📚 pdfjoiner_(4)-169 | 分段概率密度函数之连接点专题
Welcome to this focused revision lesson, coded as “pdfjoiner_(4)-169”, which targets a common type of Edexcel A‑ istics 2 problem: piecewise probability density functions and the critical conditions that “join” g real exam questions where constants must be found and probabilities calculated smoothly.
欢迎来到这节针对“pdfjoiner_(4)-169”的专项复习课。这个代号对应的是Edexcel A-Level统计学2中常见的一类题型——分段概率密度函数，以及将它们的分段正确“连接”起来的关键条件。掌握这类拼接技巧，对于求解未知常数、顺利计算概率是必不可少的。
1. Definition of a Probability Density Function | 概率密度函数的定义
A probability density function (PDF), always denoted by f(x), is a function that describes the likelihood of a continuous random variable X taking a particular value. Unlike discrete probability, f(x) does not give a probability directly; instead the area under the curve of f(x) over an interval represents the probability that X falls inside that interval.
概率密度函数（PDF），通常记作 f(x)，是用来描述连续型随机变量 X 在某个值附近出现的可能性大小的函数。与离散型概率不同，f(x) 本身不直接表示概率；而是通过曲线 f(x) 下方在某一区间上的面积，来表示 X 落入该区间的概率。
P(a ≤ X ≤ b) = ∫ₐ ᵇ f(x) dx
In Edexcel S2, you will encounter various PDFs that may be simple polynomials, trigonometric functions (though rarely), or piecewise‑defined expressions. The fundamental requirement is that f(x) ⩾ 0 in its domain and that its total area integrates to 1.
在 Edexcel S2 中，你会遇到多种概率密度函数，有些是简单的多项式，少数可能涉及三角函数，更多则是分段定义的表达式。基本要求是：在其定义域内 f(x) ⩾ 0，并且整段曲线下的总面积等于 1。
2. Two Key Properties of a PDF | 概率密度函数的两个关键性质
Every valid PDF must satisfy two non‑negotiable conditions. First, f(x) ⩾ 0 for all x in the range over which X is defined; negative densities have no physical meaning. Second, the integral of f(x) over the entire sample space must equal 1, i.e.
每一个合法的概率密度函数都必须满足两条硬性条件。第一，对 X 定义范围内的所有 x，f(x) ⩾ 0；负的密度值没有任何实际意义。第二，f(x) 在整个样本空间上的积分必须等于 1，即
∫₋ₕ₊ₕ f(x) dx = 1
or, if the support is a finite interval [k₁, k₂], then ∫ₖ₁ᵏ₂ f(x) dx = 1.
如果支撑集是一个有限区间 [k₁, k₂]，则 ∫ₖ₁ᵏ₂ f(x) dx = 1。
These properties are the tools you use to find unknown constants in a piecewise PDF—exactly the task that “pdfjoiner_(4)-169” he total area of another must sum to 1, and the function must not dip below zero anywhere.
这两个性质就是用于求解分段概率密度函数中未知常数的利器——正是“pdfjoiner_(4)-169”所强调的内容。某一段的面积加上另一段的面积总和必须为 1，且函数在任何位置都不能跌落到零以下。
3. Piecewise PDFs: Why a “Joiner”? | 分段概率密度函数：为何需要“连接点”？
In many Edexcel questions, the PDF is given in piecewise form, for example:
在许多 Edexcel 试题中，概率密度函数会以分段形式给出，例如：
f(x) = { kx, 0 ≤ x < 2
m(4 − x), 2 ≤ x ≤ 4
0, otherwise }
Here the function is “joined” at x = 2. The term “pdf joiner” uous, but its integrals on both sides must together satisfy the total probability law. The “joiner” often provides an equation that links the unknown constants k and m.
这里，函数在 x = 2 处被“连接”起来。“pdf joiner”指的就是两个分段的相接点。在这样一个连接点处，概率密度函数本身可能连续，也可能不连续，但两侧的积分必须共同满足总概率定律。连接点常常提供一个方程，将未知常数 k 和 m 联系起来。
The code “(4)-169” ght refer to a classic problem where the total interval length is 4, or there are 4 pieces? s the same: use the total area = 1 condition and any given continuity or mode information.
代号“(4)-169”可能暗指区间总长是 4 或常见题型号。但实际上 S2 中通常遇到 2 至 3 个分段。不论分段数目如何，连接点的逻辑始终不变：利用总面积等于 1 这一条件，再配合任何给定的连续性或众数等信息。
4. Finding Unknown Constants: The Joining Condition | 寻找未知常数：连接点条件
Edexcel questions often ask: “Find the value of k and m.” Start by integrating each piece over its own sub‑interval. Then set the sum of these integrals equal to 1—that is your first equation. If the pieces join smoothly (i.e. the PDF is continuous at the boundary), you get a second equation by equating the two expressions at the joining point. For the example above:
Edexcel 试题常会要求：“求 k 和 m 的值。”首先在各个子区间上对分段表达式进行积分，然后令这些积分的总和等于 1——这是第一个方程。如果分段在边界处平滑连接（即概率密度函数在连接点连续），那么就可以令两个表达式在连接点处的值相等，得到第二个方程。以上例为例：
∫₀² kx dx + ∫₂⁴ m(4 − x) dx = 1
and if continuous at x = 2, k·2 = m(4 − 2) two equations yields the unique constants. If the question does not state continuity, you must rely solely on the total area condition and additional information such as a give probability or mode.
如果在 x = 2 处连续，则 k·2 = m(4 − 2)，即 2k = 2m，所以 k = m。联立这两个方程即可得出唯一的一组常数。如果题目并未声明连续，就只能单纯依靠总面积条件，外加题目给出的概率值或众数等额外信息。
5. Calculating Probabilities | 求得概率
Once the constants are known, finding any probability is a straightforward integration. For example:
一旦常数已知，求任何概率就是简单的积分运算。例如：
P(1 < X < 3) = ∫₁² kx dx + ∫₂³ m(4 − x) dx
It is vital to split the integral at the joining point, because the functional form changes there. Failure to split is one of the most common errors in S2 exams. Always draw a vertical dotted line at the join to remind yourself where the formula changes.
关键是要在连接点处拆分积分区间，因为函数表达式在那里发生了变化。未拆分是 S2 考试中最常见的错误之一。务必在连接点处画一条垂直虚线，提醒自己公式的切换位置。
If the PDF is zero outside the defined range, any probability entirely outside that range is automatically zero. Likewise, the probability that X lies within the whole support interval is 1.
如果概率密度函数在定义范围之外取值为零，那么完全落在该范围之外的任何概率自然为零。同理，X 落在整个支撑区间内的概率就是 1。
6. Building the Cumulative Distribution Function | 构建累积分布函数
tegrating the PDF from the lower bound up to x. For a piecewise PDF, you must construct F(x) piecewise as well. For the “pdfjoiner” style function:
累积分布函数（CDF）F(x) = P(X ≤ x) 是通过从下限到 x 积分概率密度函数得到的。对于分段概率密度函数，F(x) 也需要分段构建。以“pdfjoiner”型函数为例：
For 0 ≤ x < 2: F(x) = ∫₀ˣ kt dt
For 2 ≤ x ≤ 4: F(x) = ∫₀² kt dt + ∫₂ˣ m(4 − t) dt
For x > 4: F(x) = 1
from all previous pieces when writing the formula for later ranges. The CDF is always non‑ s at 0 and ends at 1. It can be used to find median and quartiles as well as any P(X ≤ x).
必须注意的是，在写后面区间的公式时，要加总前面所有分段的累积面积。累积分布函数总是非降的，从 0 开始到 1 结束。它可用来求中位数、四分位数以及任何 P(X ≤ x)。
7. Median and Percentiles | 中位数与百分位点
ause the PDF may be asymmetric, you need to identify which piece contains the median. Set the piecewise CDF equal to 0.5 and solve, but only after checking that the solution indeed lies in the correct sub‑interval. If the median falls into the second piece, you would use the second‑ ensuring that the constant terms (the cumulative area from previous pieces) are included.
中位数 m 满足 F(m) = 0.5。由于概率密度函数可能不对称，需要先判断中位数落在哪一段。将对应的分段累积分布函数设为 0.5 并求解，但要检查解确实落在正确的子区间内。如果中位数落在第二段，就使用第二段的 F(x) 表达式，并确保常数项（之前各段的累积面积）已被包含进去。
Other percentiles, such as the lower quartile (0.25) or upper quartile (0.75), are found similarly. The key with piecewise functions is to verify the domain of the answer: reject any solution that lies outside the assumed piece.
其它百分位数，如下四分位数（0.25）或上四分位数（0.75），求法类似。对于分段函数，关键在于验证结果所在的定义域：凡是落在假定分段之外的解都要舍弃。
8. Expectation and Variance | 期望值与方差
The expected value (mean) of X is given by:
X 的期望值（均值）由下式给出：
E(X) = ∫ x f(x) dx
over the whole support. With a piecewise PDF you split the integral just as before:
在整个支撑集上计算。对于分段概率密度函数，同样要拆分积分：
E(X) = ∫₀² x·kx dx + ∫₂⁴ x·m(4 − x) dx
Similarly, E(X²) = ∫ x² f(x) dx, and
类似地，E(X²) = ∫ x² f(x) dx，而
Var(X) = E(X²) − [E(X)]²
These calculations can be lengthy, so careful algebraic simplification and working in fractions are recommended. The “joiner” does not affect the integration technique, but it does mean you must treat each segment accurately when plugging in limits. Always double‑check that you have included the entire domain of the piece.
这些计算可能十分冗长，建议细心进行代数化简并使用分数运算。连接点并不影响积分技巧，但意味着在代入上下限时，必须准确对待每一分段。务必反复检查是否已覆盖该分段的完整定义域。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
ct constant after setting the total area to 1. Another is misreading the inequality signs: “≤” vs “<” n. Yet another mistake is using the CDF piecewise without adding the prior cumulative area—this leads to a discontinuity in F(x), which is impossible for a valid CDF.
一个经典错误是：在设总面积等于 1 之后，忘记给第二段的积分乘上正确的常数。另一个是误读不等号：对于连续分布，“≤” 与 “<” 没有区别，但端点仍需包含在区间内；真正关键的是连接点处的函数形式变化。还有一个错误是，在使用分段累积分布函数时没有加上前面部分的累积面积——这会导致 F(x) 不连续，而一个合法的累积分布函数不能有跳跃。
In the exam, always sketch a rough graph of the PDF, labelling the join and the areas. Write the total area equation as early as possible. If the problem gives a “mode” at the join, this implies the highest point is there, which may provide an extra equation. And remember: “pdf joiner” problems reward method marks heavily, so even if you miscalculate an integral, show your splitting and equation‑setting process clearly.
考试中，一定要画一幅概率密度函数的草图，标出连接点和各部分面积。尽早写出总面积方程。如果题目给出连接点处为“众数”，意味着函数在该点达到最高，这就提供了一个额外方程。请记住：“pdf joiner”类题目给过程分很大方，因此即便积分算错，也要清晰展示拆分和方程建立的过程。
10. Worked Example: Case Study (4)-169 | 例题演示：案例分析 (4)-169
Let us illustrate with a model problem inspired by the tag “(4)-169”. Suppose the continuous random variable X has PDF:
让我们通过一个受“(4)-169”启发而来的模型题来演示。设连续型随机变量 X 的概率密度函数为：
f(x) = { ax², 0 ≤ x < 1
b(4 − x), 1 ≤ x ≤ 4
0, otherwise }
t the total probability is 1. Find a and b, then evaluate P(0.5 < X < 2.5) and the median.
并已知该概率密度函数在连接点 x = 1 处连续，且总概率为 1。求 a 与 b，并计算 P(0.5 < X < 2.5) 及中位数。
Step 1: join condition. Continuity at x = 1 gives a·1² = b(4 − 1) ⇒ a = 3b.
第一步：连接点条件。 由 x = 1 处的连续性，得 a·1² = b(4 − 1) ⇒ a = 3b。
Step 2: total area = 1.
第二步：总面积等于 1。
∫₀¹ ax² dx + ∫₁⁴ b(4 − x) dx = 1
ce = (8 − 3.5)b = 4.5b. So a/3 + 4.5b = 1. Substitute a = 3b: (3b)/3 + 4.5b = b + 4.5b = 5.5b = 1 ⇒ b = 2/11, then a = 6/11.
计算：第一段积分 = a[ x³/3 ]₀¹ = a/3。第二段积分：∫ b(4 − x) dx = b[4x − x²/2]，从 1 到 4：在 4 处 → 16 − 8 = 8b；在 1 处 → 4 − 0.5 = 3.5b；差值为 (8 − 3.5)b = 4.5b。于是 a/3 + 4.5b = 1。代入 a = 3b：(3b)/3 + 4.5b = b + 4.5b = 5.5b = 1 ⇒ b = 2/11，故 a = 6/11。
Step 3: probability. = 1.75/11 ≈ 0.15909. Second part: (2/11)[4x − x²/2]₁²·⁵ = (2/11)[(10 − 3.125) − (4 − 0.5)] = (2/11)[6.875 − 3.5] = (2/11)·3.375 = 6.75/11 ≈ 0.61364. Sum ≈ 0.7727. So P ≈ 0.773.
第三步：概率。 P(0.5 < X < 2.5) = ∫₀.₅¹ (6/11)x² dx + ∫₁²·⁵ (2/11)(4 − x) dx。第一部分：)[x³/3]₀.₅¹ = (6/11)(1/3 − 0.125/3) = (2/11)(1 − 0.125) = (2/11)·0.875 = 1.75/11 ≈ 0.15909。第二部分： (2/11)[(10 − 3.125) − (4 − 0.5)] = (2/11)[6.875 − 3.5] = (2/11)·3.375 = 6.75/11 ≈ 0.61364。总和 ≈ 0.7727。所以 P ≈ 0.773。
Step 4: median. 2/11 + ∫₁ᵐ (2/11)(4 − x) dx = 0.5. The integral gives (2/11)[4x − x²/2]₁ᵐ = (2/11)[(4m − m²/2) − 3.5]. Equation: 2/11 + (2/11)[4m − m²/2 − 3.5] = 0.5. Multiply through by 11: 2 + 2(4m − 0.5m² − 3.5) = 5.5 ⇒ 2 + 8m − m² − 7 = 5.5 ⇒ −m² + 8m − 5 = 5.5 ⇒ −m² + 8m − 10.5 = 0 ⇒ m² − 8m + 10.5 = 0. Solve: m = [8 ± √64 − 42]/2 = [8 ± √22]/2. √22 ≈ 4.690, so m ≈ (8 − 4.69)/2 = 1.655 (reject the larger root > 4). Thus median ≈ 1.655.
第四步：中位数。 先判断中位数落在哪一段。F(1) = ∫₀¹ (6/11)x² dx = (6/11)(1/3) = 2/11 ≈ 0.1818。由于 0.1818 < 0.5，中位数在第二段。设 F(m) = 2/11 + ∫₁ᵐ (2/11)(4 − x) dx = 0.5。该积分等于 (2/11)[4x − x²/2]₁ᵐ = (2/11)[(4m − m²/2) − 3.5]。方程：2/11 + (2/11)[4m − m²/2 − 3.5] = 0.5。两边乘 11：2 + 2(4m − 0.5m² − 3.5) = 5.5 ⇒ 2 + 8m − m² − 7 = 5.5 ⇒ −m² + 8m − 5 = 5.5 ⇒ −m² + 8m − 10.5 = 0 ⇒ m² − 8m + 10.5 = 0。求解：m = [8 ± √64 − 42]/2 = [8 ± √22]/2。√22 ≈ 4.690，所以 m ≈ (8 − 4.69)/2 = 1.655（舍去大于 4 的大根）。因此中位数 ≈ 1.655。
This worked solution perfectly illustrates why the “joiner” g process, and the split integration at the join delivered the probability and median. Such questions regularly appear under various code numbers, and “pdfjoiner_(4)-169” captures their essence.
这道完整的例题完美地说明了为什么“连接点”逻辑是分段概率密度函数问题的核心：连续性方程给出了 a 与 b 的简单关系，总面积条件完成了求解，而在连接点处拆分积分则顺利得到了概率与中位数。这类题目经常以各种编号出现在试卷中，“pdfjoiner_(4)-169”恰好抓住了其精髓。
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