Expected Value of a Discrete Random Variable | 离散随机变量的期望值

📚 Expected Value of a Discrete Random Variable | 离散随机变量的期望值

In Edexcel A-Level Statistics, the concept of expected value lies at the heart of probability distributions. It gives us a single-number summary of the 'centre' of a random variable's possible outcomes, weighted by their probabilities. Whether you are modelling the score on a biased die, the number of heads in coin tosses, or the payout from a game, the expected value tells you what you would anticipate on average if you could repeat the experiment infinitely many times. Understanding how to calculate and interpret E(X) is essential for tackling exam questions on discrete random variables.

在Edexcel A-Level统计学中，期望值的概念是概率分布的核心。它为我们提供了随机变量可能结果的一个单值“中心”总结，以概率为权重。无论你是在模拟一枚不均匀骰子的得分、抛硬币中出现正面的次数，还是游戏的赔率，期望值都告诉你在无限多次重复试验的情况下平均会得到什么。理解如何计算和解释 E(X) 对于解决离散随机变量的考试题目至关重要。

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1. What is a Discrete Random Variable? | 什么是离散随机变量？

t values. For instance, the number obtained when rolling a fair six-sided die is a discrete random variable that can be 1, 2, 3, 4, 5 or 6. Each possible value has an associated probability, and the sum of all probabilities must equal 1. The distribution is usually displayed in a table showing each value x and its corresponding P(X = x). This table is the starting point for finding the expected value.

离散随机变量是一种其值由随机事件的结果决定的变量，并且它只能取可数个不同的值。例如，掷一颗均匀六面骰子得到的点数就是一个离散随机变量，其可能取值为 1、2、3、4、5 或 6。每个可能的取值都有一个对应的概率，且所有概率之和必须等于 1。该分布通常用一个表格来展示每个值 x 及其对应的 P(X = x)。这个表格是求期望值的起点。

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2. Definition of Expected Value E(X) | 期望值 E(X) 的定义

possible value multiplied by its probability: E(X) = Σ x·P(X = x), where the sum runs over all possible values of x. This formula gives a weighted average, with the probabilities acting as weights. If you imagine repeating the random process countless times, the long-run average outcome will approach E(X).

期望值，常被称为离散随机变量 X 的均值，用 E(X) 或希腊字母 μ（mu）表示。其定义为每个可能值乘以其概率的总和：E(X) = Σ x·P(X = x)，求和涵盖 X 的所有可能取值。这个公式给出的是一个加权平均值，概率充当权重。如果你想象无数次重复这个随机过程，长期的平均结果将趋近于 E(X)。

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3. Calculating E(X) from a Probability Distribution | 从概率分布计算 E(X)

ty and add the results. For example, suppose a biased die has the following distribution: x = 1, 2, 3, 4, 5, 6 with probabilities 0.1, 0.1, 0.2, 0.2, 0.3, 0.1 respectively. Then E(X) = 1×0.1 + 2×0.1 + 3×0.2 + 4×0.2 + 5×0.3 + 6×0.1 = 0.1 + 0.2 + 0.6 + 0.8 + 1.5 + 0.6 = 3.8. Notice that the expected value need not be one of the possible outcomes - it is an average.

在实际计算 E(X) 时，首先要确保你有一个有效的概率分布，即 Σ P(X = x) = 1。然后将每个 x 值乘以其概率，把结果相加。例如，假设一个不均匀的骰子有如下分布：x = 1, 2, 3, 4, 5, 6，对应概率分别为 0.1, 0.1, 0.2, 0.2, 0.3, 0.1。则 = 1×0.1 + 2×0.1 + 3×0.2 + 4×0.2 + 5×0.3 + 6×0.1 = 0.1 + 0.2 + 0.6 + 0.8 + 1.5 + 0.6 = 3.8。请注意期望值不一定是可能的结果之一——它是一个平均值。

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4. Expected Value of a Function of X: E(g(X)) | X 的函数的期望值 E(g(X))

d extension: simply apply the function to each x value, multiply by the corresponding probability, and sum. For example, if X is the outcome of a fair die, then E(X²) = (1²×1/6) + (2²×1/6) + ... + (6²×1/6) = (1+4+9+16+25+36)/6 = 91/6 ≈ 15.167. This is different from [E(X)]², which would be (3.5)² = 12.25, illustrating that E(X²) is not the same as (E(X))². This distinction is crucial for calculating variance later.

有时你需要求 X 的某个函数的期望值，比如 X²、3X 或 1/X。公式为 E(g(X)) = Σ g(x)·P(X = x)。这是一个直接的推广：只需把每个 x 值代入函数，乘以对应的概率，再求和。例如，若 X 是一颗均匀骰子的点数，那么 E(X²) = (1²×1/6) + (2²×1/6) + ... + (6²×1/6) = (1+4+9+16+25+36)/6 = 91/6 ≈ 15.167。这与 [E(X)]² 不同，后者为 (3.5)² = 12.25，这说明 E(X²) 不等于 (E(X))²。这个区别对于之后计算方差至关重要。

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5. Linearity of Expectation | 期望的线性性质

means you can pull constants out and deal with addition separately. The proof follows directly from the definition: E(aX + b) = Σ (a x + b) P(X=x) = a Σ x P(X=x) + b Σ P(X=x) = aE(X) + b, because Σ P(X=x) = 1. This property holds regardless of whether X takes on a finite or countably infinite number of values, and it makes many calculations much simpler.

期望值最强大的性质之一是其线性性：对于任意离散随机变量 X 以及常数 a 和 b，有 E(aX + b) = aE(X) + b。这意味着你可以把常数提出来，并且分别处理加法。证明直接从定义得出：E(aX + b) = Σ (a x + 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。无论 X 取有限个还是可数无限个值，这个性质都成立，它让许多计算变得简单得多。

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6. Expected Value of a Constant | 常数的期望值

iable, the expected value shifts by that constant. For instance, if a random variable Y = X + 5, then E(Y) = E(X) + 5. This is intuitive: adding a fixed amount to every outcome simply shifts the average by that amount.

线性性质的一个特例是当 X 以概率 1 取常数 c 时，E(c) = c。更一般地，如果你给随机变量加上一个常数，期望值就偏移该常数。比如，如果随机变量 Y = X + 5，那么 E(Y) = E(X) + 5。这很直观：给每个结果加上一个固定量，平均水平也增加同一个量。

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7. Expected Value of aX + b in Practice | 实际应用中的 aX + b 的期望值

ula. For example, if E(X) = 4.2, then E(2X + 10) = 2×4.2 + 10 = 18.4. This can save time because you do not need to create a new probability distribution for the transformed variable; you just use the original E(X). However, be cautious: this property only works for linear transformations. For non-linear functions like X², you must calculate E(X²) directly from the distribution using Σ x²·P(X=x).

考试题目经常要求你先求出 E(X)，然后利用它计算如 E(3X - 2) 之类的表达式。你只需代入线性公式。例如，若 E(X) = 4.2，则 E(2X + 10) = 2×4.2 + 10 = 18.4。这可以节省时间，因为你不需要为变换后的变量创建新的概率分布；只需使用原始的 E(X)。但是要注意：这个性质只适用于线性变换。对于像 X² 这样的非线性函数，你必须用 Σ x²·P(X=x) 直接从分布计算 E(X²)。

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8. Using Expected Value to Solve for Unknown Probabilities | 利用期望值求未知概率

equations to find the unknowns. Typically, you use two facts: the sum of all probabilities equals 1, and the definition of E(X) equals the given value. For example, suppose a discrete random variable Y takes values 0, 2, 5 with probabilities p, q, 0.3 respectively, and E(Y) = 2.6. Then p + q + 0.3 = 1 → p + q = 0.7. Also, E(Y) = 0×p + 2×q + 5×0.3 = 2q + 1.5 = 2.6 → 2q = 1.1 → q = 0.55, so p = 0.15. This method of simultaneous equations is heavily tested in Edexcel A-Level Maths.

一种常见的考试题型是给出来一个含有一两个未知概率的概率分布，同时给出 E(X) 的值。你可以建立方程来求出未知数。通常，你会用到两个事实：所有概率之和等于 1，以及 E(X) 的定义等于给定的值。例如，假设离散随机变量 Y 取值为 0、2、5，对应概率分别为 p、q、0.3，且 E(Y) = 2.6。则 p + q + 0.3 = 1 → p + q = 0.7。同时，E(Y) = 0×p + 2×q + 5×0.3 = 2q + 1.5 = 2.6 → 2q = 1.1 → q = 0.55，所以 p = 0.15。这种联立方程的方法在Edexcel A-Level数学中被大量考查。

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9. Expected Value in Games of Chance | 机会游戏中的期望值

Expected value is widely applied in analysing games to determine whether they are 'fair' me is considered fair if the expected net gain is zero. For instance, suppose a game costs £2 to play, and you win £10 with probability 0.1, win £ probabilities 0.1, 0.2, 0.7. E(W) = 8×0.1 + 3×0.2 + (-2)×0.7 = 0.8 + 0.6 - 1.4 = 0. So the expected net gain is £0, making it a fair game. If E(W) were positive, the game would favour the player.

期望值被广泛应用于分析游戏，以确定它们是否“公平”或给玩家带来优势。如果期望净收益为零，该游戏就被认为是公平的。例如，假设玩一次游戏花费 2 英镑，你有 0.1 的概率赢得 10 英镑，0.2 的概率赢得 5 英镑，其余情况一无所获。设净收益为随机变量 W。则 W = (赢得金额) - 2。可能的净收益为：8, 3, -2，概率分别为 0.1, 0.2, 0.7。E(W) = 8×0.1 + 3×0.2 + (-2)×0.7 = 0.8 + 0.6 - 1.4 = 0。因此期望净收益为 0 英镑，这是一个公平的游戏。如果 E(W) 为正，游戏就对玩家有利。

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10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Students often confuse E(X²) with [E(X)]². Remember, squaring happens before taking the expectation in E(X²), whereas in [E(X)]², you first find the mean and then square it. These are generally different. Another frequent mistake is forgetting to check that probabilities sum to 1 before calculating E(X); if they don't, the distribution is invalid. Also, when dealing with transformed variables, some incorrectly create a new distribution for aX + b and compute E(aX+b) from scratch, which is error-prone and unnecessary if you use linearity. Finally, be careful with signs when net gains are involved in a game; subtracting the cost of playing affects all outcomes.

学生常常混淆 E(X²) 与 [E(X)]²。记住，在 E(X²) 中是先平方再求期望，而在 [E(X)]² 中是先求均值再平方，它们通常是不同的。另一个常见错误是在计算 E(X) 之前忘记检查概率之和是否为 1；如果概率和不等于 1，分布就是无效的。另外，在处理变换变量时，一些人错误地为 aX + b 创建新的分布，并从头计算 E(aX+b)，这容易出错，而且如果你使用线性性质，这是不必要的。最后，当游戏中涉及净收益时，要注意符号；减去游戏成本会影响所有结果。

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11. Worked Exam-Style Example | 考试风格例题详解

Question: The discrete random variable X has probability distribution shown in the table.

x	1	2	3	4
P(X=x)	0.2	a	b	0.1

Given that E(X) = 2.6, find the values of a and b. Hence find E(X²) and E(5X - 3).

问题：离散随机变量 X 的概率分布如下表所示。

x	1	2	3	4
P(X=x)	0.2	a	b	0.1

已知 E(X) = 2.6，求 a 和 b 的值。进而求 E(X²) 和 E(5X - 3)。

Solution: Since total probability is 1, 0.2 + a + b + 0.1 = 1 ⇒ a + b = 0.7. (Equation 1)
E(X) = 1×0.2 + 2a + 3b + 4×0.1 = 0.2 + 2a + 3b + 0.4 = 0.6 + 2a + 3b = 2.6 ⇒ 2a + 3b = 2.0. (Equation 2)
rom (1), a = 0.7 - b. Substitute into (2): 2(0.7 - b) + 3b = 1.4 - 2b + 3b = 1.4 + b = 2.0 ⇒ b = 0.6, so a = 0.1.
Now E(X²) = Σ x²·P(X=x) = 1²×0.2 + 2²×0.1 + 3²×0.6 + 4²×0.1 = 0.2 + 0.4 + 5.4 + 1.6 = 7.6.
E(5X - 3) = 5E(X) - 3 = 5×2.6 - 3 = 13 - 3 = 10.

解答：由于总概率为 1，0.2 + a + b + 0.1 = 1 ⇒ a + b = 0.7。（方程 1）
E(X) = 1×0.2 + 2a + 3b + 4×0.1 = 0.2 + 2a + 3b + 0.4 = 0.6 + 2a + 3b = 2.6 ⇒ 2a + 3b = 2.0。（方程 2）
联立求解：由 (1) 得 a = 0.7 - b。代入 (2)：2(0.7 - b) + 3b = 1.4 - 2b + 3b = 1.4 + b = 2.0 ⇒ b = 0.6，所以 a = 0.1。
现在 E(X²) = Σ x²·P(X=x) = 1²×0.2 + 2²×0.1 + 3²×0.6 + 4²×0.1 = 0.2 + 0.4 + 5.4 + 1.6 = 7.6。
E(5X - 3) = 5E(X) - 3 = 5×2.6 - 3 = 13 - 3 = 10。

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12. Key Takeaways and Exam Tips | 核心要点与考试技巧

X+b) = aE(X)+b to simplify calculations wherever possible. When given E(X) and unknown probabilities, set up simultaneous equations. Remember that E(X²) is needed for variance, and it cannot be deduced from E(X) alone. In game problems, define the net gain clearly as a new random variable. Practise plenty of past-paper questions to gain fluency; many exam errors come from simple arithmetic mistakes in the sum-of-products calculation. With a clear method and careful steps, expected value questions become some of the most reliable marks on the paper.

始终从确认概率分布有效（Σ P = 1）开始。熟记 E(X) = Σ x·P(X=x) 以及 E(g(X)) = Σ g(x)·P(X=x)。尽可能使用线性性质 E(aX+b) = aE(X)+b 来简化计算。当给定 E(X) 和未知概率时，建立联立方程。记住 E(X²) 是算方差所需的，不能仅由 E(X) 推出来。在游戏问题中，要清晰地定义净收益作为一个新的随机变量。大量练习历年真题以获得流利度；许多考试错误源于乘积求和计算中的简单算术错误。有了清晰的方法和仔细的步骤，期望值题目就成了试卷上最稳妥的得分点之一。

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