Mean and Variance of a Continuous Distribution | 连续分布的均值与方差

📚 Mean and Variance of a Continuous Distribution | 连续分布的均值与方差

In A-Level Mathematics, understanding the mean and variance of a continuous random variable is essential for describing the central tendency and spread of a distribution defined by a probability density function (PDF). Unlike discrete variables, where we sum probabilities, continuous variables require integration over the range of possible values. This article explains how to compute the mean (expected value) and variance for any continuous distribution, explores the effect of linear transformations, and provides worked examples to help you master these skills for the Edexcel exam.

在A-Level数学中，理解连续型随机变量的均值与方差对于描述由概率密度函数（PDF）定义的分布的中心趋势和离散程度至关重要。与离散型变量使用求和不同，连续型变量需要在可能取值范围内进行积分。本文讲解如何计算任意连续分布的均值（期望值）和方差，探讨线性变换的影响，并提供详细的例题，帮助你掌握应对Edexcel考试所需的技能。

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1. What is a Continuous Random Variable? | 什么是连续随机变量？

ity density function f(x), which satisfies f(x) ≥ 0 for all x, and the total area under the curve equals 1: ∫ f(x) dx = 1 over the entire range. The probability that X lies in an interval [a, b] is given by the definite integral P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx.

连续型随机变量可以在实数轴上的某个区间内取任意值。其概率分布由概率密度函数 f(x) 描述，满足 f(x) ≥ 0 对一切 x 成立，且曲线下的总面积为 1：在整个范围内 ∫ f(x) dx = 1。X 落在区间 [a, b] 内的概率由定积分 P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx 给出。

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2. Probability Density Function (PDF) Requirements | 概率密度函数的要求

questions begin by asking you to find an unknown constant k that makes a given function a legitimate PDF, by setting the integral equal to 1 and solving for k.

要使函数 f(x) 成为有效的概率密度函数，必须满足两个条件：第一，在定义域内 f(x) ≥ 0；第二，总面积 ∫₋∞⁺∞ f(x) dx = 1。许多考题一开始会让你求出使给定函数成为合法 PDF 的未知常数 k，方法是将积分设为 1 并解出 k 值。

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3. Expected Value (Mean) of X | X 的期望值（均值）

the PDF. It is defined as E(X) = ∫ x f(x) dx over the entire range of X. This formula is the continuous analogue of Σ x P(X=x) for discrete variables.

连续型随机变量 X 的均值或期望值，记作 E(X) 或 μ，是所有可能取值以 PDF 为权重的加权平均值。其定义为 E(X) = ∫ x f(x) dx，积分范围覆盖 X 的全部取值。该公式是离散变量中 Σ x P(X=x) 的连续对应形式。

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4. Variance of X | X 的方差

μ)²] = ∫ (x − μ)² f(x) dx. For calculation, the equivalent shortcut formula Var(X) = E(X²) − [E(X)]² is almost always used, where E(X²) = ∫ x² f(x) dx.

X 的方差记作 Var(X) 或 σ²，衡量取值偏离均值的平均平方距离。可以利用定义 Var(X) = E[(X − μ)²] = ∫ (x − μ)² f(x) dx 计算。实际计算中几乎总是使用等效的简便公式 Var(X) = E(X²) − [E(X)]²，其中 E(X²) = ∫ x² f(x) dx。

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5. Computing E(X) and Var(X) – Step-by-Step | 计算 E(X) 和 Var(X) – 分步指南

ary. (2) Identify the interval where f(x) > 0; limits of integration are restricted to this support. (3) Compute E(X) = ∫ x f(x) dx using the given limits. (4) Compute E(X²) = ∫ x² f(x) dx. (5) Then Var(X) = E(X²) − [E(X)]². Always simplify fractions and give exact answers unless asked otherwise.

要计算连续分布的均值和方差，可按以下步骤操作：(1) 确认 PDF 有效，必要时求出未知常数。 (2) 确定 f(x) > 0 的区间；积分限仅限该支撑集。 (3) 在该区间上计算 E(X) = ∫ x f(x) dx。 (4) 计算 E(X²) = ∫ x² f(x) dx。 (5) 然后 Var(X) = E(X²) − [E(X)]²。务必化简分数，除非另有要求，应给出精确值。

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6. Worked Example 1: Polynomial PDF | 例题1：多项式概率密度函数

1, ∫₀² kx² dx = 1 → k [x³/3]₀² = 1 → k (8/3) = 1 → k = 3/8. So f(x) = (3/8)x², 0 ≤ x ≤ 2.

连续型随机变量 X 的 PDF 为 f(x) = kx²，0 ≤ x ≤ 2，其他处为 0。首先求 k。因为总面积必须为 1，∫₀² kx² dx = 1 → k [x³/3]₀² = 1 → k (8/3) = 1 → k = 3/8。故 f(x) = (3/8)x²，0 ≤ x ≤ 2。

· (3/8)x² dx = (3/8)∫₀² x³ dx = (3/8)[x⁴/4]₀² = (3/8)(16/4) = (3/8)×4 = 12/8 = 3/2 = 1.5.

现在计算 x · (3/8)x² dx = (3/8)∫₀² x³ dx = (3/8)[x⁴/4]₀² = (3/8)(16/4) = (3/8)×4 = 12/8 = 3/2 = 1.5。

2/5 = 2.4. Therefore, Var(X) = E(X²) − [E(X)]² = 12/5 − (3/2)² = 2.4 − 2.25 = 0.15 = 3/20.

接着， = ∫₀² x² · (3/8)x² dx = (3/8)∫₀² x⁴ dx = (3/8)[x⁵/5]₀² = (3/8)(32/5) = 96/40 = 12/5 = 2.4。因此，Var(X) = E(X²) − [E(X)]² = 12/5 − (3/2)² = 2.4 − 2.25 = 0.15 = 3/20。

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7. Worked Example 2: Exponential-type PDF | 例题2：指数型概率密度函数

A random variable Y has PDF f(y) = 2e⁻²ʸ for y ≥ 0. Verify it is a valid PDF: ∫₀∞ 2e⁻²ʸ dy = [−e⁻²ʸ]₀∞ = 0 − (−1) = 1. The mean is E(Y) = ∫₀∞ y · 2e⁻²ʸ he known result for the exponential distribution, E(Y) = 1/2. Indeed, let u = y, dv = 2e⁻²ʸ dy, then E(Y) = 0.5.

随机变量 Y 的 PDF 为 f(y) = 2e⁻²ʸ，y ≥ 0。验证其有效性：∫₀∞ 2e⁻²ʸ dy = [−e⁻²ʸ]₀∞ = 0 − (−1) = 1。均值为 E(Y) = ∫₀∞ y · 2e⁻²ʸ dy。利用分部积分或指数分布的已知结果，E(Y) = 1/2。

For variance, first compute E(Y²) = ∫₀∞ y² · 2e⁻²ʸ r(Y) = E(Y²) − [E(Y)]² = 1/2 − (1/2)² = 1/2 − 1/4 = 1/4. So the standard deviation is 1/2.

为求方差，先计算 E(Y²) = ∫₀∞ y² · 2e⁻²ʸ dy。利用伽马积分或两次分部积分可得 E(Y²) = 2/λ² = 2/4 = 1/2。则 Var(Y) = E(Y²) − [E(Y)]² = 1/2 − (1/2)² = 1/2 − 1/4 = 1/4，标准差为 1/2。

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8. Expectation of a Function g(X) | 函数 g(X) 的期望

ers E(X) by taking g(x)=x and E(X²) with g(x)=x². It can also be used to find expectations like E(√X), E(1/X), or E(X³).

连续型随机变量的任意函数 g(X) 的期望值由 E[g(X)] = ∫ g(x) f(x) dx 在支撑集上计算得出。这个通用公式涵盖了取 g(x)=x 时的 E(X) 和取 g(x)=x² 时的 E(X²)。它还可以用来求如 E(√X)、E(1/X) 或 E(X³) 等期望。

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9. Linear Transformations: E(aX + b) and Var(aX + b) | 线性变换：E(aX + b) 与 Var(aX + b)

b) = a² Var(X); the constant b shifts the distribution but does not affect spread. These properties are identical to those for discrete variables and are extremely useful for working with scaled or translated data.

如果 X 是连续型随机变量，a 和 b 为常数，则线性函数的均值为 E(aX + b) = a E(X) + b。方差满足 Var(aX + b) = a² Var(X)；常数 b 仅平移分布而不影响离散程度。这些性质与离散型变量完全相同，在处理缩放或平移数据时非常有用。

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10. Standard Deviation | 标准差

e of outcomes from the mean and has the same units as X. For the example in Section 6, σ = √0.15 ≈ 0.3873 (exact √3/20).

X 的标准差，记作 σ 或 SD(X)，是方差的正平方根：σ = √Var(X)。它表示结果与均值的典型距离，且与 X 具有相同单位。以第6节的例题为例，σ = √0.15 ≈ 0.3873（精确值 √3/20）。

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11. Mean and Variance for the Continuous Uniform Distribution | 连续均匀分布的均值与方差

b)/2. The variance is Var(X) = (b − a)² / 12. These formulas are provided in the Edexcel formula booklet and are frequently used to solve problems without performing the integration every time.

区间 [a, b] 上的连续均匀分布的 PDF 为 f(x) = 1/(b−a)，a ≤ x ≤ b。通过直接积分可得均值为区间中点：E(X) = (a + b)/2。方差为 Var(X) = (b − a)² / 12。这些公式在 Edexcel 公式表中给出，经常直接用于解题，无需每次都进行积分。

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12. Common Mistakes and Key Takeaways | 常见错误与核心要点

, or they confuse the limits of integration. Always check that your PDF is valid before computing moments. Remember that for linear transformations, variance is multiplied by a², not a. Finally, practice integration techniques—particularly integration by parts and handling improper integrals—since they appear regularly in continuous distribution questions.

许多学生在使用 Var(X) = E(X²) − [E(X)]² 时忘记将均值平方，或混淆积分上下限。在计算矩之前，务必检查 PDF 的有效性。记住对于线性变换，方差要乘以 a² 而非 a。最后，要熟练积分技巧——特别是分部积分和处理反常积分——因为它们在连续分布题目中经常出现。

Summary of essential formulas:

核心公式总结：

• E(X) = ∫ x f(x) dx
• E(X²) = ∫ x² f(x) dx
• Var(X) = E(X²) − [E(X)]²
• Var(aX + b) = a² Var(X)
• Uniform: E(X) = (a+b)/2, Var(X) = (b−a)²/12

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