Mean and variance of the binomial distribution | 二项分布的均值与方差

📚 Mean and variance of the binomial distribution | 二项分布的均值与方差

robabilities, summarising the distribution with its mean and variance is essential for hypothesis testing and for understanding the spread of data. This article explains how to derive the formulas E(X) = np and Var(X) = np(1 − p) and how to apply them effectively in exam-style problems.

在A-Level统计中，二项分布对固定次数独立试验中的成功次数建模。除了计算概率，用均值和方差来概括分布对假设检验和理解数据离散程度至关重要。本文将解释如何推导 E(X) = np 和 Var(X) = np(1 − p) 公式，以及如何在考试题型中有效应用它们。

1. Revisiting the Binomial Distribution | 回顾二项分布

robability of success p remains constant. If the random variable X counts the number of successes, we write X ~ B(n, p). The probability of obtaining exactly r successes is P(X = r) = ⁿC_r p^r (1 − p)^n−r for r = 0, 1, 2, …, n.

当进行固定次数 n 的独立试验，每次试验结果要么成功要么失败，且成功概率 p 保持不变时，就产生二项分布。如果随机变量 X 计数成功次数，我们记作 X ~ B(n, p)。恰好获得 r 次成功的概率为 P(X = r) = ⁿC_r p^r (1 − p)^n−r，其中 r = 0, 1, 2, …, n。

2. The Binomial as a Sum of Bernoulli Trials | 二项分布作为伯努利试验之和

A binomial random variable can be viewed as the sum of n independent Bernoulli random variables. Let X_i = 1 if the i‑th trial is a success and X_i = 0 otherwise. Then X = X₁ + X₂ + … + X_n, where each X_i ~ Bernoulli(p) and the X_i are mutually independent. This representation simplifies the derivation of both the mean and the variance.

一个二项随机变量可以看作 n 个独立伯努利随机变量之和。设 X_i = 1，如果第 i 次试验成功，否则 X_i = 0。那么 X = X₁ + X₂ + … + X_n，其中每个 X_i ~ Bernoulli(p) 并且各 X_i 相互独立。这种表示能简化均值和方差的推导。

3. Deriving the Mean E(X) Using the Definition | 利用定义推导均值 E(X)

The mean, or expected value, of a discrete random variable is E(X) = Σ_r=0ⁿ r · P(X = r). Substituting the binomial probability gives E(X) = Σ_r=0ⁿ r · ⁿC_r p^r (1 − p)^n−r. Using the identity r·ⁿC_r = n·ⁿ⁻¹C_r−1, and letting m = n − 1, s = r − 1, the sum becomes n p Σ_s=0^{m m}C_s p^s (1 − p)^m−s. The remaining sum is exactly the total probability of a B(m, p) distribution, which equals 1. Hence E(X) = np.

离散随机变量的均值（期望值）为 E(X) = Σ_r=0ⁿ r · P(X = r)。代入二项概率得 E(X) = Σ_r=0ⁿ r · ⁿC_r p^r (1 − p)^n−r。利用恒等式 r·ⁿC_r = n·ⁿ⁻¹C_r−1，并令 m = n − 1, s = r − 1，该求和变为 n p Σ_s=0^{m m}C_s p^s (1 − p)^m−s。剩下的部分正好是 B(m, p) 分布的总概率，等于 1。因此 E(X) = np。

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4. Deriving the Mean Using Linearity of Expectation | 利用期望的线性性质推导均值

Using the Bernoulli sum representation, E(X) = E(X₁ + … + X_n) = E(X₁) + … + E(X_n) by linearity of expectation. For each Bernoulli trial, E(X_i p. This approach is much shorter and highlights why the mean does not depend on independence—linearity holds regardless.

利用伯努利和的表示，由期望的线性性质得 E(X) = E(X₁ + … + X_n) = E(X₁) + … + E(X_n)。对于每个伯努利试验，E(X_i) = 1 × p + 0 × (1 − p) = p。因此 E(X) = np。这种方法简洁得多，也说明均值不依赖于独立性——线性性质总是成立。

5. Deriving the Variance: E(X²) Approach | 推导方差：E(X²) 方法

ind Var(X), we can first evaluate E(X(X−1)). Using the probability function, E(X(X−1)) = Σ_r=0ⁿ r(r−1) ⁿC_r p^r (1−p)^n−r. Notice that r(r−1)ⁿC_r = n(n−1)ⁿ⁻²C_r−2 for r ≥ 2. Rewriting gives n(n−1)p² Σ_s=0^{n−2 n−2}C_s p^s (1−p)^(n−2)−s (X) = n(n−1)p² + np. Finally, Var(X) = E(X²) − [E(X)]² = n(n−1)p² + np − n²p² = np(1 − p).

为求 Var(X)，可以先计算 E(X(X−1))。利用概率函数，E(X(X−1)) = Σ_r=0ⁿ r(r−1) ⁿC_r p^r (1−p)^n−r。注意到当 r ≥ 2 时 r(r−1)ⁿC_r = n(n−1)ⁿ⁻²C_r−2。改写后得到 n(n−1)p² 乘以一个二项总和，其值为 1，故 E(X(X−1)) = n(n−1)p²。于是 E(X²) = E(X(X−1)) + E(X) = n(n−1)p² + np。最后 Var(X) = E(X²) − [E(X)]² = n(n−1)p² + np − n²p² = np(1 − p)。

6. Deriving the Variance Using Independent Bernoulli Sum | 利用独立伯努利和推导方差

ce of a sum of independent random variables equals the sum of their variances. Since X = X₁ + … + X_n and each X_i has variance p(1 − p), we obtain Var(X) = Var(X₁) + … +

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