📚 Probability Density Functions: Joining Piecewise Defined PDFs | 概率密度函数：分段函数的拼接
In Edexcel A-Level Mathematics, continuous random variables are modelled by probability density functions (PDFs). Sometimes a PDF is defined by different formulas on adjacent intervals, and these pieces must join smoothly, or at least satisfy continuity requirements, at the boundaries. Understanding how to join piecewise defined PDFs is essential for solving problems where unknown constants appear in different intervals and for calculating probabilities, cumulative distribution functions, medians, and percentiles from such models.
在 Edexcel A-Level 数学中，连续随机变量由概率密度函数 (PDF) 建模。有时一个 PDF 由相邻区间上的不同表达式定义，这些分段必须在边界处平滑连接，或至少满足连续性的要求。理解如何拼接分段定义的 PDF，对于解决不同区间上含有未知常数的问题，以及从此类模型中计算概率、累积分布函数、中位数和百分位数至关重要。
1. Introduction to Probability Density Functions | 概率密度函数简介
e probability mass functions, f(x) itself is not a probability; probability is given by the area under the curve. For any interval [a,b], P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx.
连续随机变量 X 的概率密度函数 f(x) 描述了 X 取值在 x 附近的相对可能性。与离散概率质量函数不同，f(x) 本身不是概率；概率由曲线下方区域的面积给出。对任意区间 [a,b]，P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx。
1, i.e., ∫ₐₗₗ f(x) dx = 1. When the domain is restricted to a specific interval, the integral is taken over that interval.
要使 f(x) 成为有效的 PDF，必须满足两个关键条件：第一，对于定义域内所有 x，f(x) ≥ 0；第二，曲线下的总面积必须等于 1，即 ∫全空间 f(x) dx = 1。当定义域限制在一个特定区间时，积分就在该区间上进行。
2. Conditions for a Valid PDF | 有效概率密度函数的条件
erval [α, β] (α and β can be -∞ and ∞). Then f(x) must be non-negative: f(x) ≥ 0 for all x ∈ iecewise on sub-intervals, the sum of the integrals over those sub-intervals must equal 1.
设 X 为连续随机变量，其 PDF f(x) 定义在区间 [α, β] 上（α 和 β 可为 -∞ 和 ∞）。则 f(x) 必须非负：对所有 x ∈ [α, β] 有 f(x) ≥ 0。此外，归一化条件要求 ∫αᵝ f(x) dx = 1。如果 f(x) 在子区间上分段定义，那么这些子区间上的积分之和必须等于 1。
In Edexcel exam questions, you are often given a piecewise function with one or more unknown constants, and you must use the total area condition to find them. Always check non-negativity after evaluating constants to confirm the PDF is valid over the given domain.
在 Edexcel 考试题中，常会给出一个含有一个或多个未知常数的分段函数，你必须利用总面积条件求出它们。在求出常数后，务必检查非负性，以确认 PDF 在给定定义域上是有效的。
3. Piecewise Defined Functions | 分段定义的函数
ve linearly on [0, 1] and quadratically on [1, 2], with the two pieces joining at x = 1. The pieces are typically written as f(x) = { f₁(x), α ≤ x < c; f₂(x), c ≤ x ≤ β } for some junction point c.
分段定义的 PDF 在其定义域内的不同子区间上采用不同的代数形式。例如，一个随机变量可能在 [0, 1] 上呈线性，在 [1, 2] 上呈二次函数，两段在 x = 1 处相连接。各分段通常写作 f(x) = { f₁(x), α ≤ x < c; f₂(x), c ≤ x ≤ β }，其中 c 为连接点。
At the junction point, the function may or may not be continuous, but for many practical models continuity is required, or it naturally follows from the integration conditions. Edexcel problems often specify that the pieces join to form a continuous PDF, which provides an additional equation linking the unknown parameters.
在连接点处，函数可能是连续的，也可能不连续，但在许多实际模型中连续性是被要求的，或者由积分条件自然地得到。Edexcel 的题目通常会指定各分段拼接成一个连续的 PDF，这就提供了联系未知参数的额外方程。
4. Joining PDFs at Boundaries | 在边界处拼接概率密度函数
f(x) = { ax, 0 ≤ x < 2; b(4 - x), 2 ≤ x ≤ 4 } where a and b are constants. The pieces join at x = 2. If the PDF is continuous, then the left-hand limit and right-hand limit at x = 2 must be equal: a·2 = b(4 - 2) → 2a = 2b → a = b.
拼接分段 PDF 意味着确保函数在边界点处定义明确，并满足总概率要求。设 f(x) = { ax, 0 ≤ x < 2; b(4 - x), 2 ≤ x ≤ 4 }，其中 a 和 b 为常数。各段在 x = 2 处拼接。如果 PDF 是连续的，那么在 x = 2 处的左极限和右极限必须相等：a·2 = b(4 - 2) → 2a = 2b → a = b。
ngle point which has zero probability). The normalisation integral is split into two parts: ∫₀² ax dx + ∫₂⁴ b(4 - x) dx = 1. This single equation might determine a relationship between a and b, and an extra condition (like a given probability or median) will fix them.
如果没有给定连续性条件，各分段仍然在区间划分整个定义域这个意义上拼接，彼此不重叠（除可能在单一点，该点概率为零）。归一化积分分为两部分：∫₀² ax dx + ∫₂⁴ b(4 - x) dx = 1。这个方程可以确定 a 与 b 的关系，额外的条件（如给定概率或中位数）则可求出具体数值。
5. Example: Joining Two Linear Functions | 示例：拼接两个线性函数
Consider a continuous random variable X with PDF given by:
考虑连续随机变量 X，其 PDF 为：
f(x) = { kx, 0 ≤ x ≤ 3; ½k(6 - x), 3 < x ≤ 6; 0, otherwise.
we need an adjustment: perhaps the second piece is m(6 - x) with different constants. Let' ≤ x ≤ 4 } as before, and find a and b such that the PDF is continuous and integrates to 1.
分段在 x = 3 处拼接。x = 3 处的连续性要求：k(3) = ½k(6 - 3) → 3k = 1.5k。要对非零 k 成立，我们需要调整：可能第二段是 m(6 - x) 且常数不同。让我们用一个更好的例子：f(x) = { a x, 0 ≤ x < 2; b(4 - x), 2 ≤ x ≤ 4 }，并求出 a 和 b 使得 PDF 连续且积分值为 1。
6a = 4a. Set 4a = 1 → a = 0.25. Thus f(x) = { 0.25x, 0 ≤ x < 2; 0.25(4 - x), 2 ≤ x ≤ 4 }.
由 x = 2 处的连续性得：2a = b(4 - 2) = 2b → a = b。总面积： dx = [a x²/2]₀² + [4a x - a x²/2]₂⁴ = (a·4/2) + (16a - 8a) - (8a - 2a) = 2a + 8a - 6a = 4a。令 4a = 1 → a = 0.25。因此 f(x) = { 0.25x, 0 ≤ x < 2; 0.25(4 - x), 2 ≤ x ≤ 4 }。
6. Determining Unknown Constants | 确定未知常数
egral of f(x) over the entire domain, splitting at the junction points. (2) Compute the definite integrals, keeping constants as symbols. (3) Set the sum equal to 1. (4) If continuity is specified, equate the left and right expressions at the junction to obtain another equation. (5) Solve the system for the constants. Always verify that f(x) ≥ 0.
当分段 PDF 含有未知参数时，利用总面积条件和连续性来建立方程。步骤：(1) 写出 f(x) 在整个定义域上的积分，并在连接点处分开。(2) 计算定积分，常数保留为符号。(3) 令总和等于 1。(4) 如果指定了连续性，则令连接点处左右表达式相等，得到另一个方程。(5) 求解方程组得到常数。务必验证 f(x) ≥ 0。
+ d (8/3 + 1/3) = 2c + 3d = 1. Continuity: c = d(2 - 3)² = d. So c = d, then 2c + 3c = 5c = 1 → c = 0.2, d = 0.2.
示例：f(x) = { c, 0 ≤ x < 2; d(x - 3)², 2 ≤ x ≤ 5 }，在 x = 2 处连续。条件 1：∫₀² c dx + ∫₂⁵ d(x - 3)² dx = 2c + d ∫₋₁² u² du（令 u = x - 3）= 2c + d [u³/3]₋₁² = 2c + d (8/3 + 1/3) = 2c + 3d = 1。连续性：c = d(2 - 3)² = d。所以 c = d，则 2c + 3c = 5c = 1 → c = 0.2, d = 0.2。
7. Calculating Probabilities from a Joined PDF | 由拼接PDF计算概率
near PDF f(x) = { 0.25x, 0 ≤ x < 2; 0.25(4 - x), 2 ≤ x ≤ 4 }, compute P(1 < X < 3). We split at x = 2: P = ∫₁² 0.25x dx + ∫₂³ 0.25(4 - x) dx. The first part: 0.25 [x²/2]₁² = 0.25 (2 - 0.5) = 0.375. Second part: 0.25 [4x - x²/2]₂³ = 0.25 [(12 - 4.5) - (8 - 2)] = 0.25 (7.5 - 6) = 0.375. Total probability = 0.75.
要计算 P(a < X < b)，对 f(x) 在 [a, b] 上积分，注意在区间内的任何分段连接点处划分积分。对于拼接的线性 PDF f(x) = { 0.25x, 0 ≤ x < 2; 0.25(4 - x), 2 ≤ x ≤ 4 }，计算 P(1 < X < 3)。我们在 x = 2 处分割：P = ∫₁² 0.25x dx + ∫₂³ 0.25(4 - x) dx。第一部分：0.25 [x²/2]₁² = 0.25 (2 - 0.5) = 0.375。第二部分：0.25 [4x - x²/2]₂³ = 0.25 [(12 - 4.5) - (8 - 2)] = 0.25 (7.5 - 6) = 0.375。总概率 = 0.75。
If the problem asks for P(X > k) where k is in a specific piece, integrate from k to the end of that piece then add integrals over subsequent pieces fully. Always sketch the PDF to avoid errors.
如果题目要求计算 P(X > k) 且 k 位于特定分段中，则从 k 积分至该分段末端，再加上后续完整分段的积分。一定要绘制 PDF 的草图，以避免错误。
8. Finding the Cumulative Distribution Function (CDF) | 求解累积分布函数(CDF)
The cumulative distribution function F(x) = P(X ≤ x) = ∫₋∞ˣ d of the domain up to x, splitting at junction points. For x in the first piece, F(x) = ∫αˣ f₁(t) dt. For x in the second piece, F(x) = ∫αᶜ f₁(t) dt + ∫cˣ f₂(t) dt, where c is the junction.
累积分布函数 F(x) = P(X ≤ x) = ∫₋∞ˣ f(t) dt。对于拼接的 PDF，CDF 也是分段定义的。你必须从定义域的下界积分到 x，并在连接点处分开。对于位于第一段的 x，F(x) = ∫αˣ f₁(t) dt。对于位于第二段的 x，F(x) = ∫αᶜ f₁(t) dt + ∫cˣ f₂(t) dt，其中 c 为连接点。
Using the linear example with a = 0.25: For 0 ≤ x < 2, F(x) = ∫₀ˣ 0.25t dt = 0.125x². For 2 ≤ x ≤ 4, F(x) = ∫₀² 0.25t dt + ∫₂ˣ al is 0.5 (since total area for piece one is 2a = 0.5). The second part: 0.25 [4t - t²/2]₂ˣ ct. At x=2, F(2)=2 - 0.5 - 1 = 0.5, matching the first piece limit. The CDF is continuous.
使用线性示例 a = 0.25：对于 0 ≤ x < 2，F(x) = ∫₀ˣ 0.25t dt = 0.125x²。对于 2 ≤ x ≤ 4，F(x) = ∫₀² 0.25t dt + ∫₂ˣ 0.25(4 - t) dt。已知第一个积分为 0.5（因为第一段总面积为 2a = 0.5）。第二部分：0.25 [4t - t²/2]₂ˣ = 0.25[(4x - x²/2) - (8 - 2)] = 0.25(4x - x²/2 - 6) = x - (1/8)x² - 1.5。加上 0.5 得 F(x) = x - (1/8)x² - 1，其中 2 ≤ x ≤ 4。检验 x=4：F(4)=4 - 2 - 1 = 1，正确。在 x=2 处，F(2)=2 - 0.5 - 1 = 0.5，与第一段极限相符。CDF 是连续的。
9. Median and Percentiles for Joined PDFs | 拼接PDF的中位数和百分位数
to the junction. If F(c) < 0.5, then m lies in a later piece. Solve F(m) = 0.5 using the expression for that piece. For the linear PDF, F(2) = 0.5 exactly, so the median is m = 2.
中位数 m 满足 F(m) = 0.5。由于 CDF 是分段的，你首先需比较到连接点处的累积概率，以确定 m 位于哪一段。如果 F(c) < 0.5，则 m 位于后面的一段中。利用该段的表达式解 F(m) = 0.5。对于线性 PDF，F(2) = 0.5 正好，因此中位数为 m = 2。
- 0.125x² - 1 = 0.9 → -0.125x² + x - 1.9 = 0 → multiply by -8: x² - 8x + 15.2 = 0. Solve: x = [8 ± √64 - 60.8]/2 = [8 ± √3 check if the solution falls within the assumed piece; otherwise, recalculate using the correct piece.
对于其他百分位数，如第 90 百分位数 x₀.₉，解 F(x) = 0.9。在同一 PDF 中，F(2)=0.5，F(4)=1，因此 x₀.₉ 位于 [2,4] 内。令 x - (1/8)x² - 1 = 0.9 → x - 0.125x² - 1 = 0.9 → -0.125x² + x - 1.9 = 0 → 两边乘以 -8：x² - 8x + 15.2 = 0。求解：x = [8 ± √64 - 60.8]/2 = [8 ± √3.2]/2 ≈ (8 ± 1.7889)/2。正根：(9.7889)/2 ≈ 4.894，超出了 [2,4]；另一根：(6.2111)/2 ≈ 3.1056。所以 x₀.₉ ≈ 3.11（有效）。务必检查解是否落在所假设的分段内；否则，应使用正确的分段重新计算。
10. Common Pitfalls and Examination Tips | 常见陷阱与考试技巧
umber line and mark the intervals. Another common mistake is to assume continuity where it is not given; use the total area condition first, then apply continuity only if the problem explicitly states it or if it is needed to find the constants. Do not forget to check f(x) ≥ 0; a negative constant could invalidate the PDF even if integrals sum to 1.
许多学生在计算概率或 CDF 时忘记在连接点处拆分积分，导致错误的积分限。一定要画数轴并标出区间。另一个常见错误是在题目没有给定时就假定连续性；应首先运用总面积条件，只有在题目明确说明或需要求出常数时才使用连续性。切勿忘记检查 f(x) ≥ 0；即使积分之和为 1，负的常数也可能使 PDF 失效。
5. Also, practice algebraic manipulation of quadratic equations arising from percentile problems, and be prepared to work with surds or decimals. Sketching the PDF provides a visual check on the shape and the position of the median.
在求中位数时，要验证解是否落在所用的分段内。在某些情况下，当 F(c) = 0.5 时，中位数可能就是连接点本身。此外，应练习处理由百分位数问题产生的二次方程的代数运算，并准备好处理无理数或小数。绘制 PDF 草图有助于从直观上检查形状及中位数的位置。
In Edexcel exams, these topics frequently appear in Statistics 2 (S2) or the new specification's Statistics content. Marks are awarded for setting up the integrals correctly, solving for constants, and finding probabilities and the CDF. Show full integration steps and clearly state the final piecewise PDF and CDF.
在 Edexcel 考试中，这些主题经常出现在 Statistics 2 (S2) 或新大纲的统计内容中。正确建立积分、求解常数以及计算概率和 CDF 都能得分。要展示完整的积分步骤，并清晰地写出最终的分段 PDF 和 CDF。
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