📚 Probability Density Functions and Cumulative Distribution Functions | 概率密度函数与累积分布函数
Continuous random variables form a core part of the Edexcel A-Level Statistics syllabus. Unlike discrete variables, which take a countable set of values, continuous variables can take any value within an interval. To work with them, we need two essential functions: the probability density function (PDF) and the cumulative distribution function (CDF). This article provides a thorough exploration of both, showing how they are defined, how they relate to each other, and how to use them to compute probabilities, measures of central tendency and dispersion.
连续随机变量是Edexcel A-Level统计学大纲的核心部分。与取值为可数集的离散变量不同，连续变量可以在一个区间内取任意值。为了处理连续变量，我们需要两个基本函数：概率密度函数（PDF）与累积分布函数（CDF）。本文将深入探讨这两个函数，阐述它们的定义、相互之间的关系，以及如何运用它们来计算概率、集中趋势与离散程度的度量。
1. Introduction to Continuous Random Variables | 连续随机变量简介
A continuous random variable X can take any value in a given range, typically an interval of real numbers. Examples include the height of a student, the time taken to complete a task, or the volume of liquid in a bottle. Because there are infinitely many possible outcomes, the probability that X takes any single exact value is zero. Instead, we talk about the probability that X falls within a range.
连续随机变量 X 可以在给定的范围（通常是实数区间）内取任意值。例如，学生的身高、完成一项任务所需的时间或瓶中液体的体积。由于有无限多个可能的结果，X 取某一精确值的概率为零。因此，我们讨论的是 X 落在某一区间内的概率。
Probabilities for continuous variables are described by a probability density function, often denoted f(x). The graph of f(x) is a smooth curve, and the total area under the curve and above the x‑axis equals 1. Probabilities correspond to areas under the curve between two points.
连续变量的概率由概率密度函数（通常记为 f(x)）描述。f(x) 的图像是一条光滑曲线，曲线下方与 x 轴上方之间的总面积等于 1。概率对应曲线上两点之间的面积。
2. Defining a Probability Density Function (PDF) | 概率密度函数定义
The function itself does not give a probability directly; instead, the area under the curve between two values gives the probability that X lies between them. Formally, P(c ≤ X ≤ d) = ∫cd f(x) dx.
概率密度函数 f(x) 为连续随机变量 X 在区间 [a, b] 上定义，并且对该区间内的任意 x 都有 f(x) ≥ 0。函数本身并不直接给出概率，而是曲线上两点之间的面积给出 X 介于这两个值之间的概率。形式上，P(c ≤ X ≤ d) = ∫cd f(x) dx。
nclusive does not affect the probability: P(c ≤ X ≤ d) = P(c < X < d). This is a major difference from discrete distributions.
必须记住，对于连续随机变量，对于任何特定值 c，P(X = c) = 0。因此，不等式是严格还是包含等号并不影响概率：P(c ≤ X ≤ d) = P(c < X < d)。这是与离散分布的一大区别。
3. Valid Probability Density Functions | 有效概率密度函数的条件
irst, f(x) ≥ 0 for all x in [a, b]. Second, the total area under the curve must equal 1: ∫ab f(x) dx = 1.
要使函数 f(x) 成为区间 [a, b] 上连续随机变量的概率密度函数，必须满足两个条件。第一，对于 [a, b] 内的所有 x，f(x) ≥ 0。第二，曲线下的总面积必须等于 1：∫ab f(x) dx = 1。
In examination questions, you may be asked to find an unknown constant k within a piecewise-defined PDF by using the total area condition. Always set up the integral of f(x) over the entire defined range equal to 1 and solve for k. Also check that f(x) is non‑negative everywhere once k is substituted.
在考试题中，可能会要求你利用总面积条件求出分段定义的PDF中的未知常数 k。总是将 f(x) 在整个定义域上的积分设为 1，然后解出 k。同时，在代入 k 后检查 f(x) 处处非负。
4. Understanding the Cumulative Distribution Function (CDF) | 理解累积分布函数
, F(x) = P(X ≤ x). For a continuous variable with PDF f(x), the CDF is defined as F(x) = ∫ax f(t) dt, where a is the lower bound of the domain.
累积分布函数，记作 F(x)，给出随机变量 X 取值小于或等于 x 的概率。即 F(x) = P(X ≤ x)。对于具有 PDF f(x) 的连续变量，CDF 定义为 F(x) = ∫ax f(t) dt，其中 a 是定义域的下界。
F(x) is a non‑ large x. It can be used directly to find P(X ≤ x) and, by subtraction, P(a ≤ X ≤ b) = F(b) – F(a). This is particularly helpful when the PDF is complicated but the CDF is given.
F(x) 是一个非递减函数，对于非常小的 x 从 0 开始，对于非常大的 x 以 1 结束。它可以用于直接求 P(X ≤ x)，并通过减法求 P(a ≤ X ≤ b) = F(b) – F(a)。当 PDF 复杂但给出了 CDF 时，这一点尤其有用。
5. Relationship Between PDF and CDF | PDF与CDF的关系
it follows that f(x) is the derivative of F(x) wherever the PDF exists. That is, f(x) = F'(x).
PDF 和 CDF 通过微分和积分彼此联系。由于 F(x) 是从下界到 x 的 f(x) 积分，因此在 PDF 存在的点上 f(x) 是 F(x) 的导数。即 f(x) = F'(x)。
u integrate and add the correct constant (usually 0 for the lower bound). Questions may ask you to confirm that a given F(x) is a valid CDF by checking F(lower bound) = 0 and F(upper bound) = 1.
这种关系意味着如果给出了 CDF F(x)，可以通过对其求导得到 PDF。反过来，如果要从 PDF 得到 CDF，则需进行积分并加上正确的常数（通常下界处为 0）。考题可能要求你通过检查 F(下界) = 0 和 F(上界) = 1 来验证给定的 F(x) 是否为有效的 CDF。
6. Finding Probabilities from the PDF | 由PDF求概率
t X lies in an interval [c, d], you integrate the PDF over that interval: P(c ≤ X ≤ d) = ∫cd f(x) dx. If the PDF is piecewise, be careful to only integrate over the parts of the domain where f(x) is defined.
要求 X 落在区间 [c, d] 内的概率，需要在该区间上对 PDF 积分：P(c ≤ X ≤ d) = ∫cd f(x) dx。如果 PDF 是分段函数，注意仅在有定义的那部分定义域上积分。
If using the CDF, simply evaluate F(d) – F(c). This is often quicker and less error‑prone. Always sketch the PDF graph if possible, as it helps visualise the area corresponding to the required probability and avoids sign errors.
如果使用 CDF，只需计算 F(d) – F(c) 即可。这通常更快捷，也少出错。如果可能，要画出 PDF 的草图，这有助于直观呈现对应于所求概率的面积，避免符号错误。
7. Finding the Median from the PDF | 由PDF求中位数
means the area under the PDF up to m equals 0.5. To find the median, set up the equation ∫am f(x) dx = 0.5 and solve for m, or solve F(m) = 0.5 if the CDF is known.
连续分布的中位数 m 是满足 P(X ≤ m) = 0.5 的值。这意味着 PDF 下方直到 m 的面积等于 0.5。要求中位数，建立方程 ∫am f(x) dx = 0.5 并解出 m，或者如果已知 CDF，解 F(m) = 0.5 即可。
lue that splits the distribution into two equal halves. In some symmetric distributions, the median coincides with the mode and mean. For skewed distributions, the median is a better measure of centre than the mean.
由于连续分布有 P(X = m) = 0，中位数也就是将分布分成两个相等一半的值。在某些对称分布中，中位数与众数和均值重合。对于偏态分布，中位数是比均值更好的中心度量。
8. Finding the Mode from the PDF | 由PDF求众数
The mode of a continuous random variable is the value of x where the PDF f(x) reaches its maximum. This usually involves differentiating f(x) and setting the derivative equal to zero, then checking the second derivative or boundaries to confirm a maximum. Always check the endpoints of the domain if the maximum could occur there.
连续随机变量的众数是使 PDF f(x) 达到最大值的 x 值。这通常涉及对 f(x) 求导并令导数等于零，然后检查二阶导数或端点以确认其为最大值。如果最大值可能出现在定义域端点，务必检查端点。
Some distributions have more than one local maximum; in such cases the highest peak is the mode. For a uniform distribution, the mode is not uniquely defined (every point in the interval could be considered a mode). Questions at A-Level often involve quadratic or simple polynomial PDFs, making differentiation straightforward.
某些分布有多个局部最大值；此时最高峰即为众数。对于均匀分布，众数不是唯一确定的（区间内的每一点都可以视作众数）。A-Level 考题常涉及二次函数或简单多项式的 PDF，求导过程较为直接。
9. Finding Quartiles and Percentiles | 求四分位数与百分位数
0.25 and 0.75 respectively. That is, F(Q₁) = 0.25 and F(Q₃) = 0.75. More generally, the p‑th percentile is the value x such that F(x) = p/100.
下四分位数 Q₁ 和上四分位数 Q₃ 的求法原理与中位数相同，但概率分别为 0.25 和 0.75。即 F(Q₁) = 0.25，F(Q₃) = 0.75。更一般地，第 p 个百分位数是满足 F(x) = p/100 的 x 值。
The interquartile range IQR = Q₃ – erwise your equation will give an impossible answer. Always verify that your quartile lies within the appropriate segment.
四分位距 IQR = Q₃ – Q₁ 提供了一种离散程度的度量。当 CDF 以分段函数形式给出时，务必针对所需概率选择正确的分段，否则方程将给出不可能的结果。始终验证求出的四分位数是否落在相应的分段区间内。
10. Expected Value (Mean) of a Continuous Random Variable | 连续随机变量的期望值（均值）
nuous random variable is calculated by integrating x·f(x) over the entire domain: E(X) = ∫ab x f(x) dx. This is the continuous analogue of the discrete sum of x·P(X = x).
连续随机变量的期望值 E(X) 或均值 μ 通过对整个定义域上的 x·f(x) 积分来计算：E(X) = ∫ab x f(x) dx。这是离散求和 x·P(X = x) 的连续类比。
plification. Also note that E(g(X)) = ∫ g(x) f(x) dx, a useful formula for finding expectations of functions, such as E(X²) needed for variance.
当 PDF 关于某点对称时，均值与对称轴重合。计算均值时需仔细积分并利用代数化简。还需注意 E(g(X)) = ∫ g(x) f(x) dx，这是求函数期望的有用公式，例如求方差所需的 E(X²)。
11. Variance of a Continuous Random Variable | 连续随机变量的方差
e variance Var(X) measures the spread of the distribution and is defined as Var(X) = E(X²) – he square of the mean. The standard deviation is the positive square root of the variance.
方差 Var(X) 衡量分布的离散程度，定义为 Var(X) = E(X²) – [E(X)]²。对于连续变量，E(X²) = ∫ x² f(x) dx 在定义区间上积分。然后减去均值的平方。标准差是方差的正平方根。
In exam problems, first find E(X), then E(X²), both via integration. Be meticulous with arithmetic; errors in powers or constants easily propagate. If the PDF is piecewise, split the integrals for E(X) and E(X²) accordingly and sum them.
在考试题中，首先通过积分求出 E(X)，再求 E(X²)。计算需一丝不苟；幂次或常数的错误极易扩散。如果 PDF 是分段函数，要相应地将 E(X) 和 E(X²) 的积分分段计算然后求和。
12. Worked Example and Common Mistakes | 例题与常见错误
Example: A continuous random variable X has PDF f(x) = kx(1 – x) for 0 ≤ x ≤ 1, and 0 otherwise. (a) Find k. (b) Find the median. (c) Find E(X).
例题：连续随机变量 X 的 PDF 为 f(x) = kx(1 – x)，0 ≤ x ≤ 1，其余为 0。(a) 求 k。 (b) 求中位数。 (c) 求 E(X)。
Solution: (a) Total area: ∫₀¹ kx(1 – x) dx = k ∫₀¹ (x – x²) dx = k [x²/2 – x³/3]₀¹ = k(1/2 – 1/3) = k/6 = 1, so k = 6. (b) Median m satisfies 6 ∫₀ᵐ (x – x²) dx = 0.5 → [x²/2 – x³/3]₀ᵐ = 1/12 → m²/2 – m³/3 = 1/12. Multiply by 12: 6m² – 4m³ = 1 → 4m³ – 6m² + 1 = 0. By trial, m = 0.5 works? 4(0.125) – 6(0.25) + 1 = 0.5 – 1.5 + 1 = 0, so m = 0.5. (c) E(X) = 6 ∫₀¹ x²(1 – x) dx = 6 ∫₀¹ (x² – x³) dx = 6 [x³/3 – x⁴/4]₀¹ = 6(1/3 – 1/4) = 6(1/12) = 0.5.
解答：(a) 总面积：∫₀¹ kx(1 – x) dx = k ∫₀¹ (x – x²) dx = k [x²/2 – x³/3]₀¹ = k(1/2 – 1/3) = k/6 = 1，故 k = 6。(b) 中位数 m 满足 6 ∫₀ᵐ (x – x²) dx = 0.5 → [x²/2 – x³/3]₀ᵐ = 1/12 → m²/2 – m³/3 = 1/12。两边乘 12：6m² – 4m³ = 1 → 4m³ – 6m² + 1 = 0。尝试 m = 0.5：4(0.125) – 6(0.25) + 1 = 0.5 – 1.5 + 1 = 0，故 m = 0.5。(c) E(X) = 6 ∫₀¹ x²(1 – x) dx = 6 ∫₀¹ (x² – x³) dx = 6 [x³/3 – x⁴/4]₀¹ = 6(1/3 – 1/4) = 6(1/12) = 0.5。
onfusing the PDF f(x) with a probability. Always use f(x) ≥ 0 and area = 1 to check your constants. Additionally, when finding a median or quartile, remember to set the CDF equal to the desired probability, not the PDF.
常见错误包括：在求未知常数时忘记总概率必须为 1，积分的上下限使用错误，以及将 PDF f(x) 混淆为概率。始终用 f(x) ≥ 0 和面积 = 1 来检验常数。此外，在求中位数或四分位数时，记住是令 CDF 等于目标概率，而不是令 PDF 等于目标概率。
The connection between PDF and CDF by differentiation and integration is a powerful tool. Mastery of these concepts and careful algebraic manipulation will enable you to solve a wide range of continuous distribution problems confidently.
通过微分和积分建立的 PDF 与 CDF 之间的联系是强有力的工具。掌握这些概念并谨慎进行代数运算，将使你有信心解决各种连续分布问题。
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