Expected Value: Algebraic Rules Explained | IB数学：期望代数运算规则详解

📚 Expected Value: Algebraic Rules Explained | IB数学：期望代数运算规则详解

In IB Mathematics, the concept of expected value (often denoted as E(X) or μ) is one of the most powerful tools in probability and statistics. It allows us to summarise a random variable by a single "typical" value, weighted by probability. But beyond its definition, the true power of expected value lies in its algebraic rules — operations such as E(aX + b), E(X + Y), and E(XY) appear frequently in both Analysis & Approaches (AA) and Applications & Interpretation (AI) exams.

在 IB 数学中，期望值（通常记作 E(X) 或 μ）是概率论与统计学中最强大的工具之一。它通过按概率加权的方式，将随机变量概括为一个"典型"数值。然而，期望值的真正威力不仅仅在于其定义，更在于其代数运算规则——诸如 E(aX + b)、E(X + Y) 和 E(XY) 等运算在数学分析与方法（AA）和应用与解释（AI）考试中频繁出现。

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1. Definition of Expected Value | 期望值的定义

lly, we write E(X) = Σ x · P(X = x), where the sum is taken over all possible values of x.

对于离散型随机变量 X，设其概率函数为 P(X = x)，则期望值定义为所有可能取值的加权平均。数学上写作 E(X) = Σ x · P(X = x)，其中求和遍历 x 的所有可能取值。

(x) dx over the entire support of X. In this article, we focus primarily on the algebraic rules that apply regardless of whether X is discrete or continuous.

对于连续型随机变量，求和变为积分：E(X) = ∫ x · f(x) dx，积分区间为 X 的全部取值范围。本文主要聚焦于对离散型和连续型变量均适用的代数运算规则。

E(X) = Σ x · P(X = x)    (discrete)    |    E(X) = ∫ x · f(x) dx    (continuous)

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2. The Constant Rule | 常数规则

tes that the expected value of a constant is simply that constant itself. This may seem trivial, but it forms the foundation for more complex linear transformations.

若 a 和 b 为常数，一个重要规则是 E(a) = a，即常数的期望值就是该常数本身。这看似显然，但它构成了更复杂的线性变换规则的基础。

s because the probability-weighted average of a fixed number is exactly that fixed number.

例如，若某个试验中 Y 恒等于 5，则 E(Y) = 5。原因在于固定数字的概率加权平均值恰好等于该固定数字本身。

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3. Linear Transformation: E(aX + b) | 线性变换：E(aX + b)

le works regardless of whether a is positive, negative, or zero, and it applies to both discrete and continuous distributions.

考试中最常考查的规则是线性变换规则。若 X 是随机变量，a、b 为常数，则 E(aX + b) = aE(X) + b。无论 a 为正、为负还是为零，该规则均成立，且适用于离散型和连续型分布。

Σ (ax + b) · P(X = x) = a Σ x · P(X = x) + b Σ P(X = x) = aE(X) + b, since Σ P(X = x) = 1.

要理解其原因，我们展开定义：E(aX + b) = Σ (ax + b) · P(X = x) = a Σ x · P(X = x) + b Σ P(X = x) = aE(X) + b，因为 Σ P(X = x) = 1。

E(aX + b) = aE(X) + b

t b disappears from the variance because adding a constant shifts the distribution without changing its spread.

该规则常与方差计算一同考查，因为 Var(aX + b) = a²Var(X)。注意常数 b 在方差中消失，因为加上一个常数只会平移分布，不会改变其离散程度。

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4. Addition Rule: E(X + Y) | 加法规则：E(X + Y)

ly, this rule holds for any two random variables X and Y, regardless of whether they are independent or dependent.

两个随机变量之和的期望值等于它们各自期望值之和。符号表示为 E(X + Y) = E(X) + E(Y)。值得注意的是，无论 X 和 Y 是否独立，该规则对所有随机变量均成立。

For example, if E(X) = 4 and E(Y) = 7, then E(X + Y) = 11, even if X and Y are correlated.

这个性质称为期望的线性性，是概率论中最重要的性质之一。例如，若 E(X) = 4，E(Y) = 7，则 E(X + Y) = 11，即使 X 和 Y 相关也成立。

om variables X₁, X₂, ..., Xₙ, we have E(X₁ + X₂ + ... + Xₙ) = E(X₁) + E(X₂) + ... + E(Xₙ).

加法规则自然推广到多个随机变量的情形。对于任意有限个随机变量 X₁, X₂, ..., Xₙ，有 E(X₁ + X₂ + ... + Xₙ) = E(X₁) + E(X₂) + ... + E(Xₙ)。

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5. Subtraction Rule: E(X − Y) | 减法规则：E(X − Y)

le of E(aX + b) with a = 1 and b = −E(Y), or from the addition rule applied to X and (−Y).

类似地，差的期望值等于期望值之差：E(X − Y) = E(X) − E(Y)。这可由线性变换规则令 a = 1、b = −E(Y) 直接推出，也可将加法规则应用于 X 和 (−Y) 得到。

This rule is particularly useful in IB exam problems involving profit calculations, where revenue minus cost, or score differences between two competitors, must be analysed.

此规则在 IB 考试中特别有用，常见于利润计算问题，例如收入减去成本，或者两位参赛者之间的分数差分析。

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6. The General Linear Combination | 一般线性组合

., aₙ are constants, then E(a₁X₁ + a₂X₂ + ... + aₙXₙ) = a₁E(X₁) + a₂E(X₂) + ... + aₙE(Xₙ).

综合前述规则，我们得到一般线性组合规则：若 X₁, X₂, ..., Xₙ 是随机变量，a₁, a₂, ..., aₙ 是常数，则 E(a₁X₁ + a₂X₂ + ... + aₙXₙ) = a₁E(X₁) + a₂E(X₂) + ... + aₙE(Xₙ)。

This compact formula encompasses everything from simple transformations to complex multi-variable expressions. It is the workhorse of expected value calculus in IB examinations.

这个简洁的公式涵盖了从简单变换到复杂多变量表达式的所有情形。它是 IB 考试中期望值运算的核心工具。

E(Σ aᵢXᵢ) = Σ aᵢ E(Xᵢ)

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7. Multiplication by a Constant vs. Product of Variables | 常数乘法与变量乘积

wed: E(cX) = cE(X). The latter, E(XY), is only equal to E(X)E(Y) under specific conditions — namely, when X and Y are independent.

必须区分"随机变量乘以常数"和"两个随机变量相乘"这两种情形。前者始终成立：E(cX) = cE(X)。后者 E(XY) 仅在特定条件下才等于 E(X)E(Y)——即当 X 和 Y 相互独立时。

any students mistakenly assume E(XY) = E(X)E(Y) always; this is a common trap in IB exams.

一般情况下，除非 X 与 Y 独立，否则 E(XY) ≠ E(X)E(Y)。这就是为什么乘积规则比加法规则限制更多。许多学生错误地认为 E(XY) = E(X)E(Y) 总是成立；这是 IB 考试中的常见陷阱。

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8. Independent Variables: E(XY) | 独立变量：E(XY)

orises as P(X = x, Y = y) = P(X = x) · P(Y = y), which leads directly to the product rule.

若 X 和 Y 是独立随机变量，则 E(XY) = E(X)E(Y)。独立性保证了联合概率可以分解为 P(X = x, Y = y) = P(X = x) · P(Y = y)，从而直接推出乘积规则。

ch in general is not equal to [E(X)]². The difference between E(X²) and [E(X)]² is precisely the variance of X.

例如，若 X 与 Y 独立，E(X) = 3，E(Y) = 5，则 E(XY) = 15。但如果 X = Y，那么 E(XY) = E(X²)，而 E(X²) 一般不等于 [E(X)]²。E(X²) 与 [E(X)]² 之差正好是 X 的方差。

If X ⊥ Y   then   E(XY) = E(X)E(Y)

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9. Connection to Variance | 与方差的联系

X²) − [E(X)]². This identity is one of the most frequently used formulas in IB statistics.

期望的代数规则直接支撑着方差计算。回顾 Var(X) = E[(X − μ)²]，其中 μ = E(X)。展开括号可得 Var(X) = E(X²) − [E(X)]²。这个恒等式是 IB 统计部分最常用的公式之一。

b does not affect variance. This aligns with the earlier observation that adding a constant merely shifts the location of a distribution.

利用线性规则，我们还可推导出 Var(aX + b) = a²Var(X)。常数 b 不影响方差。这与之前的观察一致——加上常数只是平移分布的位置。

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10. Worked Example: Linear Transformation | 例题：线性变换

X be the score shown on the die. The prize money Y is calculated as Y = 3X + 2. Find E(Y).

一个游戏要求玩家掷一枚均匀的六面骰子。设 X 为骰子显示的点数。奖金 Y 按 Y = 3X + 2 计算。求 E(Y)。

Y) = E(3X + 2) = 3E(X) + 2 = 3(3.5) + 2 = 12.5. The expected prize is 12.5 currency units.

对于均匀骰子，E(X) = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5。应用线性变换规则，E(Y) = E(3X + 2) = 3E(X) + 2 = 3(3.5) + 2 = 12.5。期望奖金为 12.5 个货币单位。

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11. Worked Example: Sum of Two Variables | 例题：两变量之和

Let X be the number of heads obtained when tossing two fair coins, and let Y be the number shown on a fair four-sided die (1 to 4). Calculate E(X + Y) and E(3X − 2Y + 1).

设 X 为抛两枚均匀硬币所得正面的数量，Y 为掷一枚均匀四面骰子（1 至 4）所得的点数。求 E(X + Y) 和 E(3X − 2Y + 1)。

= 2, p = 0.5, so E(X) = np = 1. For the four-sided die, E(Y) = (1 + 2 + 3 + 4) / 4 = 2.5.

对于两枚硬币，X 服从 n = 2、p = 0.5 的二项分布，因此 E(X) = np = 1。对于四面骰子，E(Y) = (1 + 2 + 3 + 4) / 4 = 2.5。

econd expression, E(3X − 2Y + 1) = 3E(X) − 2E(Y) + 1 = 3(1) − 2(2.5) + 1 = 3 − 5 + 1 = −1.

因此 E(X + Y) = E(X) + E(Y) = 1 + 2.5 = 3.5。对于第二个表达式，E(3X − 2Y + 1) = 3E(X) − 2E(Y) + 1 = 3(1) − 2(2.5) + 1 = 3 − 5 + 1 = −1。

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12. Common Mistakes and Exam Tips | 常见错误与考试提示

to compute E[g(X)] using the formula Σ g(x) · P(X = x), not by transforming E(X) directly.

一个常见错误是假设 E(1/X) = 1/E(X) 或 E(X²) = [E(X)]²。这些一般都不成立。正确的方法是使用公式 Σ g(x) · P(X = x) 计算 E[g(X)]，而不是直接对 E(X) 进行变换。

cking independence. Always verify that X and Y are independent before using this rule. If they are not independent, you must use the joint probability distribution instead.

另一个错误是不检查独立性就套用乘积规则 E(XY) = E(X)E(Y)。在使用该规则前务必验证 X 和 Y 是否独立。若不独立，则必须使用联合概率分布来计算。

nonlinear functions. When in doubt, return to the definition: E[g(X)] = Σ g(x) · P(X = x).

最后，记住期望的线性性仅适用于线性组合。它不能推广到乘积、商或其他非线性函数。如有疑问，回到定义：E[g(X)] = Σ g(x) · P(X = x)。

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