Geometric Distribution Critical Regions for A-Level Further Maths | A-Level进阶数学：几何分布临界区域的求法

📚 Geometric Distribution Critical Regions for A-Level Further Maths | A-Level进阶数学：几何分布临界区域的求法

In A-Level Further Mathematics, the geometric distribution models the number of trials needed to achieve the first success. Finding critical regions for hypothesis tests involving this distribution is a key skill for Edexcel exam success.

在 A-Level 进阶数学中，几何分布用于描述首次成功所需试验次数。求解该分布假设检验中的临界区域，是 Edexcel 考试取得高分的关键技能。

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1. The Geometric Distribution Recall | 几何分布回顾

For a random variable \(X\) with geometric distribution \(X \sim \mathrm{Geo}(p)\), where \(p\) is the probability of success on each trial, the probability mass function is:

对于随机变量 \(X\)，若 \(X \sim \mathrm{Geo}(p)\)，其中 \(p\) 是每次试验成功的概率，则其概率质量函数为：

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

The cumulative distribution function used for finding critical regions is:

用于求临界区域的累积分布函数为：

P(X ≤ x) = 1 − (1−p)^x

A useful tail probability follows from the memoryless property:

由无记忆性可得一个有用的尾概率：

P(X ≥ x) = (1−p)^x−1

These three formulas are the foundation for every critical-region calculation.

这三个公式是所有临界区域计算的基础。

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2. Hypothesis Testing with a Geometric Distribution | 几何分布的假设检验

In a hypothesis test, we start with a null hypothesis \(H_0: p = p_0\). The alternative hypothesis can be one-sided or two-sided, but for the geometric distribution critical regions are almost always one-sided.

在假设检验中，我们先给出原假设 \(H_0: p = p_0\)。备择假设可以是单侧或双侧，但几何分布临界区域几乎总是单侧的。

A critical region is a set of values of \(X\) for which we reject \(H_0\). The total probability of observing a value in this region, calculated under \(H_0\), must be less than or equal to the significance level \(\alpha\).

临界区域是使我们需要拒绝原假设 \(H_0\) 的 \(X\) 取值集合。在原假设成立的条件下，落入该区域的概率之和必须小于等于显著性水平 \(\alpha\)。

• The critical region is chosen so that it contains the most extreme values in the direction suggested by \(H_1\).
• 临界区域应包含 \(H_1\) 所暗示方向上的最极端取值。

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3. Deciding the Direction of the Rejection Region | 判断拒绝域方向

For the geometric distribution, small values of \(X\) mean success happens early, so \(p\) is likely large. Large values of \(X\) mean success is delayed, so \(p\) is likely small.

对于几何分布，较小的 \(X\) 表示较早成功，因此 \(p\) 可能较大；较大的 \(X\) 表示成功较晚，因此 \(p\) 可能较小。

• If \(H_1: p < p_0\), then we expect \(X\) to be large. Use a right-tailed critical region \(\{X \ge c\}\).
• 若 \(H_1: p < p_0\)，则我们预期 \(X\) 较大，应使用右尾临界区域 \(\{X \ge c\}\)。
• If \(H_1: p > p_0\), then we expect \(X\) to be small. Use a left-tailed critical region \(\{X \le c\}\).
• 若 \(H_1: p > p_0\)，则我们预期 \(X\) 较小，应使用左尾临界区域 \(\{X \le c\}\)。

Getting this direction wrong is the most common student error.

方向判断错误是学生最常见的错误。

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4. Right-Tailed Critical Value: Finding c | 右尾临界值：求 c

For \(H_1: p < p_0\), reject \(H_0\) when \(X\) is too large. The critical region is \(\{X \ge c\}\), where \(c\) is the smallest integer satisfying:

对于 \(H_1: p < p_0\)，当 \(X\) 过大时拒绝 \(H_0\)。临界区域为 \(\{X \ge c\}\)，其中 \(c\) 是满足下式的最小整数：

P(X ≥ c) = (1−p₀)^c−1 ≤ α

Use logarithms to solve the inequality:

使用对数求解不等式：

c − 1 ≥ \frac{\ln \alpha}{\ln(1 − p_0)}

Therefore \(c = 1 + \lceil \ln\alpha / \ln(1−p_0) \rceil\), where \(\lceil \cdot \rceil\) is the ceiling function. Remember that \(\ln(1−p_0)\) is negative, so the inequality sign reverses when dividing.

因此 \(c = 1 + \lceil \ln\alpha / \ln(1−p_0) \rceil\)，其中 \(\lceil \cdot \rceil\) 是向上取整。记住 \(\ln(1−p_0)\) 为负数，因此相除时需要变号。

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5. Left-Tailed Critical Value: Finding c | 左尾临界值：求 c

For \(H_1: p > p_0\), reject \(H_0\) when \(X\) is too small. The critical region is \(\{X \le c\}\), where \(c\) is the largest integer satisfying:

对于 \(H_1: p > p_0\)，当 \(X\) 过小时拒绝 \(H_0\)。临界区域为 \(\{X \le c\}\)，其中 \(c\) 是满足下式的最大整数：

P(X ≤ c) = 1 − (1−p₀)^c ≤ α

Rearrange to obtain:

整理得到：

c ≤ \frac{\ln(1 − \alpha)}{\ln(1 − p_0)}

So \(c = \lfloor \ln(1−\alpha) / \ln(1−p_0) \rfloor\), where \(\lfloor \cdot \rfloor\) is the floor function. If this value is less than 1, then no non-empty critical region exists at that significance level.

因此 \(c = \lfloor \ln(1−\alpha) / \ln(1−p_0) \rfloor\)，其中 \(\lfloor \cdot \rfloor\) 是向下取整。如果该值小于 1，则在该显著性水平下不存在非空临界区域。

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6. Worked Example: Right-Tailed Test | 实例：右尾检验

Let \(X \sim \mathrm{Geo}(p)\). Test \(H_0: p = 0.2\) against \(H_1: p < 0.2\) at the 5% significance level.

设 \(X \sim \mathrm{Geo}(p)\)。在 5% 显著性水平下检验 \(H_0: p = 0.2\) 对 \(H_1: p < 0.2\)。

We need the smallest \(c\) such that \((0.8)^{c−1} \le 0.05\).

我们需要最小整数 \(c\)，使得 \((0.8)^{c−1} \le 0.05\)。

(c−1)\ln(0.8) ≤ \ln(0.05) ⇒ c−1 ≥ 13.43

Thus \(c−1 = 14\) and \(c = 15\). The critical region is \(\{X \ge 15\}\).

因此 \(c−1 = 14\)，\(c = 15\)。临界区域为 \(\{X \ge 15\}\)。

The actual significance is \(P(X \ge 15) = 0.8^{14} = 0.04398\), which is below 0.05 as required.

实际显著性水平为 \(P(X \ge 15) = 0.8^{14} = 0.04398\)，如要求所料低于 0.05。

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7. Worked Example: Left-Tailed Test | 实例：左尾检验

Suppose \(X \sim \mathrm{Geo}(p)\). Test \(H_0: p = 0.05\) against \(H_1: p > 0.05\) at the 5% significance level.

假设 \(X \sim \mathrm{Geo}(p)\)。在 5% 显著性水平下检验 \(H_0: p = 0.05\) 对 \(H_1: p > 0.05\)。

We want the largest \(c\) such that \(1 − (0.95)^c \le 0.05\).

我们需要最大整数 \(c\)，使得 \(1 − (0.95)^c \le 0.05\)。

(0.95)^c ≥ 0.95 ⇒ c ≤ \frac{\ln 0.95}{\ln 0.95} = 1

So \(c = 1\). The critical region is \(\{X \le 1\}\), with \(P(X \le 1) = 0.05\) exactly.

所以 \(c = 1\)。临界区域为 \(\{X \le 1\}\)，且 \(P(X \le 1) = 0.05\)，恰好等于显著性水平。

If \(p_0 = 0.2\) instead, the floor calculation would give \(c = 0\), meaning that no critical region exists because \(P(X = 1) = 0.2 > 0.05\).

若换作 \(p_0 = 0.2\)，向下取整会得到 \(c = 0\)，意味着不存在临界区域，因为 \(P(X = 1) = 0.2 > 0.05\)。

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8. Using Tables to Verify Critical Regions | 用概率表验证临界区域

In an exam, you may be given a cumulative probability table for specific \(p\) values. To find a critical region, scan the table for the first probability that falls below or crosses \(\alpha\).

考试中，你可能会获得针对特定 \(p\) 的累积概率表。为求临界区域，扫描表格找到第一个低于或超过 \(\alpha\) 的概率。

x	1	2	3	…
P(X ≤ x)	p	1−(1−p)²	1−(1−p)³	…

For a right-tailed region, calculate \(P(X \ge x) = 1 − P(X \le x−1)\) and locate where it first drops to \(\alpha\) or below.

对于右尾区域，计算 \(P(X \ge x) = 1 − P(X \le x−1)\)，找到首次降到 \(\alpha\) 或更低的位置。

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9. Common Mistakes and Tips | 常见错误与技巧

• Mixing up the direction of the alternative hypothesis. Always ask: would a smaller or larger \(X\) support \(H_1\)?
• 弄错备择假设的方向。始终问自己：\(X\) 偏小还是偏大会支持 \(H_1\)？
• Forgetting that probabilities in a critical region must be ≤ \(\alpha\), not < \(\alpha\).
• 忘记临界区域概率必须满足 ≤ \(\alpha\)，而不是 < \(\alpha\)。
• Using \(P(X \ge c) = 1 − (1−p)^c\) for the right tail, which is wrong. The correct expression is \((1−p)^{c−1}\).
• 右尾概率误用 \(P(X \ge c) = 1 − (1−p)^c\)。正确表达式是 \((1−p)^{c−1}\)。
• Not stating the actual significance level after finding the critical region. The exam reward requires this summarising step.
• 找到临界区域后不写出实际显著性水平。考试评分要求总结这一步。

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10. Summary | 总结

To find a geometric critical region, first decide the tail direction from \(H_1\). Then use the appropriate tail formula and solve the inequality for \(c\). Always verify that the final critical region has probability no greater than \(\alpha\).

求几何分布临界区域时，首先根据 \(H_1\) 确定尾部方向，然后使用对应的尾概率公式求解不等式得到 \(c\)。务必验证最终临界区域的概率不超过 \(\alpha\)。

With practice, these calculations become routine and secure full marks in the hypothesis-testing section of Edexcel Further Mathematics.

多加练习后，这些计算会变得非常熟练，从而在 Edexcel 进阶数学的假设检验部分获得满分。

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Published by TutorHao | Further Mathematics Revision Series | aleveler.com

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