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

In Edexcel A-Level Mathematics, the geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials. Understanding its mean and variance is essential for solving problems involving expected waiting times and variability. This article explains both derivations and provides worked examples.

在 Edexcel A-Level 数学中，几何分布用于描述在一系列独立的伯努利试验中，首次成功所需的试验次数。理解其均值和方差对于解决涉及期望等待时间和变异性的问题至关重要。本文将解释两者的推导过程，并提供计算示例。

---

1. Introduction to Geometric Distribution | 几何分布简介

The geometric distribution arises when we 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 up to and including that first success. It is denoted as X ~ Geo(p).

当我们重复进行独立试验，每次试验成功的概率相同为 p，直到首次成功发生时，就产生了几何分布。随机变量 X 表示截至首次成功（包括成功的那次）的试验次数，记作 X ~ Geo(p)。

nnot succeed before the first trial. It is widely used in quality control, risk analysis and games of chance.

该分布是离散分布，取值为 x = 1, 2, 3, ...，因为在第一次试验之前不可能成功。几何分布广泛应用于质量控制、风险分析和机会游戏中。

---

2. 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:

对于 X ~ Geo(p)，首次成功发生在第 x 次试验的概率为：

P(X = x) = p (1 - p)^{x - 1},   x = 1, 2, 3, ...

Here p is the probability of success on each trial, and (1 - p) represents the probability of failure. The formula captures the idea of (x - 1) consecutive failures followed by a success on the x-th trial.

这里 p 是每次试验成功的概率，(1 - p) 表示失败的概率。该公式体现了先有 (x - 1) 次连续失败，然后在第 x 次试验成功这一过程。

It is sometimes helpful to write q = 1 - p, so P(X = x) = p q^{x - 1} The cumulative probabilities can be found using the geometric series sum: P(X ≤ x) = 1 - q^x.

有时为了方便，记 q = 1 - p，那么 P(X = x) = p q^{x - 1}。累积概率可以用几何级数求和得到：P(X ≤ x) = 1 - q^x。

---

3. Conditions and Assumptions | 条件与假设

For a geometric distribution to be appropriate, three key conditions must be satisfied. Each trial must have exactly two possible outcomes (success or failure). The trials must be independent, so the outcome of one trial does not affect another. The probability of success p must remain constant across all trials. Finally, the trials continue until the first success occurs.

要使几何分布适用，必须满足三个关键条件。每次试验只能有两种可能的结果（成功或失败）。各次试验必须独立，即一次试验的结果不会影响另一次试验。所有试验中成功的概率 p 必须保持不变。最后，试验一直持续到首次成功发生为止。

duction line until a defective one is found, provided the population is large enough for p to remain effectively constant.

典型的例子包括反复投掷一枚均匀的骰子直到出现六点 (p = 1/6)，或者从生产线上检测产品直到发现一个次品，前提是产品总量足够大，使得 p 可以认为基本恒定。

---

4. Deriving the Mean E(X) | 期望值 E(X) 的推导

To find the mean, we start from the definition of expected value for a discrete random variable:

为了求均值，我们从离散随机变量期望值的定义出发：

E(X) = Σ_x=1^∞ x P(X = x) = Σ_x=1^∞ x p (1 - p)^{x - 1}

Let q = 1 - p to simplify notation. Then E(X) = p Σ_x=1^∞ x q^{x - 1}. The sum Σ_x=1^∞ x q^{x - 1} is a known series. By differentiating the geometric series Σ_x=0^∞ q^x = 1/(1 - q), we obtain:

令 q = 1 - p 以简化记号，则 E(X) = p Σ_x=1^∞ x q^{x - 1}。和式 Σ_x=1^∞ x q^{x - 1} 是一个已知级数。对几何级数 Σ_x=0^∞ q^x = 1/(1 - q) 求导，我们得到：

Σ_x=1^∞ x q^{x - 1} = 1/(1 - q)²

Since q = 1 - p, we have 1 - q = p. Substituting back yields E(X) = p × 1/p² e mean number of trials to the first success is the reciprocal of the success probability.

因为 q = 1 - p，所以 1 - q = p。代回可得 E(X) = p × 1/p² = 1/p。因此，到首次成功的平均试验次数是成功概率的倒数。

-Level exams, you can either quote E(X) = 1/p directly or show this derivation when asked.

这个结果很直观：如果一个事件发生的概率为 0.2，那么你预期要等待大约 5 次试验。在 A-Level 考试中，你可以直接引用 E(X) = 1/p，也可以在题目要求时展示上述推导过程。

---

5. Worked Example for Mean | 均值计算示例

A biased coin shows heads with probability 0.3. The coin is tossed until a head appears. Let X be the number of tosses required. Find the expected number of tosses.

一枚有偏倚的硬币出现正面的概率为 0.3。反复抛掷该硬币直到出现正面为止。设 X 为所需的抛掷次数。求抛掷次数的期望值。

Here X ~ 3.33 tosses are needed to see the first head. This illustrates how a lower success probability increases the expected waiting time.

这里 X ~ Geo(0.3)。利用 E(X) = 1/p，得到 E(X) = 1/0.3 ≈ 3.33。平均而言，大约需要抛掷 3.33 次才能看到第一次正面。这说明较低的成功概率会延长期望等待时间。

In an exam context, remember to state clearly: 'For a geometric distribution Geo(p), E(X) = 1/p.' Substituting p = 0.3 gives the answer.

在考试中，记得要清晰地写明：‘对于几何分布 Geo(p)，E(X) = 1/p。’ 代入 p = 0.3 即可得出答案。

---

广告 / Sponsored

6. Deriving the Variance Var(X) | 方差 Var(X) 的推导

the number of trials needed to achieve the first success. It is derived from Var(X) = E(X²) - [E(X)]². First, we find E(X²).

几何分布的方差衡量了获得首次成功所需试验次数的离散程度。它由 Var(X) = E(X²) - [E(X)]² 推导而来。首先我们需要求出 E(X²)。

E(X²) = Σ_x=1^∞ x² p q^{x - 1}. A useful trick is to write x² = x(x - 1) + x, which allows us to use known sums: E(X²) = p [ Σ x(x - 1) q^{x - 1} + Σ x q^{x - 1} ].

E(X²) = Σ_x=1^∞ x² p q^{x - 1}。一个有用的技巧是将 x² 写成 x(x - 1) + x，这样我们就可以利用已知的和式：E(X²) = p [ Σ x(x - 1) q^{x - 1} + Σ x q^{x - 1} ]。

The second sum Σ x q^{x - 1} = 1/(1 - q)² = 1/p² as before. For the first sum, we differentiate the geometric series twice: Σ_x=2^∞ x(x - 1) q^{x - 2} = 2/(1 - q)³. Hence Σ x(x - 1) q^{x - 1} = 2q/(1 - q)³ = 2q/p³.

第二个和式 Σ x q^{x - 1} = 1/(1 - q)² = 1/p² 与之前相同。对于第一个和式，我们对几何级数求两次导数：Σ_x=2^∞ x(x - 1) q^{x - 2} = 2/(1 - q)³。因此 Σ x(x - 1) q^{x - 1} = 2q/(1 - q)³ = 2q/p³。

Thus E(X²) = p (2q/p³ + 1/p²) = 2q/p² + 1/p. Since E(X) = 1/p, we get Var(X) = E(X²) - (1/p)² = (2q/p² + 1/p) - 1/p² = (2q + p - 1)/p². Because q = 1 - p, 2q + p - 1 = 2(1 - p) + p - 1 = 1 - p = q, so Var(X) = q/p², or equivalently Var(X) = (1 - p)/p².

因此 E(X²) = p (2q/p³ + 1/p²) = 2q/p² + 1/p。因为 E(X) = 1/p，我们有 Var(X) = E(X²) - (1/p)² = (2q/p² + 1/p) - 1/p² = (2q + p - 1)/p²。由于 q = 1 - p，2q + p - 1 = 2(1 - p) + p - 1 = 1 - p = q，所以 Var(X) = q/p²，或等价地 Var(X) = (1 - p)/p²。

You are expected to know and be able to use the formula Var(X) = (1 - p)/p² in Edexcel A-Level examinations. A full derivation demonstrates deeper understanding but is not always required.

在 Edexcel A-Level 考试中，你需要知道并能够使用公式 Var(X) = (1 - p)/p²。完整的推导能展示更深层次的理解，但并非每次都需要。

---

7. Worked Example for Variance | 方差计算示例

A fair six-sided die is rolled repeatedly until a '5' appears. Let Y be the number of rolls. Find the variance of Y.

一枚均匀的六面骰子被反复投掷，直到出现 ‘5’ 点。设 Y 为投掷次数。求 Y 的方差。

Here p = 1/6, so Y ~ Geo(1/6). Using the variance formula, Var(Y) = (1 - 1/6) / (1/6)² = (5/6) / (1/36) = (5/6) × 36 = 30. The standard deviation is √30 ≈ 5.48 rolls.

这里 p = 1/6，因此 Y ~ Geo(1/6)。使用方差公式，Var(Y) = (1 - 1/6) / (1/6)² = (5/6) / (1/36) = (5/6) × 36 = 30。标准差为 √30 ≈ 5.48 次投掷。

This relatively large variance reflects the high uncertainty when p is small. The mean number of rolls is 6, but individual experiments can deviate substantially from that mean.

这个相对较大的方差反映了当 p 较小时不确定性很高。平均投掷次数是 6 次，但单次实验的结果可能会与均值有很大偏差。

the mean is 6, and a variance of 30 is plausible because the distribution is right-skewed.

务必检查答案的合理性：因为 p = 1/6，均值为 6，方差 30 是合理的，因为几何分布是右偏的。

---

8. Extension: Probability Generating Function (Further Maths) | 拓展：概率生成函数（进阶数学）

ability generating function (PGF). The PGF for Geo(p) with support {1,2,...} is G(t) = E(t^X) = pt / (1 - qt), for |t| < 1/q.

对于学习 Edexcel 进阶数学的学生，几何分布的均值和方差还可以利用概率生成函数 (PGF) 来推导。定义在 {1,2,...} 上的 Geo(p) 的 PGF 为 G(t) = E(t^X) = pt / (1 - qt)，其中 |t| < 1/q。

Then E(X) = G'(1). Differentiating, G'(t) = p(1 - qt + qt) / (1 - qt)² = p/(1 - qt)². Evaluating at t = 1 gives E(X) = p/(1 - q)² = p/p² = 1/p.

那么 E(X) = G'(1)。对 G 求导，G'(t) = p(1 - qt + qt) / (1 - qt)² = p/(1 - qt)²。在 t = 1 处取值得到 E(X) = p/(1 - q)² = p/p² = 1/p。

Similarly, E(X(X-1)) = G''(1), and Var(X) = G''(1) + G'(1) - [G'(1)]². Carrying out these steps confirms Var(X) = q/p². This method is elegant and links to Further Statistics 1 content.

类似地，E(X(X-1)) = G''(1)，且 Var(X) = G''(1) + G'(1) - [G'(1)]²。执行这些步骤可以验证 Var(X) = q/p²。这种方法十分巧妙，与进阶统计 1 的内容相关联。

---

9. Common Mistakes and Tips | 常见错误与注意事项

One frequent error is confusing the two common versions of the geometric distribution. In Edexcel, X counts the number of trials including the success, so P(X = x) = p (1-p)^{x - 1}. The alternative version counts failures before the first success and has mean (1-p)/p. Always check which definition is being used.

一个常见错误是混淆几何分布的两种常用形式。在 Edexcel 中，X 计算的是包括成功在内的试验次数，因此 P(X = x) = p (1-p)^{x - 1}。另一种形式计算首次成功前的失败次数，其均值为 (1-p)/p。务必检查题目使用的是哪种定义。

Another mistake is using the wrong variance formula. Remember: Var(X) = (1-p)/p², not (1-p)/p. The squared p in the denominator often catches students out. Adding units and contextualising results helps avoid careless errors.

另一个错误是使用了错误的方差公式。记住：Var(X) = (1-p)/p²，而不是 (1-p)/p。分母中 p 的平方经常让学生出错。添加单位并结合实际情况理解结果有助于避免粗心错误。

When deriving sums, do not forget that the geometric series converges only for |q| correctly adjust the starting index when differentiating (e.g., x(x-1) sum starts at x=2).

在推导级数时，不要忘记几何级数仅在 |q| < 1 时收敛，这对于概率总是成立的。此外，要确保在求导时正确调整起始索引（例如，x(x-1) 的和式从 x=2 开始）。

---

10. Summary and Key Formulae | 总结与关键公式

For a geometric distribution X ~ Geo(p) where X is the trial number of the first success, the key results are:

对于几何分布 X ~ Geo(p)，其中 X 是首次成功时的试验次数，主要结果如下：

Probability mass function	P(X = x) = p (1 - p)x - 1, x = 1,2,3,...
Mean E(X)	1/p
Variance Var(X)	(1 - p) Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com 广告 / Sponsored 培生爱德思考试教材 统计数学3 Edexcel International A Level Mathematics S… 数学 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥156.0 更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法 在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao