Geometric Distribution Mean & Variance Derivation | 几何分布的均值与方差推导

📚 Geometric Distribution Mean & Variance Derivation | 几何分布的均值与方差推导

The geometric distribution models the number of independent Bernoulli trials needed until the first success occurs. In A-Level Further Mathematics, you are often asked to use its mean and variance, but understanding where those formulas come from is essential for exam questions that test derivation skills.

几何分布用于描述在独立重复试验中，直到第一次成功出现所需的试验次数。在 A-Level 进阶数学中，你不仅要会使用它的均值与方差公式，更要理解这些公式的来源，因为考试中常常会考查推导过程。

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1. What Is the Geometric Distribution? | 什么是几何分布？

Suppose a repeated experiment has only two outcomes: success and failure. Each trial is independent, and the probability of success is the same on every trial. We let \(p\) be the probability of success, so the probability of failure is \(q = 1-p\).

假设一个重复试验只有两种结果：成功与失败。每次试验相互独立，且每次试验成功的概率保持不变。设成功概率为 \(p\)，则失败概率为 \(q = 1-p\)。

• We count the number of trials until the first success appears.

  我们记录直到第一次成功出现为止的试验次数。

• The possible values are \(1, 2, 3, \dots\), so the distribution is discrete and unbounded above.

  可能的取值为 \(1, 2, 3, \dots\)，因此这是一个离散且无上界的分布。

• A classic example is rolling a fair die until a 6 appears.

  一个经典例子是重复掷均匀骰子，直到点数 6 首次出现。

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2. Probability Mass Function and Notation | 概率质量函数与记号

If random variable \(X\) follows a geometric distribution with parameter \(p\), its probability mass function is:

若随机变量 \(X\) 服从参数为 \(p\) 的几何分布，则其概率质量函数为：

P(X = x) = p q^x−1, x = 1, 2, 3, …

The justification is simple: to get the first success on trial \(x\), the first \(x-1\) trials must all be failures, each with probability \(q\), and trial \(x\) must be a success with probability \(p\).

理由很简单：要在第 \(x\) 次试验才首次成功，前 \(x-1\) 次必须全部失败，每次概率为 \(q\)，而第 \(x\) 次必须成功，概率为 \(p\)。

We write \(X \sim \text{Geo}(p)\). In this article we use the convention where \(X\) counts the number of trials up to and including the first success.

我们记作 \(X \sim \text{Geo}(p)\)。本文采用“直到首次成功所需总试验次数”这一约定。

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3. Useful Infinite Series | 必备的无穷级数

To derive the mean and variance, we need some standard results about geometric series. Since \(0 < p < 1\), we also have \(0 < q < 1\), so the following series converge.

为了推导均值与方差，我们需要一些关于等比级数的标准结论。因为 \(0 < p < 1\)，所以 \(0 < q < 1\)，下列级数均收敛。

Σ_x=1^∞ q^x−1 = 1 / (1 − q) = 1 / p

Differentiating both sides of this series with respect to \(q\) gives another important identity:

将上式两边关于 \(q\) 求导，可以得到另一个重要恒等式：

Σ_x=1^∞ x q^x−1 = 1 / (1 − q)² = 1 / p²

These results are valid because within the radius of convergence \(|q| < 1\), term-by-term differentiation is allowed.

在收敛半径 \(|q| < 1\) 内，逐项求导是合法的，因此这些结果成立。

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4. Deriving the Mean: Direct Summation | 推导均值：直接求和法

The expected value of \(X\) is defined as:

随机变量 \(X\) 的期望定义为：

E[X] = Σ_x=1^∞ x P(X = x) = Σ_x=1^∞ x p q^x−1

Using the differentiated geometric series identity from the previous section:

利用上一节中的求导恒等式：

E[X] = p Σ_x=1^∞ x q^x−1 = p × (1 / p²) = 1 / p

Therefore the mean of the geometric distribution is \(1/p\). This matches intuition: if an event has probability \(p\), the expected waiting time until it happens is \(1/p\).

因此几何分布的均值为 \(1/p\)。这与直觉一致：若某事件发生的概率为 \(p\)，则期望等待时间就是 \(1/p\)。

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5. Deriving the Mean: Tail-Sum Method | 推导均值：尾和法

For any positive integer-valued random variable, there is a convenient identity:

对任何取正整数的随机变量，有一个便捷恒等式：

E[X] = Σ_x=1^∞ P(X ≥ x)

For the geometric distribution, \(P(X \ge x)\) is the probability that the first \(x-1\) trials are all failures, so \(P(X \ge x) = q^{x-1}\).

对几何分布而言，\(P(X \ge x)\) 表示前 \(x-1\) 次试验全部失败的概率，因此 \(P(X \ge x) = q^{x-1}\)。

E[X] = Σ_x=1^∞ q^x−1 = 1 / (1 − q) = 1 / p

This method is often quicker and is also useful in other discrete distributions.

这种方法往往更快捷，也常用于其他离散分布。

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6. Finding E[X(X − 1)] | 计算 E[X(X − 1)]

To obtain the variance, it is easier to first compute \(E[X(X-1)]\) rather than \(E[X^2]\) directly. Using the probability mass function:

为了求方差，先计算 \(E[X(X-1)]\) 比直接计算 \(E[X^2]\) 更方便。利用概率质量函数：

E[X(X−1)] = Σ_x=1^∞ x(x−1) p q^x−1

Start from the geometric series:
\[
\sum_{x=1}^{\infty} q^{x-1} = \frac{1}{1-q}.
\]
Differentiate twice with respect to \(q\):

从等比级数
\[
\sum_{x=1}^{\infty} q^{x-1} = \frac{1}{1-q}
\]
出发，对 \(q\) 求二阶导数：

Σ_x=1^∞ x(x−1) q^x−2 = 2 / (1 − q)³

Multiplying by \(q\) to convert \(q^{x-2}\) into \(q^{x-1}\):

为了将 \(q^{x-2}\) 化为 \(q^{x-1}\)，两边乘以 \(q\)：

Σ_x=1^∞ x(x−1) q^x−1 = 2q / (1 − q)³ = 2q / p³

Now multiply by the leading \(p\) from the PMF:

再乘上概率质量函数中前面的 \(p\)：

E[X(X−1)] = p × (2q / p³) = 2q / p²

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7. Deriving the Variance | 推导方差

Recall the variance formula:

回顾方差公式：

Var(X) = E[X²] − (E[X])²

We already know \(E[X] = 1/p\). To find \(E[X^2]\), use the identity:

我们已经知道 \(E[X] = 1/p\)。为求 \(E[X^2]\)，利用恒等式：

X² = X(X−1) + X

Taking expectations:

两边取期望：

E[X²] = E[X(X−1)] + E[X] = (2q / p²) + (1 / p)

Simplify by writing \(q = 1-p\):

将 \(q = 1-p\) 代入化简：

E[X²] = (2(1−p) / p²) + (1 / p) = (2 − p) / p²

Therefore:

因此：

Var(X) = (2 − p) / p² − (1 / p²) = (1 − p) / p² = q / p²

Thus the variance of the geometric distribution is \(q/p^2\).

所以几何分布的方差为 \(q/p^2\)。

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8. Alternative Convention: Number of Failures Before Success | 另一种约定：首次成功前的失败次数

Some textbooks define \(Y\) as the number of failures before the first success. Then \(Y\) takes values \(0, 1, 2, \dots\), and its PMF is:

有些教材将 \(Y\) 定义为首次成功前失败的次数。此时 \(Y\) 取值为 \(0, 1, 2, \dots\)，其概率质量函数为：

P(Y = y) = p q^y, y = 0, 1, 2, …

Since \(Y = X – 1\), where \(X\) is the number of trials up to the first success, we can immediately obtain:

因为 \(Y = X – 1\)，其中 \(X\) 是首次成功所需的总试验次数，所以可以直接得到：

E[Y] = E[X] − 1 = (1 / p) − 1 = q / p

Var(Y) = Var(X) = q / p²

Notice that the variance is the same in both conventions, while the mean differs by 1. Always check which convention your exam board uses.

注意，两种约定下方差相同，而均值相差 1。请务必确认你的考试委员会使用哪一种约定。

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9. Worked Example: Fair Die | 例题：均匀骰子

Let \(X\) be the number of rolls of a fair die until a 6 appears. Here \(p = 1/6\) and \(q = 5/6\).

设 \(X\) 为掷均匀骰子直到点数 6 出现所需的次数。此时 \(p = 1/6\)，\(q = 5/6\)。

The mean is:

均值为：

E[X] = 1 / p = 6

The variance is:

方差为：

Var(X) = q / p² = (5/6) / (1/36) = 30

The probability that the first 6 appears within three rolls is:

三次以内首次出现 6 的概率为：

P(X ≤ 3) = p + pq + pq² = 1 − q³ = 1 − (5/6)³ = 91/216

This example shows that a mean of 6 does not mean exactly 6 rolls will occur; there is still very large variability, with variance 30.

这个例子说明均值为 6 并不代表一定需要掷 6 次，其方差为 30，说明波动很大。

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10. Common Mistakes and Exam Tips | 常见错误与考试提示

• Mistake: Using \(p\) instead of \(q\) in the variance formula. The variance is \(q/p^2\), not \(1/p^2\).

  错误：在方差公式中使用 \(p\) 而不是 \(q\)。方差是 \(q/p^2\)，不是 \(1/p^2\)。

• Mistake: Forgetting to multiply by \(p\) after differentiating the geometric series.

  错误：对等比级数求导后忘记乘以 \(p\)。

• Mistake: Confusing \(E[X^2]\) with \(E[X(X-1)]\). Remember that \(E[X^2] = E[X(X-1)] + E[X]\).

  错误：混淆 \(E[X^2]\) 与 \(E[X(X-1)]\)。记住 \(E[X^2] = E[X(X-1)] + E[X]\)。

• Tip: Always state the lower limit of the summation. For trials-until-success, \(x\) starts at 1.

  提示：始终写清求和下限。对于“直到成功所需试验次数”，\(x\) 从 1 开始。

• Tip: If the question defines the distribution as failures before success, add a clear note in your working.

  提示：如果题目将分布定义为成功前失败的次数，请在解答中明确标注。

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11. Summary Table | 总结表

The table below summarises both conventions.

下表总结了两种约定的结果。

Convention	PMF	Mean	Variance
Trials until first success	p qx−1, x = 1,2,…	1/p	q/p²
Failures before first success	p qy, y = 0,1,2,…	q/p	q/p²

Remember that the derivation of the variance relies on differentiating a geometric series twice. Mastering this technique will help you answer derivation questions confidently.

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