Mean and Variance of a Geometric Distribution | 几何分布的均值与方差

📚 Mean and Variance of a Geometric Distribution | 几何分布的均值与方差

The geometric distribution models the number of independent trials needed to achieve the first success in a sequence of Bernoulli trials. This article explains how to find the mean and variance of a geometric random variable, including proofs and worked examples for Edexcel A-Level Mathematics.

几何分布用于建模在一系列独立伯努利试验中首次成功所需的试验次数。本文将解释如何求几何随机变量的均值与方差，包括证明与例题，适用于 Edexcel A-Level 数学。

---

1. What is a Geometric Distribution? | 什么是几何分布？

A geometric distribution arises when you repeat independent trials, each with the same probability of success p, until the first success occurs. The random variable X counts the number of trials needed to obtain that first success.

当你重复进行独立试验，每次成功概率为 p，直到首次成功出现时，就产生几何分布。随机变量 X 表示获得首次成功所需的试验次数。

For example, if a basketball player has a 0.3 chance of scoring from a free throw and keeps shooting until the first successful shot, the number of shots taken follows a geometric distribution with p = 0.3.

例如，一名篮球运动员罚球命中率为 0.3，一直投篮直到首次命中，那么投篮次数就服从 p = 0.3 的几何分布。

---

2. Key Conditions for a Geometric Model | 几何模型的关键条件

A trial can be modelled by a geometric distribution if three conditions hold. First, each trial is independent. Second, each trial has exactly two outcomes, usually called success and failure. Third, the probability of success p is constant for every trial.

满足以下三个条件时，试验可用几何分布建模。第一，每次试验相互独立。第二，每次试验只有两种结果，通常称为成功和失败。第三，每次试验的成功概率 p 保持不变。

The random variable is the number of trials up to and including the first success, so its possible values are 1, 2, 3, and so on.

随机变量是到首次成功为止的试验次数（包括成功的那一次），因此其可能取值为 1、2、3 等。

---

3. Probability Mass Function | 概率质量函数

For X ~ Geo(p), the probability that the first success occurs on the x-th trial is given by the probability mass function below.

对于 X ~ Geo(p)，首次成功发生在第 x 次试验的概率由下面的概率质量函数给出。

P(X = x) = p(1 − p)ˣ⁻¹, x = 1, 2, 3, …

This formula says that the first x − 1 trials must be failures, each with probability 1 − p, and the x-th trial must be a success with probability p.

这个公式表示前 x − 1 次试验必须失败，每次概率为 1 − p，而第 x 次试验必须成功，概率为 p。

---

4. Mean of a Geometric Distribution | 几何分布的均值

The expected value of X ~ Geo(p) is given by:

X ~ Geo(p) 的期望值为：

E(X) = 1/p

This result is intuitive: if the success probability is low, you expect to wait longer for the first success.

这个结果很直观：如果成功概率低，首次成功所需的平均试验次数就更多。

For example, if p = 0.2, the mean number of trials until the first success is 1/0.2 = 5.

例如，若 p = 0.2，首次成功前的平均试验次数为 1/0.2 = 5。

---

5. Proof of the Mean | 均值的证明

To prove the mean, start with the definition E(X) = ∑ x P(X = x). Substitute the geometric probability mass function into this definition.

为了证明均值，从定义 E(X) = ∑ x P(X = x) 开始。将几何概率质量函数代入该定义。

E(X) = ∑ x p(1 − p)ˣ⁻¹

Factor out p and use the standard series ∑ x rˣ⁻¹ = 1/(1 − r)² for |r| < 1. Let q = 1 − p.

提出 p，并利用标准级数 ∑ x rˣ⁻¹ = 1/(1 − r)²（|r| < 1）。令 q = 1 − p。

E(X) = p ∑ x qˣ⁻¹ = p × 1/(1 − q)² = p × 1/p² = 1/p

This completes the proof that the mean of a geometric distribution is the reciprocal of the success probability.

这就完成了几何分布均值为成功概率倒数的证明。

---

6. Variance of a Geometric Distribution | 几何分布的方差

The variance of X ~ Geo(p) is:

X ~ Geo(p) 的方差为：

Var(X) = (1 − p)/p²

The variance measures how spread out the number of trials is around the mean 1/p.

方差衡量试验次数围绕均值 1/p 的离散程度。

For a fixed success probability p, a smaller p gives both a larger mean and a much larger variance.

在固定成功概率 p 下，p 越小，均值越大，方差也越大。

---

7. Proof of the Variance | 方差的证明

A neat way to prove the variance is to first compute E[X(X − 1)]. Use the definition and the same series technique as for the mean.

证明方差的一个简洁方法是先计算 E[X(X − 1)]。利用定义和与均值相同的级数技巧。

E[X(X − 1)] = ∑ x(x − 1)p qˣ⁻¹ = p q ∑ x(x − 1) qˣ⁻²

The standard series ∑ x(x − 1) rˣ⁻² = 2/(1 − r)³ gives:

标准级数 ∑ x(x − 1) rˣ⁻² = 2/(1 − r)³ 得出：

E[X(X − 1)] = p q × 2/(1 − q)³ = 2q/p²

Then E(X²) = E[X(X − 1)] + E(X) = 2q/p² + 1/p. Finally, use Var(X) = E(X²) − [E(X)]².

然后 E(X²) = E[X(X − 1)] + E(X) = 2q/p² + 1/p。最后使用 Var(X) = E(X²) − [E(X)]²。

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

Since q = 1 − p, the variance formula is proved.

由于 q = 1 − p，方差公式得证。

---

8. Worked Example | 例题

A fair die is rolled until a six appears. Let X be the number of rolls needed. Find the mean and variance of X.

一个公平骰子被反复投掷，直到出现 6。设 X 为所需投掷次数。求 X 的均值和方差。

The probability of rolling a six is p = 1/6. Since each roll is independent and p is constant, X ~ Geo(1/6).

掷出 6 的概率为 p = 1/6。由于每次投掷独立且 p 不变，X ~ Geo(1/6)。

E(X) = 1/p = 1 ÷ (1/6) = 6

Var(X) = (1 − p)/p² = (5/6) ÷ (1/36) = 30

This means on average you expect 6 rolls to get the first six, with a variance of 30 rolls².

这意味着平均需要 6 次投掷才能得到第一个 6，方差为 30 次²。

---

9. Interpreting the Results | 结果解读

The mean 1/p tells you the expected waiting time until the first

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）