📚 Probability Generating Functions | 概率生成函数
Probability generating functions (PGFs) offer a compact and elegant way to encode the entire distribution of a discrete random variable taking non‑negative integer values. They transform convolutions into simple multiplication, turn moments into derivatives, and give a power‑series representation where coefficients are probabilities. In IB Mathematics, especially in the Higher Level Analysis and Approaches or Applications and Interpretations courses, mastery of PGFs is essential for solving problems on sums of independent variables, deriving expectation and variance, and recognising standard distributions from their generating functions.
概率生成函数（PGF）为取非负整数值的离散随机变量的完整分布提供了一种紧凑而优雅的编码方式。它们将卷积转化为简单的乘法，将矩转化为导数，并给出一个系数即为概率的幂级数表示。在IB数学中，尤其是在高级别的分析与方法或应用与解释课程里，掌握概率生成函数对解决独立变量和、推导期望与方差以及从其生成函数识别标准分布等问题至关重要。
1. Definition of a Probability Generating Function | 概率生成函数的定义
Let X be a discrete random variable taking values in {0,1,2,…}. The probability generating function (PGF) of X is defined as GX(t) = E(tX) = Σx=0∞ P(X = x) tx, for all real t such that the series converges. Typically we require |t| ≤ 1 to ensure convergence, because |GX(t)| ≤ E(|t|X) ≤ 1 when |t| The PGF is a power series whose coefficients are exactly the probability masses P(X = x).
设 X 是取值在 {0,1,2,…} 的离散随机变量。X 的概率生成函数（PGF）定义为 GX(t) = E(tX) = Σx=0∞ P(X = x) tx，其中 t 为使该级数收敛的所有实数。通常我们要求 |t| ≤ 1 以保证收敛，因为当 |t| ≤ 1 时 |GX(t)| ≤ E(|t|X) ≤ 1。该 PGF 是一个幂级数，其系数恰好就是概率质量 P(X = x)。
Since Σ P(X=x) = 1, we always have GX(1) = 1. This is a quick check for any derived PGF. The radius of convergence is at least 1, and inside this radius the PGF is infinitely differentiable. Moreover, the PGF uniquely determines the distribution of X, thanks to the identity of power series coefficients.
由于 Σ P(X=x) = 1，我们恒有 GX(1) = 1。这是检验任何推导出的 PGF 的快速方法。收敛半径至少为 1，且在此半径内该 PGF 无限次可微。此外，由于幂级数系数的唯一性，PGF 唯一确定了 X 的分布。
nerate random variable that takes the value c with probability 1, the PGF is simply G(t)=tc (1−p) + p t. These simple cases illustrate how the PGF encodes the distribution concisely.
对于一个以概率 1 取值 c 的退化的随机变量，其 PGF 就是 G(t)=tc。对于成功概率为 p 的伯努利随机变量，G(t) = (1−p) + p t。这些简单例子说明了 PGF 如何简洁地编码分布。
2. PGFs for Fundamental Discrete Distributions | 基本离散分布的 PGF
Consider a binomial random variable X ~ B(n, p). Using the binomial theorem, we obtain GX(t) = Σx=0n (n choose x) px (1−p)n−x tx = (1−p + p t)n. When n=1 this recovers the Bernoulli PGF. For a Poisson random variable X ~ Po(λ), the PGF becomes GX(t) = Σx=0∞ e−λ λx/x! tx = eλ(t−1).
考虑二项随机变量 X ~ B(n, p)。利用二项式定理，我们得到 GX(t) = Σx=0n C(n,x) px (1−p)n−x tx = (1−p + p t)n。当 n=1 时这就退化为伯努利PGF。对于泊松随机变量 X ~ Po(λ)，PGF 变为 GX(t) = Σx=0∞ e−λ λx/x! tx = eλ(t−1)。
A geometric distribution on {1,2,3,…} with success probability p has PGF G(t) = p t / (1 − q t), where q=1−p, provided |q t| < 1. The geometric distribution on {0,1,2,…} (counting failures before the first success) has PGF G(t) = p / (1 − q t). Recognising these forms is vital for solving IB problems that ask for the distribution from a given PGF.
定义在 {1,2,3,…} 上、成功概率为 p 的几何分布的 PGF 为 G(t) = p t / (1 − q t)，其中 q=1−p，且需 |q t| < 1。定义在 {0,1,2,…} 上（计数首次成功前的失败次数）的几何分布的 PGF 为 G(t) = p / (1 − q t)。识别这些形式对于解决要求根据给定 PGF 确定分布的 IB 问题至关重要。
3. PGF of a Sum of Independent Random Variables | 独立随机变量和的 PGF
If X and Y are independent discrete random variables on the non‑negative integers, then the PGF of their sum Z = X + Y is simply the product: GZ(t) = GX(t) × GY(t). This follows from GX+Y(t) = E(tX+Y) = E(tX tY) = E(tX) E(tY) by independence. This property extends to any finite sum of independent variables: GΣXi(t) = ∏ GXi(t).
如果 X 和 Y 是独立且取非负整数值的离散随机变量，那么它们之和 Z = X + Y 的 PGF 就是简单的乘积：GZ(t) = GX(t) × GY(t)。这是因为 GX+Y(t) = E(tX+Y) = E(tX tY) = E(tX) E(tY) 由独立性得到。这一性质可以推广到任意有限个独立变量的和：GΣXi(t) = ∏ GXi(t)。
Convolution of probability mass functions is thus replaced by ordinary multiplication of generating functions, a considerable simplification. For example, the sum of n independent Bernoulli(p) variables is B(n, p), and indeed (1−p + p t)n equals the product of n copies of (1−p + p t). Similarly, the sum of independent Poissons is Poisson, which is immediately seen from the exponential form of the PGF.
因此，概率质量函数的卷积被替换为生成函数的普通乘法，这大大简化了问题。例如，n 个独立伯努利(p) 变量之和为 B(n, p)，而 (1−p + p t)n 确实等于 n 个 (1−p + p t) 相乘。同样，独立泊松变量之和仍为泊松分布，这从 PGF 的指数形式可立即看出。
4. Expectation from the PGF | 由 PGF 求期望
Differentiating GX limit from below because the series may not converge at t=1 for the derivative. Formally, E(X) = G′X(1), where G′X(t) is the first derivative. More generally, the k‑th factorial moment is given by the k‑th derivative: E[X(X−1)…(X−k+1)] = G(k)X(1).
对 GX(t) 求导并令 t=1 就可得到期望值，只要从小于 1 的一侧取极限，因为导数级数在 t=1 处可能不收敛。形式上，E(X) = G′X(1)，其中 G′X(t) 是一阶导数。更一般地，第 k 阶阶乘矩由第 k 阶导数给出：E[X(X−1)…(X−k+1)] = G(k)X(1)。
For a Poisson distribution with PGF eλ(t−1), we have G′X(t) = λ eλ(t−1), so G′X on. This method avoids summation of the probability distribution altogether, which is particularly useful when the distribution is only given through its PGF.
对于 PGF 为 eλ(t−1) 的泊松分布，我们有 G′X(t) = λ eλ(t−1)，故 G′X(1)=λ，与实际期望一致。这种方法完全避免了概率分布的求和，当分布只通过其 PGF 给出时尤为有用。
5. Variance Derived from the PGF | 由 PGF 求方差
We can obtain the variance by combining the first two derivatives. Since Var(X) = E(X2) − [E(X)]2 and E(X2) = E[X(X−1)] + E(X), we compute E[X(X−1)] = G″X(1). Then Var(X) = G″X(1) + G′X(1) − [G′X(1)]2. This formula is extremely efficient, because the necessary derivatives of the PGF are often very easy to calculate.
我们可以通过前两阶导数求出方差。由于 Var(X) = E(X2) − [E(X)]2 且 E(X2) = E[X(X−1)] + E(X)，我们计算 E[X(X−1)] = G″X(1)。从而 Var(X) = G″X(1) + G′X(1) − [G′X(1)]2。这个公式非常高效，因为 PGF 的必要导数通常很容易计算。
For the binomial B(n, p), G(t) = (q + p t)n, so G′(t)= n p (q + p t)n−1 gives E(X)= n p; G″(t)= n(n−1) p2 (q + p t)n−2 gives G″(1)= n(n−1) p2. Plugging into the variance formula yields Var(X)= n p (1−p). For the Poisson, G″(1)= λ2, so Var(X)= λ2 + λ − λ2 = λ. These are exactly the textbook results.
对于二项分布 B(n, p)，G(t) = (q + p t)n，因此 G′(t)= n p (q + p t)n−1 给出 E(X)= n p；G″(t)= n(n−1) p2 (q + p t)n−2 给出 G″(1)= n(n−1) p2。代入方差公式得 Var(X)= n p (1−p)。对于泊松分布，G″(1)= λ2，因此 Var(X)= λ2 + λ − λ2 = λ。这些正是教材中的结果。
6. Higher Order Factorial Moments and Cumulants | 高阶阶乘矩与累积量
Factorial moments are natural for count data. The k‑th factorial moment is μ(k) = E[X(X−1)…(X−k+1)] = G(k)X(1). This relationship implies that all moments of X can be expressed in terms of the derivatives of the PGF at 1. For example, the third moment E(X3) can be found by first computing the factorial moments and then using Stirling numbers of the second kind, but in IB practice questions are usually limited to expectation and variance.
阶乘矩对于计数数据而言是自然的。第 k 阶阶乘矩为 μ(k) = E[X(X−1)…(X−k+1)] = G(k)X(1)。这一关系意味着 X 的所有矩都可以通过 PGF 在 1 处的导数表示。例如，三阶矩 E(X3) 可以先计算阶乘矩再利用第二类斯特林数求得，但 IB 的练习题通常只限于期望和方差。
Cumulants, though rarely required explicitly, illustrate why moments of linear combinations of independent variables behave nicely. The cumulant generating function is log GX(eθ), and its derivatives at zero give cumulants; for two independent variables, cumulants simply add. This fact underpins the simplicity of the PGF method for sums.
累积量虽然鲜有明确要求，却能说明为什么独立变量线性组合的矩性质如此良好。累积量生成函数为 log GX(eθ)，其在零处的导数给出累积量；对于两个独立变量，累积量直接相加。这一事实支撑了 PGF 方法在处理和时的简洁性。
7. Recovering Probabilities from a PGF | 从 PGF 还原概率
The most direct way to extract probabilities from a given PGF is to expand it as a power series in t. The coefficient of tx tforward via the binomial theorem or geometric series. For example, if G(t)= (0.2 + 0.8 t)5, then the coefficient of t2 is (5 choose 2) (0.8)2 (0.2)3.
从给定的 PGF 中提取概率最直接的方法是将其展开为 t 的幂级数。tx 的系数就是 P(X = x)。如果 PGF 是有理函数或可识别的闭式，这种展开通常通过二项式定理或几何级数就容易完成。例如，若 G(t)= (0.2 + 0.8 t)5，那么 t2 的系数为 C(5,2) (0.8)2 (0.2)3。
If the PGF is given as a rational expression like (a + b t)/(c − d t), one can use long division or partial fractions to find the series. More commonly, IB exam questions ask you to identify the distribution from the PGF (e.g., recognising eλ(t−1) as Poisson) and then read off probabilities from the known probability mass function, rather than expanding the entire series.
如果 PGF 以有理式给出，如 (a + b t)/(c − d t)，则可以利用长除法或部分分式来求级数。更常见的情况是，IB 考题要求你从 PGF 识别出分布（例如认出 eλ(t−1) 是泊松分布），然后从已知的概率质量函数读取概率，而非展开整个级数。
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