📚 Probability Density Functions (PDF) | 概率密度函数
Probability Density Functions form a central part of the Edexcel A-Level Statistics syllabus. Understanding PDFs enables you to model continuous random variables, compute probabilities, and derive key measures such as expectation, variance, median, and mode. This article provides a rigorous yet accessible guide to mastering PDFs, with worked examples and examination tips.
概率密度函数是爱德思 A-Level 统计课程的核心内容。理解概率密度函数能够让你对连续随机变量建模、计算概率，并推导期望、方差、中位数和众数等关键度量。本文提供一份严谨而易于掌握的 PDF 学习指南，配有详细例题和应试技巧。
1. Continuous Random Variables | 连续随机变量简介
A continuous random variable can take any value within a given interval. Unlike discrete variables, which have probability mass functions, continuous variables are described by a probability density function (PDF), denoted f(x). The probability of the variable falling in an infinitesimal interval is approximately f(x) dx.
连续随机变量可以在给定区间内取任意值。与具有概率质量函数的离散变量不同，连续变量由概率密度函数 f(x) 描述。变量落入无穷小区间的概率近似为 f(x) dx。
Common examples include the height of students, the time taken to complete a task, or the mass of a chemical sample. Because the variable is continuous, the probability of it taking any exact value is zero; we can only find probabilities over intervals.
常见的例子包括学生身高、完成任务所需时间或化学样品的质量。由于变量是连续的，取任一精确值的概率为零；我们只能求出落在区间内的概率。
2. Definition of a Probability Density Function | 概率密度函数的定义
A function f(x) is a valid PDF for a continuous random variable X if it satisfies two conditions:
函数 f(x) 是连续随机变量 X 的有效概率密度函数，需满足两个条件：
- Non‑negativity: f(x) ≥ 0 for all x in the domain.
- 非负性：定义域内所有 x 满足 f(x) ≥ 0。
- Total area equals 1: ∫all x f(x) dx = 1.
- 总面积为 1：∫全部 x f(x) dx = 1。
In the Edexcel specification, the domain is usually a finite interval [a, b]. The definite integral over that interval must equal 1. This condition is used to find unknown constants in the function.
在爱德思考试范围内，定义域通常是有限区间 [a, b]。该区间上的定积分必须等于 1。这一条件常用于求解函数中的未知常数。
3. Finding Unknown Constants in a PDF | 在 PDF 中求未知常数
2, and f(x) = 0 otherwise, find the value of k. Since it is a PDF, we have ∫₀² kx² dx = 1.
已知 f(x) = kx²，0 ≤ x ≤ 2，其他情况为零。求 k 的值。因为它是 PDF，我们有 ∫₀² kx² dx = 1。
∫₀² kx² dx = [ kx³/3 ]₀² = k(8/3) = 1 ⇒ k = 3/8
Always check non‑ ], so the PDF is valid. This technique of integrating and setting the result equal to 1 appears very often in exam questions.
始终要检查非负性：f(x) = (3/8)x² 在 [0,2] 上 ≥ 0，因此 PDF 有效。这种积分并令结果等于 1 的技巧在考试题中极为常见。
4. Calculating Probabilities Using the PDF | 用 PDF 计算概率
< X < b) = ∫ₐᵇ f(x) dx. Because the probability of a single point is zero, strict and non‑strict inequalities give the same result.
X 落在 a 和 b 之间的概率由 PDF 曲线下的面积给出：P(a < X < b) = ∫ₐᵇ f(x) dx。由于单点的概率为零，严格与非严格不等式结果相同。
Using the previous example with f(x) = (3/8)x² on [0,2], find P(1 < X < 1.5):
使用前例 f(x) = (3/8)x²，区间 [0,2]，求 P(1 < X < 1.5)：
∫₁₁.₅ (3/8)x² dx = (3/8)[ x³/3 ]₁₁.₅ = (1/8)(1.5³ – 1³) = (1/8)(3.375 – 1) = 2.375/8 = 0.2969 (4 d.p.)
Thus the probability is about 0.2969. Exam questions will often ask you to express the answer exactly as a fraction or a decimal to a specified accuracy.
因此概率约为 0.2969。考试题常常要求将答案表示为精确分数或指定精确度的小数。
5. The Cumulative Distribution Function (CDF) | 累积分布函数
The cumulative distribution function, F(x), is defined as F(x) = P(X ≤ x) = ∫₋∞ˣ PDF defined on [a,b], F(x) = 0 for x < a, F(x) = 1 for x > b, and for a ≤ x ≤ b, F(x) = ∫ₐˣ f(t) dt.
累积分布函数 F(x) 定义为 F(x) = P(X ≤ x) = ∫₋∞ˣ f(t) dt。对于定义在 [a,b] 上的 PDF，当 x < a 时 F(x) = 0，当 x > b 时 F(x) = 1，当 a ≤ x ≤ b 时 F(x) = ∫ₐˣ f(t) dt。
For our f(x) = (3/8)x² on [0,2], the CDF for 0 ≤ x ≤ 2 is:
对于我们在 [0,2] 上的 f(x) = (3/8)x²，当 0 ≤ x ≤ 2 时 CDF 为：
F(x) = ∫₀ˣ (3/8)t² dt = (3/8)[ t³/3 ]₀ˣ = x³/8
nd F(x) = 1 for x > 2. The CDF is particularly useful for finding medians and percentiles.
因此 F(x) = 0 (x < 0)，F(x) = x³/8 (0 ≤ x ≤ 2)，F(x) = 1 (x > 2)。CDF 在求中位数和百分位数时尤其有用。
6. Median and Mode of a Continuous Distribution | 连续分布的中位数与众数
Median (m) . the point where the cumulative probability reaches ½. Solve ∫ₐᵐ f(x) dx = 0.5 to find m.
中位数 (m)：满足 F(m) = 0.5，即累积概率达到 ½ 的点。求解 ∫ₐᵐ f(x) dx = 0.5 可得 m。
Using F(x) = x³/8, set x³/8 = 0.5 ⇒ x³ = 4 ⇒ m = ³√4 ≈ 1.587. So the median is ³√4.
使用 F(x) = x³/8，令 x³/8 = 0.5 ⇒ x³ = 4 ⇒ m = ³√4 ≈ 1.587。因此中位数为 ³√4。
Mode x). For f(x) = (3/8)x², the function is increasing on [0,2], so the mode is at the upper bound, x = 2.
众数：众数是 f(x) 在定义域内取得最大值的 x 点。通常通过微分或观察 f(x) 的形状求得。对于 f(x) = (3/8)x²，函数在 [0,2] 上递增，因此众数在上限 x = 2 处。
7. Expectation E(X) of a Continuous Random Variable | 连续随机变量的期望 E(X)
The expected value (mean) of X is given by μ = E(X) = ∫all x x f(x) dx. For a finite interval [a,b], E(X) = ∫ₐᵇ x f(x) dx.
X 的期望值（均值）为 μ = E(X) = ∫全部 x x f(x) dx。对于有限区间 [a,b]，E(X) = ∫ₐᵇ x f(x) dx。
For f(x) = (3/8)x² on [0,2]:
对于 [0,2] 上的 f(x) = (3/8)x²：
(X) = ∫₀² x · (3/8)x² dx = (3/8) ∫₀² x³ dx = (3/8)[ x⁴/4 ]₀² = (3/8)(16/4) = (3/8)×4 = 1.5
So the mean is 1.5. Notice that expectation is the 'balance point' of the distribution.
因此均值为 1.5。注意期望值是分布的“平衡点”。
8. Variance and Standard Deviation | 方差与标准差
Variance is defined as Var(X) = E(X²) – [E(X)]², where E(X²) = ∫ x² f(x) dx. The standard deviation is σ = √Var(X).
方差定义为 Var(X) = E(X²) – [E(X)]²，其中 E(X²) = ∫ x² f(x) dx。标准差为 σ = √Var(X)。
Compute E(X²) for our example:
计算本例中的 E(X²)：
(X²) = ∫₀² x² · (3/8)x² dx = (3/8) ∫₀² x⁴ dx = (3/8)[ x⁵/5 ]₀² = (3/8)(32/5) = 96/40 = 2.4
Then Var(X) = 2.4 – (1.5)² = 2.4 – 2.25 = 0.15. So σ = √0.15 ≈ 0.3873.
然后 Var(X) = 2.4 – (1.5)² = 2.4 – 2.25 = 0.15，因此 σ = √0.15 ≈ 0.3873。
Exam questions frequently ask for both variance and standard deviation, sometimes requiring the exact form (e.g. 3/20) rather than a decimal.
考试题常同时要求方差与标准差，有时需要保留精确形式（如 3/20）而非小数。
9. The Continuous Uniform Distribution | 连续均匀分布
A random variable X follows a uniform distribution on [a,b] if its PDF is f(x) = 1/(b–a) for a ≤ x ≤ b, and 0 otherwise. The distribution is 'flat', meaning all sub‑intervals of equal length have the same probability.
如果随机变量 X 在区间 [a,b] 上服从均匀分布，其 PDF 为 f(x) = 1/(b–a)，a ≤ x ≤ b，其他为零。该分布是“平坦的”，意味着等长子区间具有相同的概率。
For the uniform distribution, E(X) = (a+b)/2, Var(X) = (b–a)²/12. The CDF is F(x) = (x–a)/(b– ormulas are provided in the Edexcel formulae booklet, but you must know how to apply them.
对于均匀分布，E(X) = (a+b)/2，Var(X) = (b–a)²/12。CDF 为 F(x) = (x–a)/(b–a) (a ≤ x ≤ b)。这些公式均在爱德思公式表中提供，但必须掌握如何应用。
Example: The waiting time T (minutes) for a bus is uniformly distributed between 0 and 20. Find P(T > 15) and the median waiting time.
示例：等候公交车的时间 T（分钟）在 0 到 20 间均匀分布。求 P(T > 15) 和等候时间的中位数。
P(T > 15) = (20–15)/(20–0) = 5/20 = 0.25. Median m satisfies (m–0)/20 = 0.5 ⇒ m = 10 minutes.
10. Mixed Practice and Typical Exam Questions | 混合练习与典型考题
A common exam structure: you are given a PDF with an unknown constant, asked to (a) find the constant, (b) specify the CDF, (c) find the median, (d) calculate E(X) and Var(X), and (e) determine probabilities over an interval. Always show clear integration steps and state the final exact form where possible.
常见的试题结构：给出含未知常数的 PDF，要求 (a) 求常数，(b) 写出 CDF，(c) 求中位数，(d) 计算 E(X) 和 Var(X)，(e) 求某个区间的概率。始终要写出清晰的积分步骤，并尽可能给出精确结果。
Let’s work through a compact example: f(x) = k(3x – x²) for 0 ≤ x ≤ 3. (i) Find k. (ii) Find P(X > 2). (iii) Find E(X).
我们通过一个简练的例子来练习：f(x) = k(3x – x²)，0 ≤ x ≤ 3。(i) 求 k。(ii) 求 P(X > 2)。(iii) 求 E(X)。
(i) ∫₀³ k(3x – x²) dx = k[ (3/2)x² – x³/3 ]₀³ = k( (3/2)×9 – 27/3 ) = k(13.5 – 9) = 4.5k = 1 ⇒ k = 2/9.
(ii) P(X > 2) = ∫₂³ (2/9)(3x – x²) dx = (2/9)[ (3/2)x² – x³/3 ]₂³ = (2/9)[ (13.5 – 9) – ( (3/2)×4 – 8/3 ) ] = (2/9)[4.5 – (6 – 2.6667)] = (2/9)[4.5 – 3.3333] = (2/9)×1.1667 ≈ 0.2593.
(iii) E(X) = ∫₀³ x·(2/9)(3x – x²) dx = (2/9)∫₀³ (3x² – x³) dx = (2/9)[ x³ – x⁴/4 ]₀³ = (2/9)(27 – 81/4) = (2/9)(27 – 20.25) = (2/9)×6.75 = 1.5. (Check with symmetry? f(x) is not symmetric, but the mean is 1.5 coincidentally.)
(ii) 精确：P(X>2) = (2/9)[(3/2)x² – x³/3]₂³ = (2/9)[(27/2 – 9) – (6 – 8/3)] = (2/9)[(9/2) – (10/3)] = (2/9)(7/6) = 7/27 ≈ 0.2593.
11. Key Points and Exam Tips | 关键点与应试技巧
- Always write the PDF with its domain, e.g. f(x)=... for a ≤ x ≤ b, and zero otherwise.
始终要注明 PDF 的定义域，例如 f(x)=...，a ≤ x ≤ b，其他为零。 - When integrating, double‑check limits – probabilities are areas between limits.
积分时仔细核对上下限 —— 概率就是上下限之间的面积。 - Use the CDF for probabilities like P(X < c) or P(X > c) when it is easier than integrating the PDF repeatedly.
当累积分布函数已知时，利用它求 P(X < c) 或 P(X > c) 比反复积分 PDF 更便捷。 - Remember that median solves F(m)=0.5, mode is the maximum of f(x) (check endpoints).
记住中位数由 F(m)=0.5 解出，众数则是 f(x) 的最大值点（检查端点）。 - Variance formula Var(X)=E(X²)–[E(X)]² is quicker than ∫(x–μ)²f(x)dx.
使用方差公式 Var(X)=E(X²)–[E(X)]² 比直接积分 ∫(x–μ)²f(x)dx 更快。 - Exact fractions are preferred; if a decimal is required, give at least 3 significant figures.
优先使用精确分数；若需小数，至少保留三位有效数字。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图