📚 PDF Joiner: Combining Probability Density Functions in Edexcel A-Level Maths | Edexcel A-Level 数学中的概率密度函数合并技巧
Probability density functions (PDFs) are the backbone of continuous random variables in A-Level Statistics. But what happens when we need to combine two or more independent random variables? This article explains how to 'join' PDFs through linear combinations, focusing on the rules of expectation, variance and the special role of the normal distribution. Mastering these tools will sharpen your problem-solving skills for Edexcel exam questions on sums and differences of random variables.
概率密度函数（PDF）是 A-Level 统计中连续随机变量的核心。当我们需要将两个或多个独立的随机变量合并时，会发生什么？本文将讲解如何通过线性组合来“合并”PDF，重点介绍期望、方差的运算规则以及正态分布的特殊性质。掌握这些工具将使你更熟练地应对 Edexcel 考试中关于随机变量和与差的题目。
1. What Is a Probability Density Function? | 什么是概率密度函数？
a under the curve is always 1, i.e. ∫₋∞⁺∞ f(x) dx = 1, and the probability that X lies between a and b is given by ∫ₐᵇ f(x) dx.
概率密度函数 f(x) 描述了连续随机变量 X 取值在某个范围内的可能性。曲线下的总面积恒为 1，即 ∫₋∞⁺∞ f(x) dx = 1，而 X 落在 a 和 b 之间的概率等于 ∫ₐᵇ f(x) dx。
Unlike discrete distributions, individual points have zero probability. We always work with intervals when calculating PDF probabilities, making integration a critical skill.
与离散分布不同，单点的概率为零。使用 PDF 计算概率时我们总是处理区间，因此积分是一项关键技能。
2. Independent Random Variables and New Variables | 独立随机变量与新变量
iable Z = aX + bY + c, where a, b, c are constants. This is the fundamental operation of a 'PDF joiner' – creating a new distribution from existing ones.
当 X 和 Y 是相互独立的连续随机变量时，我们经常需要构造新变量 Z = aX + bY + c，其中 a、b、c 为常数。这就是“PDF 合并器”的基本操作——从现有分布生成新分布。
Independence is crucial because it allows us to combine variances additively. If X and Y are not independent, an extra covariance term would be needed, which is beyond the scope of standard A-Level Maths.
独立性至关重要，因为它使得方差可以线性相加。如果 X 和 Y 不独立，就需要加入协方差项，这超出了标准 A-Level 数学的范围。
3. Expectation and Variance of Linear Combinations | 线性组合的期望与方差
E(aX + bY + c) = aE(X) + bE(Y) + c. This holds regardless of the underlying distribution.
对于独立的随机变量 X 和 Y，期望总是线性的：E(aX + bY + c) = aE(X) + bE(Y) + c。无论基础分布是什么，这一性质均成立。
onstant c disappears, and the coefficients a and b are squared. The standard deviation is the square root of this total variance.
方差的运算规则为：Var(aX + bY + c) = a²Var(X) + b²Var(Y)。注意常数 c 消失，系数 a 和 b 需要平方。标准差是总方差的平方根。
- If Z = X + Y, then Var(Z) = Var(X) + Var(Y)
- If Z = X – Y, then Var(Z) = Var(X) + Var(Y) (still a sum!)
- 若 Z = X + Y，则 Var(Z) = Var(X) + Var(Y)
- 若 Z = X – Y，则 Var(Z) = Var(X) + Var(Y)（仍然是相加！）
The fact that variance adds even when subtracting is a common trap; many students mistakenly subtract the variances. Always add variances for independent variables.
即使在做减法时方差仍然相加，这是一个常见的陷阱；许多学生错误地减去方差。务必记住：对于独立变量，方差总是相加。
4. The Normal Distribution – The Ultimate PDF Joiner | 正态分布——终极 PDF 合并器
If X ~ N(μ₁, σ₁²) and Y ~ n Z = aX + bY + c also follows a normal distribution. This is a uniquely powerful property – the normal distribution is closed under linear combination.
若 X ~ N(μ₁, σ₁²) 与 Y ~ N(μ₂, σ₂²) 独立，则任何线性组合 Z = aX + bY + c 也服从正态分布。这是一个极其强大的性质——正态分布在线性组合下是封闭的。
Z ~ N(aμ₁ + bμ₂ + c, a²σ₁² + b²σ₂²)
This means we can immediately write down the PDF of Z without performing any integration. The new distribution is still bell-shaped, and all probabilities can be found using the standardised normal table.
这意味着我们可以立即写出 Z 的 PDF，而无需进行任何积分。新分布仍然是钟形曲线，所有概率都可以借助标准正态分布表求得。
5. Working with Sums and Differences of Normals | 正态分布的和与差运算
Consider two independent normal variables: X ~ N(50, 4²) and Y ~ N(30, 3²). For Z = X + Y, we have μ = 50 + 30 = 80 and σ² = 4² + 3² = 25, so σ = 5. Thus Z ~ N(80, 5²).
考虑两个独立的正态变量：X ~ N(50, 4²) 和 Y ~ N(30, 3²)。对于 Z = X + Y，均值 μ = 50 + 30 = 80，方差 σ² = 4² + 3² = 25，于是 σ = 5。因此 Z ~ N(80, 5²)。
If we need W = X – Y, then μ = 50 – 30 = 20, but σ² = 4² + 3² = 25 (variance adds again), so σ = 5 and W ~ N(20, 5²).
若我们需要 W = X – Y，则 μ = 50 – 30 = 20，但 σ² = 4² + 3² = 25（方差再次相加），所以 σ = 5，W ~ N(20, 5²)。
The key is to ensure the variance is never subtracted. A quick check: the combined uncertainty must increase, not decrease, when we subtract independent variables.
关键在于确保方差从不做减法。一个快速检验：当我们对独立变量做减法时，合并的不确定性必定增加而非减少。
6. Example: Exam-Style Problem with Combined Normals | 例题：合并正态分布的考试题型
Problem: The weight of apples from farm A is normally distributed with mean 150 g and standard deviation 18 g. The weight of apples from farm B is normally distributed with mean 130 g and standard deviation 12 g. Two apples are chosen at random, one from each farm. Find the probability that the total weight exceeds 310 g.
题目：A 农场苹果的重量服从均值为 150 g、标准差为 18 g 的正态分布。B 农场苹果的重量服从均值为 130 g、标准差为 12 g 的正态分布。从每个农场各随机挑选一个苹果。求总重量超过 310 g 的概率。
Solution: Let X be weight from farm A, Y from farm B. X ~ N(150, 18²), Y ~ N(130, 12²). Define T = X + Y. Then T ~ N(150+130, 18²+12²) = N(280, 468). Standard deviation σ = √468 ≈ 21.63 g.
解答：设 X 为 A 农场重量，Y 为 B 农场重量。X ~ N(150, 18²)，Y ~ N(130, 12²)。定义 T = X + Y。则 T ~ N(280, 468)，标准差 σ = √468 ≈ 21.63 g。
Standardise: P(T > 310) = P(Z > (310 – 280)/21.63) = P(Z > 1.386) ≈ 1 – 0.9171 = 0.0829. There is about an 8.3% chance the total weight exceeds 310 g.
标准化：P(T > 310) = P(Z > (310 – 280)/21.63) = P(Z > 1.386) ≈ 1 – 0.9171 = 0.0829。总重量超过 310 g 的概率约为 8.3%。
7. Joining More Than Two Variables | 合并两个以上的变量
ₙ are independent, the sum S = X₁ + X₂ + ... + Xₙ has E(S) = ΣE(Xᵢ) and Var(S) = ΣVar(Xᵢ).
线性组合的规则可以自然地推广到 n 个独立随机变量。若 X₁, X₂, ..., Xₙ 相互独立，则总和 S = X₁ + X₂ + ... + Xₙ 满足 E(S) = ΣE(Xᵢ)，Var(S) = ΣVar(Xᵢ)。
s makes it easy to handle large sample totals in quality control or repeated measurements.
如果每个 Xᵢ 都服从正态分布，S 仍为正态分布，其均值为 Σμᵢ，方差为 Σσᵢ²。这使得在质量控制或重复测量中处理大量样本总和变得非常简单。
8. Mixing Distributions – A Different Kind of Join | 混合分布——另一种合并方式
Sometimes we encounter a 'mixture' of PDFs: a single random variable comes from one of several distributions with certain probabilities. This is different from linear combination, but it is another form of PDF joiner.
有时我们会遇到几种 PDF 的“混合”：单个随机变量以一定概率来自某个分布之一。这与线性组合不同，但也是 PDF 合并的另一种形式。
X₁) + (1-p)E(X₂). However, the variance is not just a simple weighted average; you must use E(X²) – [E(X)]².
对于混合分布，总期望是加权平均：E(X) = pE(X₁) + (1-p)E(X₂)。然而方差并非简单的加权平均；你必须使用 E(X²) – [E(X)]² 来计算。
0.4×f₁(x) + 0.6×f₂(x). Recognising this as a mixture is essential for correct integration.
Edexcel 的题目有时会给出分段定义的 PDF，例如 f(x) = 0.4×f₁(x) + 0.6×f₂(x)。将其识别为混合分布对于正确积分至关重要。
9. Common Pitfalls When Joining PDFs | 合并 PDF 的常见误区
Assuming variance subtracts: Repeatedly emphasised – Var(X – Y) = Var(X) + Var(Y). Never subtract variances for independent variables.
误以为方差可以相减：反复强调—— Var(X – Y) = Var(X) + Var(Y)。绝不要对独立变量做方差减法。
Forgetting to square coefficients: In Var(aX + bY), coefficients a and b must be squared. For instance, if Z = 3X – 2Y, then Var(Z) = 9Var(X) + 4Var(Y), not 3² – 2².
忘记将系数平方：在 Var(aX + bY) 中，系数 a 和 b 必须平方。例如，若 Z = 3X – 2Y，则 Var(Z) = 9Var(X) + 4Var(Y)，而不是 3² – 2²。
Using wrong standard deviation: Always calculate variance first, then take square root for standard deviation. Many students mistakenly combine standard deviations directly.
误用标准差：始终保持先算方差，再开方求得标准差。许多学生错误地直接对标准差进行运算。
Misapplying normality: The closure under linear combination applies strictly to independent normal variables. Do not assume sums of other distributions remain the same family unless the question tells you.
误用正态性：线性组合下的封闭性严格适用于独立正态变量。除非题目明确说明，否则不要假设其他分布的和仍属于同一分布族。
| Correct step | 正确步骤 | Wrong practice | 错误做法 |
|---|---|
| Var(X – Y) = Var(X) + Var(Y) | Var(X – Y) = Var(X) – Var(Y) |
| Var(2X + 3Y) = 4Var(X) + 9Var(Y) | Var(2X + 3Y) = 2Var(X) + 3Var(Y) |
| Calculate σ after variance | 先方差后标准差 | Add SDs directly | 直接加标准差 |
10. Using Technology and Tables for Joined Normals | 使用计算器与表格处理合并正态分布
Edexcel exams allow the use of calculators with statistical functions. You can directly input the normal distribution with the new mean and standard deviation after linear combination, making probability calculations quick and accurate.
Edexcel 考试允许使用带有统计功能的计算器。你可以直接输入线性组合后的新均值和标准差来计算正态分布概率，既快速又准确。
When standardising manually, remember the formula z = (x – rly: state the combined parameters, write the standardisation line, and then look up Φ(z).
手动标准化时，记住公式 z = (x – μ)/σ。清晰地组织你的书写：写出组合参数，写下标准化方程，然后查表求 Φ(z)。
If the probability involves lower tail, use symmetry, e.g. P(Z < –1.2) = 1 – Φ(1.2). Always sketch the bell curve to avoid sign errors.
如果涉及左尾概率，利用对称性，例如 P(Z < –1.2) = 1 – Φ(1.2)。始终画一个钟形草图以避免符号错误。
11. Practice Questions to Sharpen Your Joiner Skills | 巩固合并技巧的练习题
1. The lengths of metal rods from two machines are normally distributed: Machine 1 ~ N(25.0, 0.15²), Machine 2 ~ N(30.0, 0.20²). One rod from each machine is selected. Find the probability that the rod from Machine 2 is at least 5.5 cm longer than the rod from Machine 1.
1. 两台机器生产的金属棒长度服从正态分布：机器 1 ~ N(25.0, 0.15²)，机器 2 ~ N(30.0, 0.20²)。从每台机器各选一根。求机器 2 的棒长至少比机器 1 长 5.5 cm 的概率。
2. X ~ N(100, 15²) and Y ~ N(70, 10²) independent. Find the distribution of W = 2X – 3Y + 5 and hence find P(W > 120).
2. X ~ N(100, 15²) 和 Y ~ N(70, 10²) 相互独立。求 W = 2X – 3Y + 5 的分布，并由此计算 P(W > 120)。
3. A random variable T is a mixture: with probability 0.3, T ~ N(20,4²); with probability 0.7, T ~ N(28,5²). Find E(T) and Var(T).
3. 随机变量 T 是一混合分布：以概率 0.3 来自 N(20,4²)，以概率 0.7 来自 N(28,5²)。求 E(T) 和 Var(T)。
Attempt these before checking the worked solutions. The mixture problem reminds you that not all PDF joins are simple linear combinations.
在查看答案详解之前，请尝试完成这些练习。混合问题提醒我们，并非所有的 PDF 合并都是简单的线性组合。
12. Summary – Become a PDF Joiner Expert | 总结——成为 PDF 合并专家
(aX + bY + c) = aE(X) + bE(Y) + c, Var(aX + bY + c) = a²Var(X) + b²Var(Y), and the knowledge that normal variables stay normal after any linear join. Watch out for the hidden traps – iance always adds, coefficients are squared, and mixtures need special variance treatment.
A-Level 数学中，概率密度函数的合并以独立随机变量的线性组合为中心。核心工具包是：E(aX + bY + c) = aE(X) + bE(Y) + c，Var(aX + bY + c) = a²Var(X) + b²Var(Y)，并且知道正态变量在任何线性合并后仍保持正态。谨防隐藏陷阱——方差永远相加，系数要平方，混合分布需要特殊的方差处理。
With consistent practice, you will handle any Edexcel question that asks you to join distributions, whether they involve simple sums or more complex mixtures. Your PDF joiner skills will also build a strong foundation for further statistics topics beyond A-Level.
通过持续练习，你将能应对任何要求你合并分布的 Edexcel 考题，无论是简单求和还是更复杂的混合。你的 PDF 合并技能也将为后续的进阶统计专题打下坚实基础。
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