IB Mathematics: Linear Combinations of Normal Variables | IB 数学：正态变量的线性组合

📚 IB Mathematics: Linear Combinations of Normal Variables | IB 数学：正态变量的线性组合

The normal distribution is one of the most important ideas in IB Mathematics. After learning how to standardise a single normal variable, the next step is to combine two or more independent normal variables. In IB exam questions, you are often asked to find probabilities involving a sum or difference of independent normal variables. The key idea is simple: because each variable is normal, their linear combination is also normal.

正态分布是 IB 数学中最重要的概念之一。在学习完如何标准化单个正态变量之后，下一步就是处理多个独立正态变量的组合。IB 考题经常要求我们求两个独立正态变量之和或差的概率。核心思想很简单：既然每个变量都是正态的，它们的线性组合也仍然是正态的。

1. Why Linear Combinations Matter | 为什么线性组合很重要

In real life, many measurements are made by adding or subtracting independent quantities. For example, the total weight of two boxes is the sum of two individual weights. The difference in response times between two tests is the difference of two normal variables. If both quantities can be modelled by normal distributions, we can find probabilities about the total or difference using one normal model.

在实际问题中，很多测量值都是由独立量相加或相减得到的。例如，两个箱子的总重量是各自重量之和；两次测试的反应时间之差是两个正态变量之差。如果两个量都可以用正态分布建模，那么我们就能用一个正态模型来求总和或差值的概率。

2. Review: The Normal Distribution | 回顾：正态分布

A continuous random variable X is normally distributed if its distribution is bell-shaped and symmetric about its mean. We write X ~ N(μ, σ²), where μ is the mean and σ² is the variance. The standard deviation is σ, so σ = √σ². To calculate probabilities, we standardise X by converting it into a standard normal variable Z = (X − μ)/σ, where Z ~ N(0, 1).

连续随机变量 X 服从正态分布时，它的分布曲线呈钟形，并且关于均值对称。我们记为 X ~ N(μ, σ²)，其中 μ 是均值，σ² 是方差。标准差是 σ，即 σ = √σ²。为了计算概率，我们通过标准化公式 Z = (X − μ)/σ 将 X 转化为标准正态变量 Z，其中 Z ~ N(0, 1)。

3. The Key Theorem | 核心定理

Let X₁ ~ N(μ₁, σ₁²) and X₂ ~ N(μ₂, σ₂²) be independent normal random variables. For any constants a, b and c, the linear combination W = aX₁ + bX₂ + c is also a normal random variable.

设 X₁ ~ N(μ₁, σ₁²) 和 X₂ ~ N(μ₂, σ₂²) 是独立的正态随机变量。对于任意常数 a、b 和 c，线性组合 W = aX₁ + bX₂ + c 仍然是正态随机变量。

mean(W) = aμ₁ + bμ₂ + c

Var(W) = a²σ₁² + b²σ₂²

The constant c shifts the centre of the distribution but does not affect the spread. This theorem is extremely powerful because it reduces a problem about two normal variables into a problem about a single normal variable.

常数 c 只会移动分布的中心，而不会影响分布的高低胖瘦。这个定理非常强大，因为它把涉及两个正态变量的问题转化为只涉及单个正态变量的问题。

4. The Mean of a Linear Combination | 线性组合的均值

The expectation operator is linear. This means that, whether or not the variables are independent, we can calculate the mean by direct multiplication and addition:

期望运算具有线性性质。这意味着，无论变量是否独立，我们都可以直接进行乘法和加法来计算均值：

E(aX₁ + bX₂ + c) = aE(X₁) + bE(X₂) + c

For example, if X₁ ~ N(10, 4) and X₂ ~ N(20, 9), then for W = 2X₁ − 3X₂ + 5, the mean is 2×10 − 3×20 + 5 = −35. The negative coefficient simply means that subtraction of X₂ lowers the mean.

例如，若 X₁ ~ N(10, 4)，X₂ ~ N(20, 9)，则对于 W = 2X₁ − 3X₂ + 5，均值为 2×10 − 3×20 + 5 = −35。负系数意味着减去 X₂ 会使均值下降。

5. The Variance of a Linear Combination | 线性组合的方差

When X₁ and X₂ are independent, the variance of a linear combination does not include a covariance term. The coefficients are squared:

当 X₁ 和 X₂ 独立时，线性组合的方差中不含协方差项。系数要平方：

Var(aX₁ + bX₂ + c) = a²Var(X₁) + b²Var(X₂)

It is a common error to write Var(2X₁ − 3X₂) = 4Var(X₁) − 9Var(X₂). This is false. The variance must always add positive contributions. Using the same example, Var(W) = 4×4 + 9×9 = 16 + 81 = 97.

一个常见错误是写成 Var(2X₁ − 3X₂) = 4Var(X₁) − 9Var(X₂)。这是错误的。方差中每一项都必须是正的。沿用上面的例子，Var(W) = 4×4 + 9×9 = 16 + 81 = 97。

Expression	Mean	Variance
X₁ + X₂	μ₁ + μ₂	σ₁² + σ₂²
X₁ − X₂	μ₁ − μ₂	σ₁² + σ₂²
aX₁ + bX₂ + c	aμ₁ + bμ₂ + c	a²σ₁² + b²σ₂²

6. Worked Example 1: Sum and Difference | 例题 1：和与差

Let X ~ N(100, 25) and Y ~ N(120, 36) be independent. Define T = X + Y and D = X − Y.

设 X ~ N(100, 25) 和 Y ~ N(120, 36) 相互独立。定义 T = X + Y，D = X − Y。

For T, the mean is 100 + 120 = 220 and the variance is 25 + 36 = 61. Therefore T ~ N(220, 61).

对于 T，均值为 100 + 120 = 220，方差为 25 + 36 = 61。因此 T ~ N(220, 61)。

For D, the mean is 100 − 120 = −20 and the variance is 25 + 36 = 61. Therefore D ~ N(−20, 61). Notice that the variance of X − Y is exactly the same as the variance of X + Y when the variables are independent.

对于 D，均值为 100 − 120 = −20，方差为 25 + 36 = 61。因此 D ~ N(−20, 61)。请注意，当变量独立时，X − Y 的方差与 X + Y 的方差完全相同。

7. Standardising a Combined Normal Variable | 对组合正态变量进行标准化

Once we know that W is normal with a particular mean and variance, probability calculations are exactly the same as for a single normal variable. We compute a z-score:

一旦我们知道 W 服从正态分布并具有确定的均值和方差，概率计算就与单个正态变量完全一样。我们计算 z 分数：

Z = (W − mean(W)) / SD(W)

Here SD(W) is the standard deviation of W, which is the square root of its variance. For W = aX₁ + bX₂ + c, we have SD(W) = √(a²σ₁² + b²σ₂²). Once Z is found, use a calculator or the standard normal table.

其中 SD(W) 是 W 的标准差，也就是方差的算术平方根。对于 W = aX₁ + bX₂ + c，有 SD(W) = √(a²σ₁² + b²σ₂²)。求出 Z 之后，就可以使用计算器或标准正态分布表进行计算。

8. Worked Example 2: Probability with Two Independent Normals | 例题 2：双正态变量的概率计算

Let X ~ N(10, 4) and Y ~ N(15, 6) be independent. Find P(X + Y > 30).

设 X ~ N(10, 4) 和 Y ~ N(15, 6) 相互独立。求 P(X + Y > 30)。

Let W = X + Y. Then:

令 W = X + Y。则：

W ~ N(10 + 15, 4 + 6) = N(25, 10)

The required probability is P(W > 30). Standardise:

所求概率为 P(W > 30)。标准化：

Z = (30 − 25) / √10 ≈ 1.581

Therefore P(W > 30) = P(Z > 1.581) ≈ 0.0569. The answer is about 5.69%.

因此 P(W > 30) = P(Z > 1.581) ≈ 0.0569。答案约为 5.69%。

9. Common Pitfalls: Independence and Signs | 常见误区：独立性与符号

The theorem above only applies when the normal variables are independent. If independence is not stated, you cannot use the simple variance addition rule without also considering covariance. In IB questions, watch the wording carefully.

上述定理只适用于变量相互独立的情况。如果题目没有说明独立，就不能简单地使用方差相加规则，除非同时考虑协方差。在 IB 题目中，请仔细阅读题干。

Pitfall 1: Do not ignore independence. The linear combination of normal variables is normal only when the original variables are independent.

误区一：不要忽略独立性。只有在原变量相互独立时，正态变量的线性组合才是正态的。
Pitfall 2: A minus sign does not reduce variance. Var(X − Y) = Var(X) + Var(Y), not Var(X) − Var(Y).

误区二：减号不会减少方差。Var(X − Y) = Var(X) + Var(Y)，而不是 Var(X) − Var(Y)。
Pitfall 3: If the parameters are given as standard deviations, square them before combining variances.

误区三：如果题中给出的是标准差，要先平方得到方差，再进行组合。
Pitfall 4: Do not forget that a constant c changes the mean but not the variance.

误区四：不要忘记常数 c 只改变均值，不改变方差。

10. Subtraction: Var(X − Y) = Var(X) + Var(Y) | 相减时方差相加

Many students find subtraction counterintuitive. If X ~ N(30, 9) and Y ~ N(25, 4) are independent, then X − Y has mean 30 − 25 = 5. The variance is 9 + 4 = 13, not 9 − 4.

许多同学觉得减法难以理解。若 X ~ N(30, 9) 和 Y ~ N(25, 4) 相互独立，则 X − Y 的均值为 30 − 25 = 5。方差为 9 + 4 = 13，而不是 9 − 4。

Why? The coefficient of Y in X − Y is −1. When we square this coefficient, we get (−1)² = 1. Therefore Y contributes 1 × Var(Y) = 4 units of variance. The variability is added.

这是为什么呢？因为 X − Y 中 Y 的系数是 −1。把这个系数平方后得到 (−1)² = 1。因此 Y 贡献 1 × Var(Y) = 4

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IB Mathematics: Linear Combinations of Normal Variables | IB 数学：正态变量的线性组合