之和
One of the most powerful applications of PGFs is finding the distribution of the sum of independent random variables. If \(X\) and \(Y\) are independent, then
概率生成函数最强大的应用之一是求独立随机变量之和的分布。若 \(X\) 与 \(Y\) 独立，则
GX+Y(t) = GX(t) · GY(t)
This follows from the fact that \(E(t^{X+Y}) = E(t^X t^Y) = E(t^X)E(t^Y)\) because of independence.
这由 \(E(t^{X+Y}) = E(t^X t^Y) = E(t^X)E(t^Y)\) 以及独立性而得。
For example, if \(X \sim \text{Poisson}(\lambda_1)\) and \(Y \sim \text{Poisson}(\lambda_2)\) independently, then
例如，若 \(X \sim \text{Poisson}(\lambda_1)\) 与 \(Y \sim \text{Poisson}(\lambda_2)\) 独立，则
GX+Y(t) = eλ1(t−1) eλ2(t−1) = e(λ1+λ2)(t−1)
Hence \(X+Y \sim \text{Poisson}(\lambda_1+\lambda_2)\). Similarly, the sum of independent binomial variables with the same success probability is again binomial.
因此 \(X+Y \sim \text{Poisson}(\lambda_1+\lambda_2)\)。类似地，具有相同成功概率的独立二项变量之和仍服从二项分布。
10. Worked Example | 典型例题
Problem: Let \(X\) be a discrete random variable whose PGF is \(G(t)=\frac{1}{4}(1+t)^2\). Find \(P(X=2)\), \(E(X)\) and \(\text{Var}(X)\).
例题：设离散型随机变量 \(X\) 的概率生成函数为 \(G(t)=\frac{1}{4}(1+t)^2\)。求 \(P(X=2)\)、\(E(X)\) 和 \(\text{Var}(X)\)。
First expand the PGF:
首先展开概率生成函数：
G(t) = 1/4 + 1/2 t + 1/4 t2
Thus \(P(X=0)=1/4\), \(P(X=1)=1/2\), \(P(X=2)=1/4\). Therefore \(P(X=2)=1/4\).
因此 \(P(X=0)=1/4\), \(P(X=1)=1/2\), \(P(X=2)=1/4\)。所以 \(P(X=2)=1/4\)。
Next, differentiate: \(G'(t)=\frac{1}{2}+\frac{1}{2}t\), so \(E(X)=G'(1)=1\). The second derivative is \(G”(t)=\frac{1}{2}\), so \(E[X(X-1)] = G”(1)=1/2\). Hence \(\text{Var}(X) = 1/2 + 1 – 1^2 = 1/2\).
接着求导：\(G'(t)=\frac{1}{2}+\frac{1}{2}t\)，所以 \(E(X)=G'(1)=1\)。二阶导数为 \(G”(t)=\frac{1}{2}\)，因此 \(E[X(X-1)] = G”(1)=1/2\)。所以 \(\text{Var}(X) = 1/2 + 1 – 1^2 = 1/2\)。
This example illustrates how easily PGFs recover the full distribution and moments.
此例说明概率生成函数如何轻松地恢复完整分布与矩。
11. Summary and Exam Tips | 总结与考试建议
Probability generating functions provide a unified way to handle discrete distributions. Memorise the PGFs of binomial, Poisson, geometric, negative binomial and uniform, because they are frequently asked in exam questions.
概率生成函数为处理离散分布提供了一种统一的方法。请熟记二项、泊松、几何、负二项和均匀分布的概率生成函数，因为它们在考试中频繁出现。
Always state the definition \(G(t)=E(t^X)\), and remember the derivative facts for mean and variance. When finding probabilities from a given PGF, differentiate and divide by the appropriate factorial.
始终写出定义 \(G(t)=E(t^X)\)，并牢记求均值与方差的导数公式。当从给定概率生成函数求概率时，进行求导并除以相应的阶乘。
Check whether \(G(1)=1\); this is a quick and useful verification of your PGF. Also pay attention to the exact convention used for geometric and negative binomial distributions in the question.
检查 \(G(1)=1\)，这是对概率生成函数快速有效的验证。同时注意题目中几何分布和负二项分布所采用的具体约定。
Finally, remember that for independent variables, PGFs multiply. This can save time in multi-part questions involving sums of random variables.
最后，记住独立变量的概率生成函数相乘。在处理涉及随机变量之和的多步题目时，这可以节省时间。
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