PDF Joiner (4)–175: Mastering Integration of Probability Density Functions | 概率密度函数合并(4)–175：掌握概率密度函数积分

📚 PDF Joiner (4)–175: Mastering Integration of Probability Density Functions | 概率密度函数合并(4)–175：掌握概率密度函数积分

In Edexcel A‑Level Mathematics, particularly within the Statistics component, continuous random variables are described by probability density functions (PDFs). The concept of a 'PDF joiner' often appears in exam questions where a piecewise‑defined function must be treated as a single valid PDF. This article breaks down all the essential skills you need to master such problems, using a classic style of question labelled (4)–175 to illustrate each step. We will integrate, differentiate, and reason about areas under curves to ensure that any joined‑up PDF behaves exactly as probability theory demands.

在爱德思A‑Level数学中，尤其是在统计部分，连续随机变量由概率密度函数（PDF）来描述。“PDF合并”的概念常出现在考试题中，需要将分段定义的函数视为一个有效的整体PDF。本文将分解你掌握这类问题所需的所有核心技能，并以一道标注为(4)–175的经典题型为例，逐步演示。我们将通过积分、微分以及对曲线下面积的推理，确保任何拼接的PDF都完全符合概率理论的要求。

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1. Understanding the Problem: What is a PDF Joiner? | 理解问题：什么是PDF合并？

A 'PDF joiner' refers to a probability density function that is defined by two or more separate formulas on adjacent intervals. The challenge is to 'join' these pieces together so that the overall function satisfies the two fundamental properties of every PDF: the total area under the graph equals 1, and the function is never negative. Typically, one or both pieces contain an unknown constant, often labelled k, which you must find. This process integrates the separate pieces over their respective domains and sets their sum equal to 1.

“PDF合并”指的是一个概率密度函数由两个或多个在相邻区间上分别定义的公式组成。挑战在于将这些片段“合并”起来，使整体函数满足每个PDF的两个基本性质：曲线下的总面积等于1，且函数永不取负值。通常，其中一个或两个片段包含一个未知常数（常记为k），你需要求出它。这一过程将各个片段在其各自定义域上积分，并令其总和等于1。

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2. Ensuring Total Probability Equals 1 | 确保总概率等于1

The cornerstone condition for any PDF f(x) is ∫_–∞^∞ f(x) dx = 1. When the function is non‑zero only on a finite interval [a, b], the integral reduces to ∫_a^b pieces, say f(x) = g(x) for a ≤ x < c and f(x) = h(x) for c ≤ x ≤ b, you must calculate ∫_a^c g(x) dx + ∫_c^b h(x) dx = 1. This equation is the key to finding any unknown constants.

任何PDF f(x) 的基石条件是∫_–∞^∞ f(x) dx = 1。当函数仅在有限区间 [a, b] 上非零时，该积分简化为∫_a^b f(x) dx = 1。对于一个由两段拼接而成的PDF，例如 f(x) = g(x) 当 a ≤ x < c，且 f(x) = h(x) 当 c ≤ x ≤ b，你必须计算∫_a^c g(x) dx + ∫_c^b h(x) dx = 1。这个方程就是求取所有未知常数的关键。

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3. Calculating the Constant k | 计算常数k

Consider a problem where f(x) = kx² for 0 ≤ x < 2 and f(x) = (3/8)(4 – x) for 2 ≤ x ≤ 4, with f(x) = 0 otherwise. To find k, integrate each piece: ∫₀² kx² dx = k [x³/3]₀² = 8k/3. The second piece gives ∫₂⁴ (3/8)(4 – x) dx = (3/8)[4x – x²/2]₂⁴. Evaluate the bracket: at x=4, 16 – 8 = 8; at x=2, 8 – k/3 + 3/4 = 1, solve to get 8k/3 = 1/4, so k = 3/32. This systematic approach never fails.

考虑一个具体问题：f(x) = kx² 当 0 ≤ x < 2，f(x) = (3/8)(4 – x) 当 2 ≤ x ≤ 4，其余地方 f(x) = 0。为求 k，需对各段积分：∫₀² kx² dx = k [x³/3]₀² = 8k/3。第二段给出∫₂⁴ (3/8)(4 – x) dx = (3/8)[4x – x²/2]₂⁴。计算括号值：x=4 时，16 – 8 = 8；x=2 时，8 – 2 = 6；差值为 2。乘以 3/8 得 6/8 = 3/4。令 8k/3 + 3/4 = 1，解得 8k/3 = 1/4，故 k = 3/32。这套系统的方法万无一失。

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4. Verifying the Function is Non‑Negative | 验证函数非负性

at f(x) ≥ 0 everywhere on its domain. For 0 ≤ x < 2, f(x) = (3/32)x², which is clearly non‑negative. For 2 ≤ x ≤ 4, (3/8)(4 – x) is also non‑negative because 4 – e joined PDF could dip below zero, the function would be invalid, and you would need to re‑examine the given forms or the values of the constants. Fortunately, well‑designed exam questions always yield a valid PDF once k is correct.

求出 k 后，需检查在其定义域内各处是否满足 f(x) ≥ 0。当 0 ≤ x < 2 时，f(x) = (3/32)x²，显然非负。当 2 ≤ x ≤ 4 时，(3/8)(4 – x) 也非负，因为在此区间上 4 – x ≥ 0。如果拼接后的PDF有任何部分降到零以下，该函数即为无效，你需要重新审视给定的表达式或常数的值。好在精心设计的考题在k值正确时总会给出一个有效的PDF。

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5. Finding Probabilities over Intervals | 求区间的概率

join point, you must split the integral at the junction. For our example, P(1 < X < 3) = ∫₁² (3/32)x² dx + ∫₂³ (3/8)(4 – x) dx. The first part: (3/32)[x³/3]₁² = (1/32)(8 – 1) = 7/32. The second: (3/8)[4x – x²/2]₂³; at x=3, 12 – 4.5 = 7.5; at x=2, 8 – 18/32 = 25/32. Getting into the habit of splitting at the join is essential for accuracy.

要求 P(a < X < b)，需在 a 与 b 之间对拼接的PDF积分。如果区间跨越了拼接点，你必须在交界处将积分分开。以我们的例子，P(1 < X < 3) = ∫₁² (3/32)x² dx + ∫₂³ (3/8)(4 – x) dx。第一部分：(3/32)[x³/3]₁² = (1/32)(8 – 1) = 7/32。第二部分：(3/8)[4x – x²/2]₂³；x=3 时，12 – 4.5 = 7.5；x=2 时，8 – 2 = 6；差值为 1.5，乘以 3/8 得 4.5/8 = 9/16。相加：7/32 + 9/16 = 7/32 + 18/32 = 25/32。养成在拼接处分开积分的习惯对准确性至关重要。

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6. Deriving the Cumulative Distribution Function | 推导累积分布函数

F), F(x) = P(X ≤ x), is built piece by piece. For x < 0, F(x) = 0. For 0 ≤ x < 2, F(x) = ∫₀^x (3/32)t² dt = (3/32)(x³/3) = x³/32. For 2 ≤ x ≤ 4, F(x) = F(2) + ∫₂^x (3/8)(4 – t) dt. F(2) = 2³/32 = 8/32 = 1/4. The integral from 2 to x: (3/8)[4t – t²/2]₂^x = (3/8)[(4x – x²/2) – (8 – 2)] = (3/8)(4x – x²/2 – 6). Thus for 2 ≤ x ≤ 4, F(x) = 1/4 + (3/8)(4x – x²/2 – 1. The CDF must be continuous at the join point; check that the two formulas meet at x=2.

累积分布函数（CDF）F(x) = P(X ≤ x) 需逐段构建。对于 x < 0，F(x) = 0。当 0 ≤ x < 2 时，F(x) = ∫₀^x (3/32)t² dt = (3/32)(x³/3) = x³/32。当 2 ≤ x ≤ 4 时，F(x) = F(2) + ∫₂^x (3/8)(4 – t) dt。F(2) = 2³/32 = 8/32 = 1/4。从 2 到 x 的积分：(3/8)[4t – t²/2]₂^x = (3/8)[(4x – x²/2) – (8 – 2)] = (3/8)(4x – x²/2 – 6)。因此在 2 ≤ x ≤ 4 上，F(x) = 1/4 + (3/8)(4x – x²/2 – 6)。当 x > 4 时，F(x) = 1。CDF 必须在拼接点处连续；请自行验证在 x=2 处两个公式是否衔接一致。

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7. Determining the Median and Quartiles | 确定中位数和四分位数

0.5. Here F(2) = 0.25 and F(4) = 1, so 0.5 lies in the second piece. Solve 1/4 + (3/8)(4m – m²/2 – 6) = 0.5. Subtract 1/4: (3/8)(4m – ½m² – 6) = 0.25. Multiply by 8/3: 4m – ½m² – 6 = 2/3. Multiply by 2: 8m – m² – 12 = 4/3, so –m² + 8m – 12 – 4/3 = 0 → m² – 8m + 12 + 4/3 = 0 → m² – 8m + 40/3 = 0. Multiply by 3: 3m² – 24m + 40 = 0. Use the quadratic formula: m = [24 ± √576 – 480]/6 = [24 ± √96]/6 = [24 ± 4√6]/6 = 4 ± (2√6)/3. The value must be between 2 and 4; 4 – (2√6)/3 ≈ 4 – 1.633 = 2.367, which is valid. Thus the median is 4 – (2√6)/3.

中位数 m 满足 F(m) = 0.5。首先确定 0.5 落在 CDF 的哪一段。此处 F(2) = 0.25，F(4) = 1，故 0.5 位于第二段。解方程 1/4 + (3/8)(4m – m²/2 – 6) = 0.5。减去 1/4：(3/8)(4m – ½m² – 6) = 0.25。两边乘以 8/3：4m – ½m² – 6 = 2/3。乘以 2：8m – m² – 12 = 4/3，移项得 –m² + 8m – 12 – 4/3 = 0 → m² – 8m + 12 + 4/3 = 0 → m² – 8m + 40/3 = 0。乘以 3：3m² – 24m + 40 = 0。使用求根公式：m = [24 ± √576 – 480]/6 = [24 ± √96]/6 = [24 ± 4√6]/6 = 4 ± (2√6)/3。该值必须介于 2 与 4 之间；4 – (2√6)/3 ≈ 4 – 1.633 = 2.367，符合要求。故中位数为 4 – (2√6)/3。

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8. Working with Piecewise‑Defined PDFs | 处理分段定义的PDF

Some joined PDFs use more than two pieces or employ different function types (linear, quadratic, constant). The same principles apply: integrate each piece over its own interval, sum to 1, and ensure non‑negativity. Always sketch the graph if possible. Visualising the 'joiner' as a literal joining of two curves helps you see whether the pieces connect smoothly or have a jump. In Edexcel exams, a jump at the join point is allowed because PDFs need not be continuous, only the total area matters. However, the CDF must always be continuous, so if the PDF has a jump, the CDF changes slope but does not break.

有些拼接的PDF使用多于两段，或包含不同类型的函数（线性、二次、常数）。适用的原理相同：将每一段在各自区间上积分，总和等于1，并确保非负性。尽可能画出草图。将“拼接”形象地看作两条曲线的实际连接，有助于你观察片段之间是平滑连接还是存在跳跃。在爱德思考试中，拼接点处的跳跃是允许的，因为PDF无需处处连续，只要总面积符合要求即可。然而，CDF 必须始终连续，因此若 PDF 存在跳跃，CDF 会改变斜率但不会断开。

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9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

main of each piece, especially when the definition uses inequalities like 0 ≤ x < 2 and 2 ≤ x ≤ 4 – forget to verify that f(x) ≥ 0; if k comes out negative, the PDF is invalid. Always double‑ lies within the correct interval. Practise with varied examples to build a robust routine.

一个常见错误是，在计算跨越拼接点的概率时忘记将积分分开。另一个错误是弄错每段的定义域，尤其在定义中使用诸如 0 ≤ x < 2 和 2 ≤ x ≤ 4 这样的不等式时——若你用第二段表达式从 0 积分到 2，就会出错。学生也常忽略验证 f(x) ≥ 0；如果算出的 k 为负数，该PDF即无效。务必再次检查 CDF 在上界处是否达到 1，以及中位数是否落在正确的区间内。通过多样化的练习来建立起稳固的解题流程。

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10. Exam‑Style Practice: (4)–175 Walkthrough | 考试风格练习：(4)–175 详细解答

Let's walk through a complete problem modelled on the (4)–175 style. A continuous random variable X has PDF given by f(x) = kx(3 – x) for 0 ≤ x < 1, and f(x) = (3/4)(2 – 1/2. (b) Find P(0.5 < X < 1.5). (c) Obtain the CDF F(x). (d) Determine the upper quartile.

我们来完整解答一道仿照(4)–175风格的题目。连续随机变量 X 的PDF为：f(x) = kx(3 – x) 当 0 ≤ x < 1，f(x) = (3/4)(2 – x) 当 1 ≤ x ≤ 2，其余为零。(a) 证明 k = 1/2。(b) 求 P(0.5 < X < 1.5)。(c) 求 CDF F(x)。(d) 确定上四分位数。

(a) Integrate piece 1: ∫₀¹ k(3x – x²) dx = k[(3/2)x² – x³/3]₀¹ = k(3/2 – 1/3) = k(9/6 – 2/6) = (7/6)k. Piece 2: ∫₁² (3/4)(2 – x) dx = (3/4)[2x – x²/2]₁² = (3/4)[(4 – 2) – (2 – 0.5)] = (3/4)(2 – ait, this does not give 1/2. I need to adjust the function so that k=1/2 indeed works. Let's design a problem where k=1/2. Let first piece be f(x) = kx(2 – x) for 0 ≤ x < 1, then ∫₀¹ k(2x – x²) dx = k[x² – x³/3]₀¹ = k(1 – 1/3) = (2/3)k. Second piece: f(x) = (3/4)(2 – integral 3/8. Then (2/3)k + 3/8 = 1 → (2/3)k = 5/8 → k = (5/8)×(3/2) = 15/16, not 1/2. Let's set second piece differently. Perhaps use linear second piece: f(x) = m(x – 2)? No. Simpler: use constant second piece? he ? We want to demonstrate the method, not necessarily get k=1/2 from real numbers unless I calculate appropriately. I'll construct a clean problem: f(x) = kx² for 0 ≤ x < 2, and f(x) = h(4 – x) for 2 ≤ x ≤ 4. Then compute. But the walkthrough should be consistent. I' the functions are appropriately chosen. I can fabricate numbers that work: Let f(x) = k(1 – x) for 0 ≤ x < 1? alkthrough, I can continue with the example I started but adjust numbers to get k=1/2. Let's define f(x) = kx(2 – cond piece on [1,2] is something like a constant c, then c*1 = 2/3, so c=2/3. But need non‑negative, okay. Then overall PDF: f(x)= (1/2)x(2 – x)= 2/3 for 1 ≤ x ≤ 2. That works. But then part (a) would be to show k=1/2 given the form? That's not interesting. I'll create a problem where k is in the second piece maybe. Let's do: f(x)= (1/2)x(2 – x) for 0 ≤ x < 1, f(x)= k(2 – l: using total area=1 to find k. In that case, piece1 integral = 1/3, piece2 integral = k∫₁² (2 – x) dx = k[2x – x²/2]₁² = k[(4 – 2) – (2 – 0.5)] = k(2 – .5) = 0.5k. So 1/3 + 0.5k =1 → 0.5k = 2/3 → k = 4/3. That is valid. Then (a) show k=4/3. I'll use that. So the problem: f(x)= (1/2)x(2 – x) for 0 ≤ x < 1, and f(x)= k(2 – x) for 1 ≤ x ≤ 2. (a) Find k. (b) Find P(0.5

Now construct section 10 with this specific example. I'll present it as a worked solution.

ields k=4/3. (b) Probability across join: split at 1. (c) CDF piecewise: for 0≤x<1, F(x)=∫₀^x (1/2)t(2 – t) dt = (1/2)[t² – t³/3]₀^x = (1/2)(x² – x³/3). For 1≤x≤2, F(x)=F(1)+∫₁^x (4/3)(2 – t) dt. F(1) from first piece: (1/2)(1 – 1/3)= (1/2)(2/3)=1/3. The integral: (4/3)[2t – t²/2]₁^x = (4/3)[(2x – x²/2) – (2 – 0.5)] = (4/3)(2x – x²/2 – 1.5). So F(x)=1/3 + (4/3)(2x – x²/2 – satisfies F(Q3)=0.75. Check F(2)=1, so likely Q3 is in second piece. Solve 1/3 + (4/3)(2q – q²/2 – 1.5) = 0.75. Multiply by 3: 1 + 4(2q – 0.5q² – 1.5) = 2.25 → 4(2q – 0.5q² – 1.5) = 1.25 → 2q – 0.5q² – 1.5 = 0.3125 → multiply by 2: 4q – q² – 3 = 0.625 → –q² + 4q – 3.625 = 0 → q² – 4q + 3.625 =0. Solve: q = [4 ± √16 – 14.5]/2 = [4 ± √1.5]/2 = 2 ± (√1.5)/2. Only 2 – (√1.5)/2 ≈ 2 – 0.612 = 1.388 is within [1,2]. So upper quartile ≈ 1.388. Provide exact form: 2 – √3/8 maybe? √1.5/2 = √3/2/2 = √3/(2√2) = √6/4. So Q3 = 2 – √6/4. Good.

Now write the section.

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10. Exam‑Style Practice: (4)–175 Walkthrough | 考试风格练习：(4)–175 详细解答

Let' e everything with a typical joiner question. The random variable X has PDF f(x) = (1/2)x(2 – x) for 0 ≤ x < 1, and f(x) = k(2 – ). (c) Derive the cumulative distribution function F(x). (d) Find the upper quartile of X.

我们通过一道典型的拼接题来巩固所有内容。随机变量 X 的 PDF 为 f(x) = (1/2)x(2 – x) 当 0 ≤ x < 1，f(x) = k(2 – x) 当 1 ≤ x ≤ 2，其余为零。(a) 证明 k = 4/3。(b) 计算 P(0.5 < X < 1.5)。(c) 推导累积分布函数 F(x)。(d) 求 X 的上四分位数。

(a) Integrate the first piece: ∫₀¹ (1/2)(2x – x²) dx = (1/2)[x² – x³/3]₀¹ = (1/2)(1 – 1/3) = 1/3. Second piece: ∫₁² k(2 – x) dx = k[2x – x²/2]₁² = k[(4 – 2) – (2 – 0.5)] = k(2 – 1.5) = 0.5k. Total area: 1/3 + 0.5k = 1 → 0.5k = 2/3 → k = 4/3.

(b) P(0.5 < X < 1.5) = ∫_0.5¹ (1/2)x(2 – x) dx + ∫₁ Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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