📚 Probability Density Functions and Cumulative Distribution Functions | 概率密度函数与累积分布函数
In A‑Level Statistics, understanding the link between a probability density function (PDF) and its cumulative distribution function (CDF) is essential. The process that 'joins' a PDF to its CDF is integration, while differentiation takes you back. This article unpacks both concepts, explores how they are validated, and shows how to locate key features such as the median, mode, expectation and variance. By working through a uniform distribution example and a typical exam‑style problem, you will gain the confidence needed for Edexcel S2 or the continuous random variables topic in the new specification.
在A‑Level统计学中，理解概率密度函数（PDF）与累积分布函数（CDF）之间的联系至关重要。将PDF“连接”到CDF的是积分过程，而微分则让你返回。本文将剖析这两个概念，探讨如何验证它们，并展示如何定位中位数、众数、期望与方差等关键特征。通过一个均匀分布的例子和一道典型考题，你将获得应对Edexcel S2或新大纲连续随机变量主题所需的信心。
1. Continuous Random Variables and Why We Need PDFs | 连续随机变量与为什么需要PDF
A continuous random variable can take any value within an interval or union of intervals. Unlike a discrete variable, you cannot list all possible outcomes, so we describe its behaviour with a probability density function f(x). The function f(x) itself does not give a probability directly; instead, the area under the curve f(x) between two points gives the probability that the variable lies in that interval.
连续随机变量可以在某个区间或区间的并集内取任意值。与离散变量不同，你无法列出所有可能的结果，因此我们用概率密度函数 f(x) 来描述其行为。函数 f(x) 本身并不直接给出概率；相反，f(x) 曲线下两点之间的面积给出了变量落在那段区间内的概率。
2. Probability Density Function (PDF) | 概率密度函数（PDF）
omain, and the total area under the curve over the entire domain equals 1. Symbolically, ∫all x f(x) dx = 1. The probability that X lies between a and b is P(a < X < b) = ∫ab f(x) dx. Because continuous probabilities are areas, P(X = a) = 0 for any single value a.
一个PDF f(x) 必须满足两个条件：在定义域内所有 x 上 f(x) ≥ 0，且整个定义域上曲线下的总面积等于1。用符号表示就是 ∫所有 x f(x) dx = 1。X 落在 a 与 b 之间的概率为 P(a < X < b) = ∫ab f(x) dx。由于连续概率是面积，对任何单个值 a 都有 P(X = a) = 0。
3. Cumulative Distribution Function (CDF) | 累积分布函数（CDF）
value less than or equal to x: F(x) = P(X ≤ x). For a continuous random variable, F(x) = ∫–∞x f(t) dt. As x increases, F(x) is a non‑decreasing function that starts at 0 (as x → –∞) and ends at 1 (as x → ∞).
累积分布函数 F(x) 给出了 X 取值小于或等于 x 的概率：F(x) = P(X ≤ x)。对于连续随机变量，F(x) = ∫–∞x f(t) dt。随着 x 增大，F(x) 是一个非递减函数，始于 0（当 x→–∞）并终于 1（当 x→∞）。
4. The 'Joiner' Between PDF and CDF | PDF与CDF之间的“连接者”
, you obtain the CDF by integrating: F(x) = ∫ f(x) dx (with the constant chosen so that F(– ). Conversely, if you know the CDF F(x), the PDF is recovered by differentiation: f(x) = F'(x), provided F is differentiable. This dual relationship is the statistical 'pdf joiner' – it ties the density to the accumulated probability.
这两个函数通过微积分连接在一起：微分与积分。给定 PDF f(x)，你可以通过积分得到 CDF：F(x) = ∫ f(x) dx（并选择常数使得 F(–∞) = 0）。反之，如果已知 CDF F(x)，则可通过微分恢复 PDF：f(x) = F'(x)，前提是 F 可导。这种双重关系就是统计上的“PDF连接者”——它将密度与累积概率联系在一起。
5. Validating a PDF | 验证一个PDF
To check whether a function f(x) can serve as a PDF, first confirm that it is non‑ 1, it is a valid PDF. Often you must find an unknown constant k by solving ∫ f(x) dx = 1.
要检查一个函数 f(x) 是否可用作 PDF，首先确认它在指定范围内非负。然后计算 f(x) 在其整个定义域上的积分。如果积分等于 1，那么它就是一个有效的 PDF。通常你需要通过求解 ∫ f(x) dx = 1 来求出未知常数 k。
6. Finding the Median and Percentiles | 求中位数与百分位数
The median m is the value such that F(m) = 0.5, i.e. ∫–∞m f(x) dx = 0.5. More generally, the 100pth percentile (0 < p < 1) satisfies F(xp ctly increasing where f(x) > 0, the solution is usually unique. Always ensure the answer lies inside the defined domain.
中位数 m 是满足 F(m) = 0.5 的值，即 ∫–∞m f(x) dx = 0.5。更一般地，第100p 百分位数（0 < p < 1）满足 F(xp) = p。由于 F 在 f(x) > 0 处连续且严格递增，解通常是唯一的。务必确保答案落在定义域内。
7. Mode of a Continuous Distribution | 连续分布的众数
is the value of x that maximises the PDF f(x). To find it, differentiate f(x) and solve f' ivative is negative. Remember that the mode is about the density, not the CDF. Some distributions can have more than one local maximum, leading to multimodal situations.
众数是使 PDF f(x) 最大化的 x 值。要求出众数，可对 f(x) 求导并解 f'(x) = 0，然后检查二阶导数为负。请记住众数是关于密度而非 CDF 的。有些分布可能有多个局部最大值，从而导致多峰情况。
8. Expectation and Variance | 期望与方差
e expectation (mean) is E(X) = ∫ x f(x) dx over the domain. The variance is Var(X) = E(X²) – given as a CDF, you can differentiate to get the PDF first, or use the formula E(X) = ∫ [1 – F(x)] dx for non‑negative X (provided the integral converges).
对于具有 PDF f(x) 的连续随机变量 X，期望（均值）为 E(X) = ∫ x f(x) dx，积分区间为整个定义域。方差为 Var(X) = E(X²) – [E(X)]²，其中 E(X²) = ∫ x² f(x) dx。当分布以 CDF 的形式给出时，你可以先微分得到 PDF，或者对于非负 X 使用 E(X) = ∫ [1 – F(x)] dx（只要积分收敛）。
9. The Uniform Distribution – A Worked Example | 均匀分布——一个实例
Suppose X is uniformly distributed over the interval [2, 5]. The PDF is f(x) = 1/(5‑2) = 1/3 for 2 ≤ x ≤ 5, and 0 elsewhere. The CDF is F(x) = (x‑2)/3 for 2 ≤ x ≤ 5, 0 before 2 and 1 after 5. The median is the solution to (m‑2)/3 = 0.5, giving m = 3.5. The expectation is (2+5)/2 = 3.5, and the variance is (5‑2)²/12 = 9/12 = 0.75.
假设 X 在区间 [2, 5] 上服从均匀分布。PDF 为 f(x) = 1/(5‑2) = 1/3，当 2 ≤ x ≤ 5，其他位置为 0。CDF 为 F(x) = (x‑2)/3，当 2 ≤ x ≤ 5，在 2 之前为 0，5 之后为 1。中位数是方程 (m‑2)/3 = 0.5 的解，得到 m = 3.5。期望为 (2+5)/2 = 3.5，方差为 (5‑2)²/12 = 9/12 = 0.75。
10. Exam‑Style Problem: From PDF to CDF and Beyond | 考试题型：从PDF到CDF及其后续
Consider the PDF given by f(x) = kx(4 – e the cumulative distribution function F(x). (c) Determine the median. (d) Calculate E(X).
考虑 PDF：f(x) = kx(4 – x)，0 ≤ x ≤ 4，其他为 0。(a) 求常数 k。(b) 推导累积分布函数 F(x)。(c) 求中位数。(d) 计算 E(X)。
Solution walkthrough
(a) Integrate f(x) from 0 to 4: ∫₀⁴ kx(4 – x) dx = k ∫₀⁴ (4x – x²) dx = k [2x² – x³/3]₀⁴ = k (32 – 64/3) = k (32/3). Set equal to 1: k = 3/32. (b) For 0 ≤ x ≤ 4, F(x) = ∫₀ˣ (3/32) t(4 – t) dt = (3/32)[2t² – t³/3]₀ˣ = (3/32)(2x² – x³/3) = (6x² – x³)/32. So F(x) = 0 for x < 0, (6x² – x³)/32 for 0 ≤ x ≤ 4, and 1 for x > 4. (c) Median m satisfies (6m² – m³)/32 = 0.5 → 6m² – m³ = 16 → m³ – 6m² + 16 = 0. Trial: m = 2 gives 8 – Because the function is increasing, m = 2 is the unique median. (d) E(X) = ∫₀⁴ x·(3/32)x(4 – x) dx = (3/32) ∫₀⁴ (4x² – x³) dx = (3/32) [ (4/3)x³ – x⁴/4 ]₀⁴ = (3/32) (256/3 – 64) = (3/32)(64/3) = 2. Note how the pdf joiner integration recovers the CDF seamlessly.
(a) 对 f(x) 从 0 到 4 积分：∫₀⁴ kx(4 – x) dx = k ∫₀⁴ (4x – x²) dx = k [2x² – x³/3]₀⁴ = k (32 – 64/3) = k (32/3)。令其等于1，得 k = 3/32。(b) 对于 0 ≤ x ≤ 4，F(x) = ∫₀ˣ (3/32) t(4 – t) dt = (3/32)[2t² – t³/3]₀ˣ = (3/32)(2x² – x³/3) = (6x² – x³)/32。因此 F(x) = 0 当 x < 0，(6x² – x³)/32 当 0 ≤ x ≤ 4，1 当 x > 4。(c) 中位数 m 满足 (6m² – m³)/32 = 0.5 → 6m² – m³ = 16 → m³ – 6m² + 16 = 0。试根：m = 2 得 8 – 24 + 16 = 0，所以 m = 2 是一个根。由于函数递增，m = 2 是唯一的中位数。(d) E(X) = ∫₀⁴ x·(3/32)x(4 – x) dx = (3/32) ∫₀⁴ (4x² – x³) dx = (3/32) [ (4/3)x³ – x⁴/4 ]₀⁴ = (3/32) (256/3 – 64) = (3/32)(64/3) = 2。注意这里的pdf连接者积分是如何无缝地还原CDF的。
11. Avoiding Common Pitfalls | 避免常见错误
egration constant yields F(x) = 0 at the lower boundary. Also, when computing variance, use the formula E(X²) – µ² rather than trying to integrate (x – µ)² f(x) directly if it looks messy. Finally, ensure that the PDF is non‑negative everywhere – a negative k or an incorrectly chosen range will invalidate the model.
许多学生忘记对于连续变量 P(X = a) = 0，从而导致不一致的积分限。在推导 CDF 时，务必检查积分常数是否在下边界处使得 F(x) = 0。此外，计算方差时，使用公式 E(X²) – µ² 比在看起来繁琐时直接积分 (x – µ)² f(x) 更好。最后，确保 PDF 处处非负——一个负的 k 或者选择不正确的范围将使模型无效。
12. Connecting Theory to Edexcel Exam Success | 将理论与Edexcel考试成功相连接
Typical exam questions ask you to find an unknown constant, derive a CDF, then determine the median, quartiles or expectations. The key is to treat the PDF as the gradient of the CDF, and the CDF as the area accumulator. By mastering the integral 'joiner' and practising with functions involving polynomials, exponentials, or piecewise definitions, you will be well equipped. Always sketch the PDF and CDF when possible to visualise the areas and check your answers.
典型的考试题目会要求你找出未知常数，推导出 CDF，然后确定中位数、四分位数或期望。关键是把 PDF 视为 CDF 的梯度，把 CDF 视为面积累加器。通过掌握积分“连接者”，并练习涉及多项式、指数函数或分段定义的函数，你会准备充分。只要有可能，就画出 PDF 和 CDF 的草图，以可视化面积并检查答案。
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