📚 1C Expectation and Variance of the Sample Mean and Sample Sum | 样本均值与样本总和的期望和方差
When we draw a random sample from a population, the statistics we calculate from it - such as the sample mean and the sample sum - are themselves random variables. Their expectation and variance determine how accurate our estimates are and how much they fluctuate from sample to sample, which is essential for parameter estimation and for constructing confidence intervals. This chapter develops the fundamental properties of the expectation and variance of the sample sum and the sample mean under the assumption of independent and identically distributed (i.i.d.) random variables.
当我们从总体中抽取随机样本时，我们所计算的统计量——例如样本均值和样本总和——本身就是随机变量。它们的期望和方差决定了估计的准确性，以及不同样本之间波动的大小，这对于估计总体参数和构建置信区间至关重要。本章将在独立同分布（i.i.d.）随机变量的假设下，推导样本总和与样本均值的期望和方差的基本性质。
1. Introduction to Sampling Distributions | 抽样分布简介
When we take a sample of size n from a population, the value of the sample mean X̄ and the sample sum T will change from one sample to the next. The probability distribution of a statistic over all possible samples of size n is called its sampling distribution. The sampling distribution is the bridge between a single observed sample and the unknown population.
当我们从总体中抽取容量为 n 的样本时，样本均值 X̄ 和样本总和 T 会随样本的不同而变化。一个统计量在容量为 n 的所有可能样本上的概率分布，称为它的抽样分布。抽样分布是连接单个观测样本与未知总体之间的桥梁。
Knowing the expected value and variance of the sampling distribution allows us to predict the long-run behaviour of these estimators - how close they tend to land to the true population parameter, and how much they scatter around it.
了解抽样分布的期望值和方差，可以让我们预测这些估计量的长期表现——它们平均而言会落在距离真实总体参数多近的位置，以及围绕它波动的幅度有多大。
Three quantities are central to this chapter:
本章的核心是三个量：
- An individual observation Xᵢ - a single draw from the population, with mean μ and variance σ².
- The sample sum T = X₁ + X₂ + ... + Xₙ - the total of all n observations.
- The sample mean X̄ = T / n - the average of the n observations.
- 单个观测值 Xᵢ——从总体中抽取的一个个体，其均值为 μ、方差为 σ²。
- 样本总和 T = X₁ + X₂ + ... + Xₙ——n 个观测值的总和。
- 样本均值 X̄ = T / n——n 个观测值的平均数。
2. Independent and Identically Distributed (i.i.d.) Random Variables | 独立同分布随机变量
In most introductory settings we assume that the sample observations X₁, X₂, ..., Xₙ are independent and identically distributed. "Identically distributed" means every Xᵢ has the same distribution as the population, so each has the same mean μ and the same variance σ². "Independent" means that the value taken by one observation tells us nothing about the values taken by the others.
在大多数入门情形中，我们假设样本观测值 X₁, X₂, ..., Xₙ 是独立同分布的。"同分布"意味着每个 Xᵢ 都与总体具有相同的分布，因此都具有相同的均值 μ 和相同的方差 σ²。"独立"意味着一个观测值取什么值，不会提供关于其他观测值取值的任何信息。
mply adds. Without independence, a covariance term would appear and the formulas in this chapter would no longer hold.
独立性对方差计算有一个非常有用的推论。如果 X 与 Y 独立，则 Cov(X, Y) = 0，它们之和的方差可以直接相加。若没有独立性，就会出现协方差项，本章的公式也将不再成立。
We also assume that the population variance σ² is finite. This rules out heavy-tailed distributions with infinite variance, for which the sample variance has no useful limiting behaviour.
我们还假设总体方差 σ² 是有限的。这排除了方差无穷的重尾分布，因为对这类分布而言，样本方差没有有用的极限行为。
3. The Sample Sum as a Random Variable | 样本总和作为随机变量
Define the sample sum as T = X₁ + X₂ + ... + Xₙ = ΣXᵢ, and the sample mean as X̄ = T / n. Because each Xᵢ is random, both T and X̄ t with a different sample and you obtain a different value of T and a different value of X̄.
定义样本总和为 T = X₁ + X₂ + ... + Xₙ = ΣXᵢ，样本均值为 X̄ = T / n。由于每个 Xᵢ 都是随机的，因此 T 和 X̄ 也都是随机变量：换一个样本重复试验，就会得到不同的 T 值和不同的 X̄ 值。
There is an important asymmetry between the expectation and the variance of a sum. The expectation of a sum is always the sum of the expectations, whether or not the variables are independent. The variance of a sum, however, is the sum of the variances only when the variables are independent (or at least uncorrelated).
和式的期望与方差之间有一个重要的不对称性。和式的期望永远等于期望之和，无论变量是否独立。然而，和式的方差只有在变量相互独立（或至少不相关）时，才等于方差之和。
This is why every result in this chapter is stated under an explicit independence assumption, and why exam questions almost always tell you that the observations are independent.
这正是本章所有结论都必须在明确的独立假设下陈述的原因，也是考试题目几乎总会告诉你观测值相互独立的原因。
4. Expectation of the Sample Sum | 样本总和的期望
The expectation of the sample sum is n times the population mean. Using the linearity of expectation, which needs no independence assumption, we obtain the result directly.
样本总和的期望等于总体均值的 n 倍。利用期望的线性性质（该性质不需要独立性假设），我们可以直接得到这个结果。
E(T) = E(X₁ + X₂ + ... + Xₙ) = E(X₁) + E(X₂) + ... + E(Xₙ) = nμ
So the expected total of n observations is exactly n times the population mean, regardless of the sample size chosen. In other words, the sample sum is an unbiased estimator of nμ.
因此，n 个观测值的期望总和恰好是总体均值的 n 倍，与所选择的样本容量无关。换句话说，样本总和是 nμ 的无偏估计量。
e total of fifty observations will be 600, though any particular sample may give a total well above or below that value.
例如，若总体均值 μ = 12，抽取容量为 n = 50 的样本，则 E(T) = 50 × 12 = 600。平均而言，五十个观测值的总和为 600，尽管任何特定样本的总和可能远高于或远低于该值。
5. Variance of the Sample Sum | 样本总和的方差
Because the observations are independent, their variances add. The variance of the sample sum therefore grows linearly with the sample size.
由于观测值相互独立，它们的方差可以相加。因此样本总和的方差随样本容量线性增长。
Var(T) = Var(X₁) + Var(X₂) + ... + Var(Xₙ) = nσ²
Taking the square root gives the standard deviation of the sample sum:
取平方根即得样本总和的标准差：
SD(T) = √nσ² = σ√n
Notice that the variance of the sum grows like n, while the standard deviation grows only like √n. This is a recurring theme: totals become more variable as the sample grows, but not as fast as the sample size itself.
注意，总和的方差按 n 增长，而标准差只按 √n 增长。这是一个反复出现的主题：随着样本增大，总和的变异性增大，但增速慢于样本容量本身。
For example, with σ² = 9 (so σ = 3) and n = 50, we get Var(T) = 50 × 9 = 450 and SD(T) = 3√50 ≈ 21.21.
例如，当 σ² = 9（即 σ = 3）、n = 50 时，Var(T) = 50 × 9 = 450，SD(T) = 3√50 ≈ 21.21。
6. Expectation of the Sample Mean | 样本均值的期望
The sample mean is obtained by dividing the sample sum by n. Since n is a constant, the expectation simply divides by n as well.
样本均值由样本总和除以 n 得到。由于 n 是常数，期望也同样除以 n。
E(X̄) = E(T / n) = (1/n) E(T) = (1/n)(nμ) = μ
This is the single most important property of the sample mean: it is an unbiased estimator of the population mean μ. On average, the sample mean hits the true population mean exactly, for every sample size n, no matter how small.
这是样本均值最重要的性质：它是总体均值 μ 的无偏估计量。平均而言，无论样本容量 n 多小，样本均值都恰好命中真实的总体均值。
Crucially, this unbiasedness does not require independence. It follows purely from the linearity of expectation. What independence buys us is not a correct centre, but a small spread.
关键在于，这一无偏性并不需要独立性，它完全由期望的线性性质推出。独立性带给我们的不是正确的中心位置，而是较小的离散程度。
Unbiasedness does not mean that X̄ equals μ in any particular sample. It only means that if we repeated the sampling procedure many times, the average of all the X̄ values obtained would converge to μ.
无偏性并不意味着任何一个具体样本中的 X̄ 都等于 μ。它只意味着，如果我们把抽样过程重复很多次，所得到的所有 X̄ 值的平均数会收敛到 μ。
7. Variance of the Sample Mean and the Standard Error | 样本均值的方差与标准误
Dividing a random variable by n divides its variance by n², not by n. Applying this to X̄ = T / n gives the central result of this chapter.
把一个随机变量除以 n，其方差要被除以 n²，而不是除以 n。将这一点应用于 X̄ = T / n，就得到本章的核心结论。
Var(X̄) = Var(T / n) = (1/n²) Var(T) = (1/n²)(nσ²) = σ² / n
The standard deviation of the sample mean has its own name - the standard error of the mean - and is written SE(X̄).
样本均值的标准差有专门的名称——均值的标准误，记作 SE(X̄)。
SE(X̄) = σ / √n
Because the variance of X̄ is σ²/n, the sample mean is a far more stable quantity than a single observation. Averaging is the reason that larger samples give more precise estimates of μ, and the reason that the standard error shrinks as n grows.
由于 X̄ 的方差为 σ²/n，样本均值比单个观测值稳定得多。正是"取平均"这个动作，使得更大的样本能给出对 μ 更精确的估计，也使得标准误随 n 增大而减小。
For example, if σ = 8 and n = 25, then Var(X̄) = 64/25 = 2.56 and SE(X̄) = 8/√25 = 8/5 = 1.6.
例如，若 σ = 8、n = 25，则 Var(X̄) = 64/25 = 2.56，SE(X̄) = 8/√25 = 8/5 = 1.6。
8. Summary Table and Key Formulas | 公式汇总表
The table below collects everything derived so far. It is worth memorising, since almost every question in this topic reduces to one of these six entries.
下表汇总了迄今为止推导出的全部结果，值得背诵，因为本专题的几乎所有题目最终都归结为这六项中的某一项。
| Quantity 量 | Expectation 期望 | Variance 方差 | Standard deviation 标准差 |
| Single observation Xᵢ | μ | σ² | σ |
| Sample sum T = ΣXᵢ | nμ | nσ² | σ√n |
| Sample mean X̄ = T/n | μ | σ²/n | σ/√n |
A very common exam follow-up is to standardise the sample mean. Subtracting the mean and dividing by the standard error produces a variable with mean 0 and variance 1.
考试中非常常见的后续问题是把样本均值标准化。减去均值再除以标准误，就得到一个均值为 0、方差为 1 的变量。
Z = (X̄ − μ) / (σ / √n), E(Z) = 0, Var(Z) = 1
Under the Central Limit Theorem, when n is large this standardised variable is approximately N(0, 1), which is the foundation of large-sample confidence intervals and hypothesis tests.
根据中心极限定理，当 n 很大时，这个标准化变量近似服从 N(0, 1)，这正是大样本置信区间与假设检验的基础。
9. Worked Examples | 典型例题
Example 1. eviation σ = 8 g. A random sample of n = 25 bags is weighed. Find E(T), Var(T), SD(T), E(X̄), Var(X̄) and SE(X̄).
例 1。 某机器灌装白糖。每袋重量是一个随机变量，均值 μ = 500 克，标准差 σ = 8 克。随机抽取 n = 25 袋称重。求 E(T)、Var(T)、SD(T)、E(X̄)、Var(X̄) 和 SE(X̄)。
- E(T) = nμ = 25 × 500 = 12500 g
- Var(T) = nσ² = 25 × 64 = 1600, so SD(T) = 40 g
- E(X̄) = μ = 500 g
- Var(X̄) = σ²/n = 64/25 = 2.56, so SE(X̄) = √2.56 = 1.6 g
- E(T) = nμ = 25 × 500 = 12500 克
- Var(T) = nσ² = 25 × 64 = 1600，故 SD(T) = 40 克
- E(X̄) = μ = 500 克
- Var(X̄) = σ²/n = 64/25 = 2.56，故 SE(X̄) = √2.56 = 1.6 克
Note that the standard deviation of a single bag is 8 g, but the standard error of the mean of 25 bags is only 1.6 g. Averaging has reduced the uncertainty by a factor of 5, which is exactly √25.
注意：单袋重量的标准差是 8 克，而 25 袋均值的标准误只有 1.6 克。取平均把不确定性缩小了 5 倍，恰好等于 √25。
Example 2. ent samples are taken: sample 1 of size n₁ = 16 and sample 2 of size n₂ = 25, with means X̄₁ and X̄₂. Find the expectation, variance and standard deviation of D = X̄₁ − X̄₂.
例 2。 某总体均值 μ = 40，方差 σ² = 100。抽取两个相互独立的样本：样本 1 容量 n₁ = 16，样本 2 容量 n₂ = 25，均值分别为 X̄₁ 和 X̄₂。求 D = X̄₁ − X̄₂ 的期望、方差和标准差。
E(D) = 40 − 40 = 0
Var(D) = 100/16 + 100/25 = 6.25 + 4 = 10.25
SD(D) = √10.25 ≈ 3.2016
ot mean subtraction of the variances; the minus sign disappears when it is squared in Var(aX + bY) = a²Var(X) + b²Var(Y) with a = 1 and b = −1.
方差是相加而不是相减的，因为 D 是两个独立随机变量之差。随机变量相减并不意味着方差相减；当在 Var(aX + bY) = a²Var(X) + b²Var(Y) 中取 a = 1、b = −1 时，负号会在平方后消失。
