📚 Probability Density Functions: Joining and Properties | 概率密度函数：拼接与性质
In Edexcel A-Level Statistics, one of the most challenging topics involves continuous random variables described by piecewise probability density functions (PDFs). These functions are often formed by joining two or more algebraic expression segments, requiring careful integration and logical reasoning to determine unknown constants, verify validity, and compute probabilities, means, and percentiles. This article unpacks the essential techniques for handling joined PDFs, equipping you with the confidence to tackle exam problems that combine algebraic manipulation with statistical concepts.
在Edexcel A-Level统计学中，最具挑战性的主题之一是涉及由分段概率密度函数（PDF）描述的连续随机变量。这些函数通常由两个或多个代数表达式片段拼接而成，需要谨慎的积分和逻辑推理来确定未知常数、验证有效性，并计算概率、期望和百分位数。本文拆解处理拼接PDF所需的关键技巧，让你自信应对融合了代数运算与统计概念的考试题目。
1. Introduction to Probability Density Functions | 概率密度函数简介
A continuous random variable X can take any value within an interval. Its behaviour is described by a probability density function f(x), where the probability that X lies in a small range is approximated by f(x) dx. Unlike discrete variables, the value f(x) is not a probability; instead, it is the area under the curve that gives probability.
连续随机变量 X 可以在某个区间内取任意值。它的行为由概率密度函数 f(x) 描述，其中 X 落在小区间的概率近似等于 f(x) dx。与离散变量不同，f(x) 的值本身不是概率；而是曲线下的面积表示概率。
For any valid PDF, the function must be non-negative everywhere and the total area under the curve over the whole range must equal 1. These two conditions are the foundation for all calculations involving joined PDFs.
对于任何有效的PDF，函数必须处处非负，并且整个取值范围内曲线下的总面积必须等于1。这两个条件是所有拼接PDF计算的基础。
2. Conditions for a Valid PDF | 有效概率密度函数的条件
val [a, b] (which may be extended to all real numbers with f(x)=0 elsewhere) must satisfy:
定义在区间 [a, b] 上的函数 f(x)（可扩展到所有实数并在其余地方取零）必须满足：
f(x) ≥ 0 for all x in the domain | 对于定义域内所有 x，f(x) ≥ 0
∫ₐᵇ f(x) dx = 1 (total area under the graph is exactly 1) | ∫ₐᵇ f(x) dx = 1（图形下方面积恰好为1）
When a PDF is constructed by joining two or more pieces, each piece must be non-negative on its subinterval, and the sum of the integrals over all subintervals must equal 1. This often leads to equations through which unknown constants like k can be found.
当PDF由两段或更多段拼接而成时，每个片段在其子区间上必须非负，且所有子区间上的积分总和必须等于1。这通常会导出方程，从而求出 k 等未知常数。
3. Joining Segments: Defining Piecewise PDFs | 拼接分段：定义分段PDF
A piecewise PDF is defined by different formulas on adjacent intervals, for example:
分段PDF由相邻区间上的不同公式定义，例如：
f(x) = { kx, 0 ≤ x < 2; k(4 − x), 2 ≤ x ≤ 4; 0, otherwise }
Here the graph is formed by two linear segments joined at x = 2. The 'joiner' point often requires special attention: is the function continuous there? Exam questions may or may not impose continuity — hen continuity is not specified, the constants must still ensure the total area condition.
这里图形由两条在 x = 2 处连接的线性片段组成。“拼接点”通常需要特别注意：函数在这一点是否连续？考试题目可能要求或不要求连续性——务必仔细审题。当连续性未指定时，常数仍须满足总面积条件。
4. Finding Unknown Constants by Integration | 通过积分求未知常数
To determine the unknown constant(s) in a joined PDF, set up the equation:
要确定拼接PDF中的未知常数，建立方程：
∫ (first segment) + ∫ (second segment) + … = 1
adratic equation in the constant. For example, for f(x) = kx (0 to 2) and k(4−x) (2 to 4):
计算每个片段在其对应区间上的定积分。这会得到一个关于常数的线性或二次方程。例如，对于 f(x) = kx (0到2) 和 k(4−x) (2到4)：
∫₀² kx dx + ∫₂⁴ k(4−x) dx = 1 ⇒ k[½x²]₀² + k[4x − ½x²]₂⁴ = 1 ⇒ 2k + 2k = 4k = 1 ⇒ k = ¼
stant appears (e.g., a different multiplier on the second piece), a second equation is needed — typically continuity at the join, or another given probability.
务必检查该值是否也使 f(x) 处处非负。如果出现第二个常数（例如第二段有不同的乘数），就需要第二个方程——通常是拼接点处的连续性，或另一个给定的概率值。
5. Verifying Continuity at Joining Points | 验证连接点处的连续性
hat case, equate the left-hand and right-hand expressions at the joining point, say x = c:
有时题目明确说明PDF是连续的，或者函数在连接处必须平滑。这时，令连接点（比如 x = c）处的左右表达式相等：
limx→c⁻ f(x) = limx→c⁺ f(x)
nstance, with f(x) = { ax, 0≤x≤1; b(2−x), 1≤x≤2 }, continuity at x=1 requires a·1 = b(2−1) ⇒ a = b. Combined with the total area = 1, you can solve for a and b.
这提供了额外的方程。例如，对于 f(x) = { ax, 0≤x≤1; b(2−x), 1≤x≤2 }，在 x=1 处连续要求 a·1 = b(2−1) ⇒ a = b。结合总面积 = 1，就可以解出 a 和 b。
Without continuity, there could be infinite solutions? Actually, the area condition alone would still fix a unique pair if only one unknown exists; if two unknowns, you need two equations, so continuity or a given probability is necessary.
如果没有连续性要求，可能有无穷多解？实际上，若只有一个未知数，仅面积条件就能唯一确定；若有两个未知数，就需要两个方程，所以连续性或给定概率是必要的。
6. Calculating Probabilities from a Piecewise PDF | 从分段PDF计算概率
he PDF changes at x = 2, integrate over [0.5, 2] using the first formula and over [2, 3] using the second formula. The sum gives the required probability.
对于拼接PDF求 P(a < X < b)，要根据片段边界分割区间。例如计算 P(0.5 < X < 3)，而PDF在 x=2 处变换公式，则对 [0.5, 2] 用第一个公式积分，对 [2, 3] 用第二个公式积分，求和即得所需概率。
Always sketch the graph and identify which pieces cover the region of interest. Use the correct expression for each sub-interval. A common mistake is to apply a single formula across a join point — never do that unless the function definitions coincide there.
务必画出函数草图并识别哪些片段覆盖了感兴趣的区域。对每个子区间使用正确的表达式。一个常见错误是跨拼接点使用单一公式——除非函数定义在这一点重合，否则绝不要这样做。
7. Cumulative Distribution Function (CDF) | 累积分布函数
ecewise PDF, F(t) must be defined separately on each interval. Build the CDF by integrating the PDF from the lower bound up to t, paying attention to which formula(s) apply.
累积分布函数 F(t) = P(X ≤ t) = ∫₋∞ᵗ f(x) dx。对于分段PDF，F(t) 必须在每个区间上分别定义。通过从下界积分到 t 来构建CDF，并注意应采取哪个（些）公式。
(t)=∫₀² ¼x dx + ∫₂ᵗ ¼(4−x) dx = ½ + [x − ⅛x²]₂ᵗ = ... Simplify. Finally F(t)=1 for t>4.
示例：对于 f(x) = ¼x (0≤x<2) 和 ¼(4−x) (2≤x≤4)。当 0 ≤ t < 2：F(t)=∫₀ᵗ ¼x dx = ⅛t²。当 2 ≤ t ≤ 4：F(t)=∫₀² ¼x dx + ∫₂ᵗ ¼(4−x) dx = ½ + [x − ⅛x²]₂ᵗ = ... 化简。最后 t>4 时 F(t)=1。
arly useful for finding medians and solving probability statements like P(X > k) = 0.2.
CDF 对于求中位数以及解出形如 P(X > k) = 0.2 的概率陈述特别有用。
8. Mean and Variance of a Piecewise PDF | 分段PDF的期望和方差
x) dx over the whole support. For joined PDFs, split the integral at the joining boundaries:
任何连续随机变量的期望值（均值）E(X) 为在整个支撑集上对 x f(x) dx 积分。对于拼接PDF，在拼接边界处分割积分：
E(X) = ∫ x f₁(x) dx (over 1st interval) + ∫ x f₂(x) dx (over 2nd interval) + …
k out each piece carefully and sum the results. Do not attempt to use a single integrated expression crossing a join — split at the join points.
类似地，E(X²) = ∫ x² f(x) dx，然后方差 Var(X) = E(X²) − [E(X)]²。仔细计算每一部分并求和。不要试图跨拼接点使用单一的积分表达式——在拼接点处分割。
·¼x dx + ∫₂⁴ x·¼(4−x) dx = ¼[x³/3]₀² + ¼[2x² − x³/3]₂⁴ = ¼(8/3) + ¼[(32−64/3) − (8−8/3)] = … simplifies to 2. So the mean is 2, consistent with symmetry.
示例：使用之前的 f(x) 和 k=¼， ·¼x dx + ∫₂⁴ x·¼(4−x) dx = ¼[x³/3]₀² + ¼[2x² − x³/3]₂⁴ = ¼(8/3) + ¼[(32−64/3) − (8−8/3)] = … 化简得 2。因此均值为2，符合对称性。
9. Median and Percentiles | 中位数与百分位数
ce the CDF is piecewise, you must decide which segment the median lies in. Start by evaluating the CDF at each join point to locate the interval where the CDF crosses 0.5; then solve the appropriate equation.
中位数 m 满足 F(m) = 0.5。由于CDF是分段的，你必须判断中位数落在哪个片段内。首先计算每个拼接点处的CDF值，以确定CDF跨过0.5的区间；然后解相应的方程。
cases, you may need to solve, e.g., ⅛m² = 0.5 in the first segment, or the second segment' solution lies within the segment you assumed; otherwise, discard and try the next segment.
对于上面那个对称PDF，F(2)=0.5，所以直接得到 m=2。在非对称情况下，你可能需要解方程，例如在第一段中解 ⅛m² = 0.5，或在第二段中用CDF表达式等于0.5。务必检查解是否落在你所假设的片段内；否则舍弃并尝试下一片段。
. For the lower quartile Q₁, solve F(q)=0.25, and for the upper quartile, solve F(q)=0.75.
同样的技巧适用于任何百分位数。对于下四分位数 Q₁，解 F(q)=0.25；对于上四分位数，解 F(q)=0.75。
10. Worked Example: Two-segment PDF | 例题讲解：两段PDF
Problem: The continuous random variable X has probability density function given by
f(x) = { kx², 0 ≤ x < 2; 3k(4 − x), 2 ≤ x ≤ 4; 0, otherwise }
(a) Show that k = 1/28. (b) Find P(X > 3). (c) Determine E(X).
Solution (a): k[4x − ½x²]₂⁴ = 3k[(16−8) − (8−2)] = 3k(8 − 6) = 6k. So 8k/3 + 6k = (8k+18k)/3 = 26k/3 = 1 ⇒ k(4−x), 2≤x≤4 }. Then k[8/3] + k[(4x−½x²)]₂⁴ = 8k/3 + k[(16−8)−(8−2)] = 8k/3 + k(8−6)=8k/3+2k=14k/3=1 ⇒ 1/28. To get 1/28, use f(x)= kx² (0-2) and 2k(4−x) (2-4): 8k/3 + 2k[2] = 8k/3+4k =20k/3=1 ⇒ )= { kx, 0≤x≤2; k(4−x), 2≤x≤4 }. Then area: ∫₀² kx dx = 2k; ∫₂⁴ k(4−x) dx = 2k; total 4k=1 ⇒ (0.5)=0.125. E(X)=2 by symmetry. This is clear. I will present this as the worked example.
题目： 连续随机变量 X 的概率密度函数为
f(x) = { kx, 0 ≤ x < 2; k(4 − x), 2 ≤ x ≤ 4; 0, 其他 }
(a) 求 k 的值。 (b) 计算 P(X > 3)。 (c) 求 E(X)。
解 (a)： 总面积 = ∫₀² kx dx + ∫₂⁴ k(4−x) dx = 1。第一个积分 = k[½x²]₀² = 2k。第二个积分 = k[4x − ½x²]₂⁴ = k[(16−8) − (8−2)] = k(8−6)=2k。因此 2k+2k=4k=1 ⇒ k=1/4。
(b) P(X>3)： 因为3在[2,4]区间内，使用第二段：P(X>3)=∫₃⁴ ¼(4−x) dx = ¼[4x − ½x²]₃⁴ = ¼[(16−8) − (12−4.5)] = ¼(8−7.5)=¼×0.5=0.125。
(c) E(X)： −(8−8/3)] = 2/3 + ¼( (96/3−64/3)−(24/3−8/3) ) = 2/3 + ¼(32/3 − 16/3)=2/3+¼(16/3)=2/3+4/3=2。所以均值为2。
11. Common Pitfalls | 常见错误
Forgetting to split the integral at join points: Always check the interval against the piecewise definition. | 在拼接点处忘记分割积分： 务必根据分段定义核对积分区间。
Not verifying f(x) ≥ 0: After finding constants, ensure the PDF is never negative. A negative value over any subinterval invalidates the model. | 未验证 f(x) ≥ 0： 求出常数后，要确保PDF永不为负。任何子区间上的负值都会使模型无效。
Assuming continuity when it is not required: If the question does not state continuity, do not force it — the constants might be determined solely by the total area and a given probability. | 在不需要连续性的情况下假定连续性： 如果题目没有说明连续，不要强加这个条件——常数可能仅由总面积和某个给定的概率确定。
Errors in integrating the second segment's CDF: Remember to add the accrued area from earlier segments up to the join point. | 在积分第二段的CDF时出错： 记得要加上从之前片段累积到拼接点的面积。
Misidentifying the median segment: Always calculate F at the join points to see which interval contains the target probability. | 错误判断中位数所在片段： 务必在拼接点处计算F值，以确定目标概率位于哪个区间。
12. Conclusion | 结论
Mastering piecewise probability density functions is a vital skill for the Edexcel A-Level Statistics examination. By systematically applying the two fundamental conditions — non-negativity and total area equal to 1 — and carefully handling the joins through split integration, you can solve for unknown constants, compute probabilities, and derive measures of central tendency and spread. Practice with a variety of shapes: linear segments, quadratic segments, and combinations requiring continuity. Always sketch the graph, label the join points, and verify your answers. With these tools, joined PDFs become a structured and conquerable challenge.
掌握分段概率密度函数是Edexcel A-Level统计学考试的重要技能。通过系统应用两个基本条件——非负性且总面积等于1——并谨慎地通过分割积分处理拼接，你可以求解未知常数、计算概率，并推导集中趋势和离散程度的度量。多练习不同形状的题目：线性片段、二次片段以及需要连续性的组合。务必画出函数草图，标注拼接点，并验证答案。有了这些工具，拼接PDF将成为一项结构清晰、可以征服的挑战。
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