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, …, so the distribution is discrete and unbounded above.

  可能的取值为 1, 2, 3, …，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 ∼ 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 ∼ 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 ≥ x) is the probability that the first x-1 trials are all failures, so P(X ≥ x) = q^x-1.

对几何分布而言，P(X ≥ x) 表示前 x-1 次试验全部失败的概率，因此 P(X ≥ 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[X²] directly. Using the probability mass function:

为了求方差，先计算 E[X(X-1)] 比直接计算 E[X²] 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: ∑_x=1^∞ q^x-1 = 11-q. Differentiate twice with respect to q:

从等比级数 ∑_x=1^∞ q^x-1 = 11-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[X²], use the identity:

我们已经知道 E[X] = 1/p。为求 E[X²]，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/p².

所以几何分布的方差为 q/p²。

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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, …, 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, …，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/p², not 1/p².

  错误：在方差公式中使用 p 而不是 q。方差是 q/p²，不是 1/p²。

• 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[X²] with E[X(X-1)]. Remember that E[X²] = E[X(X-1)] + E[X].

  错误：混淆 E[X²] 与 E[X(X-1)]。记住 E[X²] = 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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