📚 pdfjoiner_(4)-247: Piecewise PDFs & Joining Constants | 分段概率密度函数与连接常数
In Edexcel A-Level Mathematics Statistics 2, you will frequently encounter continuous random variables defined by piecewise probability density functions (PDFs). Often, the question involves an unknown constant that 'joins' the separate pieces, ensuring the function satisfies the conditions for a valid PDF. Mastering the technique of finding joining constants and using them to compute probabilities, the cumulative distribution function, and measures of centre and spread is essential for exam success. This article will guide you through the key concepts, worked examples, and common pitfalls.
在Edexcel A-Level数学S2中，你会经常遇到由分段概率密度函数（PDF）定义的连续随机变量。题目通常包含一个未知常数，它将不同部分“连接”起来，使函数满足有效PDF的条件。掌握求连接常数的技巧，并利用它计算概率、累积分布函数以及集中趋势和离散程度的指标，对考试成功至关重要。本文将带你深入关键概念、例题解析和常见陷阱。
1. Introduction to Probability Density Functions | 概率密度函数简介
A probability density function (PDF) is a function f(x) used to specify the probability of a continuous random variable X falling within a particular range of values. For a continuous variable, the probability that X takes any exact value is zero; therefore, probabilities are determined by areas under the PDF curve.
概率密度函数（PDF）是用来描述连续随机变量X落在某个值区间内的概率的函数。对于连续变量，X取任何确切值的概率为零；因此，概率由PDF曲线下的面积决定。
The graph of a PDF can take many shapes, but it must always satisfy two fundamental conditions.
PDF对所有实数x有定义，且X在a与b之间的概率为P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx。PDF的图像可以有多种形状，但必须满足两个基本条件。
2. Valid PDF Requirements | 有效PDF的条件
∞ f(x) dx = 1. If a function is defined piecewise, these conditions must hold across all pieces, and the integrals over the defined intervals must sum to 1.
要使f(x)成为有效的概率密度函数，须满足：(i) 对所有x有f(x) ≥ 0；(ii) 曲线下的总面积为1，即∫₋∞⁺∞ f(x) dx = 1。若函数是分段定义的，则这些条件必须在所有段上成立，且定义区间上的积分之和必须等于1。
3. Piecewise PDFs – The Joining Concept | 分段PDF——连接的概念
≤ 4; 0 otherwise }. Here, the two non-zero pieces meet at x = 2. The constant k acts as a 'joiner' – ined, but a PDF does not have to be continuous; only the integral condition is compulsory.
分段PDF在不同区间由不同公式定义。Edexcel S2的典型例子为：f(x) = { kx, 0 ≤ x < 2; k(4 − x), 2 ≤ x ≤ 4; 0 其他 }。这里两个非零部分在x=2处相接。常数k充当“连接器”——通过设置k值使得总面积为1。通常相接点会被用来考察连续性，但PDF并不要求连续；只有积分条件是必须的。
4. Setting Up the 'Joiner' Equation | 建立“连接”方程
l area condition: ∫(first piece) dx + ∫(second piece) dx = 1. For the above PDF, we write:
∫₀² kx dx + ∫₂⁴ k(4 − x) dx = 1
当PDF含有未知常数k时，连接各段的方程来源于总面积条件：∫(第一段) dx + ∫(第二段) dx = 1。对上述PDF，可写出：
∫₀² kx dx + ∫₂⁴ k(4 − x) dx = 1
Solving this will yield the value of k that successfully 'joins' the two sections into a single valid PDF.
求解该方程将得出成功“连接”两段成为一个有效PDF的k值。
5. Using Total Area = 1 to Find Constants | 利用总面积为1求常数
s 2k + 2k = 4k. Set equal to 1 → 4k = 1 → k = ¼. Always check that f(x) ≥ 0 for all x. With k=¼, f(x)=0.25x on [0,2) and 0.25(4−x) on [2,4], both non-negative, so the PDF is valid.
分别计算每段积分：∫₀² kx dx = k [½ x²]₀² = 2k。然后∫₂⁴ k(4 − x) dx = k [4x − ½ x²]₂⁴ = k[(16−8)−(8−2)] = 2k。相加得4k，令其等于1 → 4k = 1 → k = ¼。务必检查对所有x有f(x) ≥ 0。由k=¼可得两段均非负，因此该PDF有效。
6. Calculating Probabilities from a Piecewise PDF | 由分段PDF计算概率
For instance, to find P(1 ≤ X ≤ 3), the interval must be split at the joining point x = 2:
P(1 ≤ X ≤ 3) = ∫₁² 0.25x dx + ∫₂³ 0.25(4 − x) dx
已知k后，可通过积分对应段求出概率。例如求P(1 ≤ X ≤ 3)，必须在连接点x=2处分割区间：
P(1 ≤ X ≤ 3) = ∫₁² 0.25x dx + ∫₂³ 0.25(4 − x) dx
− ½ x²]₂³ = 0.25[(12 − 4.5) − (8 − 2)] = 0.25(7.5 − 6) = 0.375. Total probability = 0.75.
计算：第一段积分 = 0.25[½ x²]₁² = 0.375。第二段 = 0.25[4x − ½ x²]₂³ = 0.25(7.5 − 6) = 0.375。总概率为0.75。
7. Cumulative Distribution Function (CDF) | 累积分布函数
be expressed in pieces. Using the same f(x) with k=¼, we define F(x) for different ranges.
累积分布函数F(x) = P(X ≤ x)由PDF从下界积分至x得到。对于分段PDF，F(x)需分段表达。以k=¼的f(x)为例，我们为不同区间定义F(x)。
For x < 0: F(x) = 0. For 0 ≤ x < 2: F(x) = ∫₀ˣ 0.25t dt = 0.125x². For 2 ≤ x ≤ 4: F(x) = ∫₀² 0.25t dt + ∫₂ˣ 0.25(4 − t) dt = 0.5 + 0.25[4t − ½ t²]₂ˣ For x > 4: F(x) = 1. Always check that F(x) is increasing and satisfies F(2)=0.5, F(4)=1.
当x < 0时：F(x)=0。当0 ≤ x < 2时：F(x) = ∫₀ˣ 0.25t dt =
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