📚 Unbiased Estimation of Mean and Variance | 均值与方差的无偏估计
In statistics, we often use samples to draw conclusions about an entire population. Estimating the population mean μ and variance σ² from sample data is a fundamental task. However, the natural formulas for estimating variance from a sample require careful adjustment to be correct. This article explains the concept of unbiased estimators, why the sample mean is unbiased for the population mean, and why we divide by n−1 when estimating the population variance.
在统计学中，我们经常利用样本来推论总体。用样本数据估计总体均值 μ 和总体方差 σ² 是一项基础任务。然而，样本估计方差的自然公式需要经过仔细调整才能正确。本文将解释无偏估计量的概念、为什么样本均值是总体均值的无偏估计量，以及在估计总体方差时为什么要除以 n−1。
1. What is an Unbiased Estimator? | 什么是无偏估计？
An estimator is a rule or formula used to estimate an unknown parameter from sample data. Because samples vary, an estimator is a random variable. It has a sampling distribution with its own mean and variance. We say an estimator θ̂ is unbiased for a parameter θ if the expected value of the estimator equals the true parameter value:
E(θ̂) = θ
The absence of bias means that, if we repeatedly took many samples and computed the estimate each time, the average of those estimates would converge to the true parameter. Unbiasedness does not guarantee that a single estimate is close to the true value; it speaks to the long-run average.
无偏性意味着，如果我们反复抽取多个样本并每次计算估计值，这些估计值的平均值将收敛到真实参数。无偏性并不保证单次估计一定接近真实值；它描述的是长期平均行为。
In IB statistics questions, this property matters because we often make point estimates from small samples. Using a biased formula can lead to systematic errors that do not disappear even when many samples are averaged.
在 IB 统计题目中，这一性质非常重要，因为我们经常从小样本进行点估计。使用有偏公式会导致系统性误差，即使对多个样本取平均也不会消失。
2. Population and Sample Statistics | 总体与样本统计量
Let x₁, x₂, …, xₙ be a random sample of size n from a population with mean μ and variance σ². The sample mean is defined as:
x̄ = (1/n)Σxᵢ
For a finite population of size N, the population variance is σ² = (1/N)Σ(xᵢ − μ)². The sample variance we commonly use is:
s² = (1/(n−1)) Σ(xᵢ − x̄)²
The denominator n−1 is called the number of degrees of freedom. The reason for this denominator is the central focus of this article.
设 x₁, x₂, …, xₙ 是来自均值为 μ、方差为 σ² 的总体的容量为 n 的随机样本。样本均值定义为：
x̄ = (1/n)Σxᵢ
对于大小为 N 的有限总体，总体方差为 σ² = (1/N)Σ(xᵢ − μ)²。而我们常用的样本方差为：
s² = (1/(n−1)) Σ(xᵢ − x̄)²
分母 n−1 称为自由度。这个分母的由来正是本文的核心内容。
3. Unbiasedness of the Sample Mean | 样本均值的无偏性
Consider the expected value of the sample mean:
E(x̄) = E[(1/n)Σxᵢ] = (1/n)ΣE(xᵢ) = (1/n)(nμ) = μ
Since E(x̄) = μ, the sample mean is an unbiased estimator of the population mean. On average, over many samples, the sample mean is correct.
考虑样本均值的期望：
E(x̄) = E[(1/n)Σxᵢ] = (1/n)ΣE(xᵢ) = (1/n)(nμ) = μ
由于 E(x̄) = μ，所以样本均值是总体均值的无偏估计量。在多次抽样的平均意义下，样本均值是准确的。
This result holds regardless of the underlying distribution, provided the population has a finite mean. It also explains why x̄ is the standard estimator for μ in confidence intervals and hypothesis tests.
只要总体具有有限的均值，无论其分布形式如何，这个结果都成立。这也解释了为什么 x̄ 是置信区间和假设检验中估计 μ 的标准量。
4. Why Do We Need n−1? | 为什么需要 n−1？
If we calculate the average squared deviation from the sample mean using a denominator of n, the resulting statistic is:
v = (1/n) Σ(xᵢ − x̄)²
Its expected value is not σ²; it is slightly smaller. The reason is that the sample mean is itself chosen to minimise the sum of squared deviations. As a result, the deviations from x̄ are systematically smaller than deviations from the true mean μ.
如果我们用分母 n 来计算围绕样本均值的平均平方偏差，得到的统计量为：
v = (1/n) Σ(xᵢ − x̄)²
它的期望并不是 σ²，而是略微偏小。原因在于样本均值本身是使平方偏差和达到最小的取值。因此，围绕 x̄ 的偏差平均上会小于围绕真实均值 μ 的偏差。
Because x̄ is computed from the same data, it automatically lies near the center of the observed points. This makes the squared deviations from x̄ underestimate the variability around the true population mean.
由于 x̄ 是从同一数据计算得到的，它自动位于观测点中心附近。这使得围绕 x̄ 的平方偏差低估了围绕真实总体均值的变异性。
5. Exact Derivation: E(s²) = σ² | 精确推导：E(s²) = σ²
Start by rewriting the sum of squared deviations from the sample mean:
Σ(xᵢ − x̄)² = Σ(xᵢ − μ + μ − x̄)²
= Σ(xᵢ − μ)² + 2Σ(xᵢ − μ)(μ − x̄) + Σ(μ − x̄)²
The cross term simplifies because Σ(xᵢ − μ) = n(x̄ − μ). Therefore:
Σ(xᵢ − x̄)² = Σ(xᵢ − μ)² − n(x̄ − μ)²
Taking expectations:
E[Σ(xᵢ − x̄)²] = ΣE[(xᵢ − μ)²] − nE[(x̄ − μ)²]
The first term is nσ². The variance of the sample mean is Var(x̄) = σ²/n, so E[(x̄ − μ)²] = σ²/n. Thus:
E[Σ(xᵢ − x̄)²] = nσ² − n(σ²/n) = (n−1)σ²
Therefore, dividing by n−1 gives:
E(s²) = E[(1/(n−1)) Σ(xᵢ − x̄)²] = σ²
从改写样本偏差平方和开始：
Σ(xᵢ − x̄)² = Σ(xᵢ − μ + μ − x̄)²
= Σ(xᵢ − μ)² + 2Σ(xᵢ − μ)(μ − x̄) + Σ(μ − x̄)²
交叉项化简，因为 Σ(xᵢ − μ) = n(x̄ − μ)。因此：
Σ(xᵢ − x̄)² = Σ(xᵢ − μ)² − n(x̄ − μ)²
两边取期望：
E[Σ(xᵢ − x̄)²] = ΣE[(xᵢ − μ)²] − nE[(x̄ − μ)²]
第一项为 nσ²。样本均值的方差为 Var(x̄) = σ²/n，所以 E[(x̄ − μ)²] = σ²/n。因此：
E[Σ(xᵢ − x̄)²] = nσ² − n(σ²/n) = (n−1)σ²
因此，除以 n−1 后有：
E(s²) = E[(1/(n−1)) Σ(xᵢ − x̄)²] = σ²
6. The Biased Estimator Dividing by n | 以 n 作分母的有偏估计量
From the derivation above, it is clear that:
E(v) = [(n−1)/n] σ²
Since (n−1)/n < 1, the estimator v systematically underestimates σ². Its bias is:
Bias(v) = E(v) − σ² = −σ²/n
For large n, the bias is small, but it never becomes exactly zero. This is why we say that dividing by n instead of n−1 produces a biased estimator.
由上述推导可知：
E(v) = [(n−1)/n] σ²
因为 (n−1)/n < 1，所以估计量 v 会系统性地低估 σ²。其偏差为：
Bias(v) = E(v) − σ² = −σ²/n
当 n 很大时偏差很小，但永远不会恰好为零。这就是为什么我们说以 n 而不是 n−1 作为分母会得到有偏估计量。
7. Bessel’s Correction | 贝塞尔修正
The factor n/(n−1) is known as Bessel’s correction. Multiplying the biased estimator by this factor yields the unbiased estimator:
s² = [n/(n−1)] × (1/n) Σ(xᵢ − x̄)² = (1/(n−1)) Σ(xᵢ − x̄)²
Notice that Bessel’s correction affects only the variance estimate, not the mean estimate. The standard deviation is the square root: s = √(s²). Although s itself is not mathematically unbiased for σ, it is the standard value commonly reported.
因子 n/(n−1) 称为贝塞尔修正。用这个因子乘以有偏估计量即可得到无偏估计量：
s² = [n/(n−1)] × (1/n) Σ(xᵢ − x̄)² = (1/(n−1)) Σ(xᵢ − x̄)²
注意，贝塞尔修正只影响方差估计，
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