Finding Critical Regions for a Geometric Distribution | 寻找几何分布的临界区域

📚 Finding Critical Regions for a Geometric Distribution | 寻找几何分布的临界区域

In A-Level hypothesis testing, one of the more challenging scenarios involves the geometric distribution. Unlike the binomial distribution, the geometric distribution has an infinite sample space, and its shape is always positively skewed. Finding the critical region requires careful handling of cumulative probabilities and a clear understanding of how the alternative hypothesis translates into an extreme set of values for the test statistic. This article explains the step‑by‑step process, covers both one‑tailed and two‑tailed tests, and provides worked examples aligned with the Edexcel specification.

在 A‑Level 假设检验中，涉及几何分布的题目往往具有一定挑战性。与二项分布不同，几何分布的样本空间是无限的，其分布形态始终为正偏。寻找临界区域需要仔细处理累积概率，并清楚理解备择假设如何转化为检验统计量的极端值集合。本文按步骤讲解这一过程，涵盖单侧检验与双侧检验，并提供符合 Edexcel 考纲的详细例题。

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1. Recalling the Geometric Distribution | 回顾几何分布

The geometric distribution models the number of independent trials up to and including the first success. If the probability of success on each trial is p, we write X ~ Geo(p). The probability mass function is given by

几何分布模拟的是在一系列独立试验中，直到首次成功为止所进行的试验次数。如果每次试验成功的概率为 p，我们记作 X ~ Geo(p)。其概率质量函数为

P(X = x) = p × (1 – p)^{x – 1}, x = 1, 2, 3, …

The cumulative distribution function (CDF) is particularly useful when finding critical regions. Two key formulae are

累积分布函数在寻找临界区域时尤其有用。两个关键公式为

P(X ≤ x) = 1 – (1 – p)^x

P(X ≥ x) = (1 – p)^{x – 1}

These allow us to compute tail probabilities quickly without summing individual terms.

利用这些公式，我们可以快速计算尾部概率，而无需逐项相加。

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2. Hypothesis Testing with Discrete Distributions | 离散分布的假设检验

A hypothesis test begins with a null hypothesis H₀: p = p₀ and an alternative hypothesis H₁. The test statistic is the observed number of trials until the first success, x. Because the geometric distribution is discrete, we can rarely achieve the exact significance level α. Instead, we find the largest set of values whose total probability under H₀ does not exceed α. This set is called the critical region (or rejection region). Any observed value falling inside the critical region leads to rejection of H₀.

假设检验从原假设 H₀: p = p₀ 和备择假设 H₁ 开始。检验统计量是观测到的直到首次成功的试验次数 x。由于几何分布是离散的，我们极少能恰好达到显著性水平 α。因此，我们寻找的是在 H₀ 下总概率不超过 α 的最大值集合，这个集合称为 临界区域（或拒绝域）。若观测值落入临界区域，则拒绝原假设。

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3. Direction of the Alternative Hypothesis in Geometric Tests | 几何检验中备择假设的方向

One of the most common pitfalls is linking the direction of H₁ to the correct tail. For the geometric distribution, a smaller success probability p means you expect to wait longer for the first success. Therefore:

• If H₁: p < p₀, then under H₁ the number of trials tends to be larger, so we reject H₀ for unusually large values of x. This yields a critical region in the upper tail.
• If H₁: p > p₀, then under H₁ success occurs earlier, so we reject H₀ for unusually small values of x. This yields a critical region in the lower tail.

这个部分最容易出错：需要将 H₁ 的方向与正确的尾部对应起来。对于几何分布，成功概率 p 越小，意味着等待首次成功的时间越长。因此：

• 如果 H₁: p < p₀，则在备择假设下试验次数倾向于更大，因此当 x 异常大时拒绝原假设，临界区域位于上尾。
• 如果 H₁: p > p₀，则在备择假设下成功发生得更早，因此当 x 异常小时拒绝原假设，临界区域位于下尾。

This reasoning is the opposite of what some students instinctively apply, so it is essential to internalise it.

这一推理与一些学生直觉相反，因此必须牢固掌握。

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4. Upper-Tail Critical Regions for Testing p < p₀ | 上尾临界区域（检验 p < p₀）

To construct an upper‑tail critical region at significance level α for H₁: p < p₀, we need the smallest integer c such that P(X ≥ c | p = p₀) ≤ α. The critical region is then {x : x ≥ c}. Because P(X ≥ c) = (1 – p₀)^{c – 1} is a decreasing function of c, we can start from small values of c and increase until the inequality holds. The first c that satisfies the condition becomes the lower bound of the critical region.

要构建 H₁: p < p₀ 下显著性水平为 α 的上尾临界区域，我们需要满足 P(X ≥ c | p = p₀) ≤ α 的最小整数 c。此时临界区域为 {x : x ≥ c}。因为 P(X ≥ c) = (1 – p₀)^{c – 1} 是 c 的递减函数，我们可以从较小的 c 值开始尝试，逐步增大，直至不等式成立。最先满足条件的 c 即为临界区域的下界。

For example, suppose H₀: p = 0.3, H₁: p < 0.3, α = 0.05. Then P(X ≥ c) = 0.7^{c – 1}. Checking:

• c = 6: 0.7⁵ = 0.16807 > 0.05
• c = 7: 0.7⁶ ≈ 0.1176 > 0.05
• c = 8: 0.7⁷ ≈ 0.0824 > 0.05
• c = 9: 0.7⁸ ≈ 0.0576 > 0.05
• c = 10: 0.7⁹ ≈ 0.0404 < 0.05

Thus the smallest c with probability ≤ 0.05 is c = 10. The critical region is X ≥ 10. The actual significance level is about 0.0404.

例如，设 H₀: p = 0.3，H₁: p < 0.3，α = 0.05。则 P(X ≥ c) = 0.7^{c – 1}。检验：

• c = 6：0.7⁵ = 0.16807 > 0.05
• c = 7：0.7⁶ ≈ 0.1176 > 0.05
• c = 8：0.7⁷ ≈ 0.0824 > 0.05
• c = 9：0.7⁸ ≈ 0.0576 > 0.05
• c = 10：0.7⁹ ≈ 0.0404 < 0.05

因此，满足概率 ≤ 0.05 的最小 c 为 10，临界区域为 X ≥ 10，实际显著性水平约为 0.0404。

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5. Lower-Tail Critical Regions for Testing p > p₀ | 下尾临界区域（检验 p > p₀）

For H₁: p > p₀, we look at the lower tail. We need the largest integer c such that P(X ≤ c | p = p₀) ≤ α. The critical region is then {x : x ≤ c}. Since P(X ≤ c) = 1 – (1 – p₀)^c increases with c, we can test increasing values of c and stop just before the probability exceeds α. The largest c still satisfying the inequality gives the upper bound of the critical region.

对于 H₁: p > p₀，我们关注下尾。需要找到满足 P(X ≤ c | p = p₀) ≤ α 的最大整数 c，临界区域为 {x : x ≤ c}。因为 P(X ≤ c) = 1 – (1 – p₀)^c 随 c 增大而增大，我们可以从 c = 1 开始试，直到概率即将超过 α 为止。最后一个仍满足不等式的 c 就是临界区域的上界。

Using the same null value p₀ = 0.3 and α = 0.05, H₁: p > 0.3:

• c = 1: P(X ≤ 1) = 0.3 (≤ 0.05? No)
• Actually P(X ≤ 1) = 0.3 > 0.05, so even c = 1 exceeds α. Therefore there is no c ≥ 1 making P(X ≤ c) ≤ 0.05? Let's check carefully: we need a c such that the total probability is ≤ 0.05. c = 1 gives 0.3, already too large. So no lower‑ alues can accumulate a probability ≤ 0.05. In this case the test cannot reject H₀ at the 5% level for any observation? Wait — this example shows that with p₀ = 0.3, a lower‑tail test at 5% itical region is empty for α = 0.05. This illustrates that the test may have limited power at very small samples; we must choose a larger α or acknowledge that we cannot find a critical region. For a more illustrative example, use p₀ = 0.1, α = 0.05, H₁: p > 0.1. Then P(X ≤ c) = 1 – 0.9^c.

仍用 p₀ = 0.3 和 α = 0.05，H₁: p > 0.3：

• c = 1：P(X ≤ 1) = 0.3，已超过 0.05。因此不存在满足条件的 c，即无法构造出概率 ≤ α 的拒绝域。这说明，当原假设成功概率较大时，下尾检验可能无法在 5% 水平下拒绝任何观测值。此时只能调大 α，或承认无法找到临界区域。为了清晰展示下尾临界区域的求法，我们改用 p₀ = 0.1，α = 0.05，H₁: p > 0.1。此时 P(X ≤ c) = 1 – 0.9^c。

Check integers:

c	P(X ≤ c) = 1 – 0.9c
1	0.1
2	0.19
3	0.271

All of these exceed 0.05, so again no critical region exists. This shows that for small p₀, lower‑tail tests often require a larger α. However, in exam questions the parameters will be chosen so that a critical region does exist; the technique remains the same.

这些值均大于 0.05，因此仍然不存在临界区域。这说明，对于较小的 p₀，下尾检验往往需要更大的 α。但在考试题目中，通常会选择合适的参数使得临界区域存在；求解方法不会改变。

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6. Two-Tailed Tests and Splitting the Significance Level | 双侧检验与显著性水平分配

For a two‑tailed test with H₁: p ≠ p₀, we usually split the significance level equally between the two tails, allocating α/2 to each tail. We then find two critical values, c₁ and c₂, such that

• P(X ≤ c₁) ≤ α/2, with c₁ as large as possible,
• P(X ≥ c₂) ≤ α/2, with c₂ as small as possible.

The critical region becomes {x : x ≤ c₁} ∪ {x : x ≥ c₂}. Because the geometric distribution is skewed, the two tails are not symmetric; the tail probabilities are computed directly from the CDF rather than using a symmetric range.

对于双侧检验 H₁: p ≠ p₀，通常将显著性水平均分给两侧，每侧各分配 α/2。我们需要找到两个临界值 c₁ 和 c₂，使得：

• P(X ≤ c₁) ≤ α/2，且 c₁ 尽可能大，
• P(X ≥ c₂) ≤ α/2，且 c₂ 尽可能小。

此时临界区域为 {x : x ≤ c₁} ∪ {x : x ≥ c₂}。由于几何分布是偏态的，两侧并不对称；尾部概率直接使用 CDF 计算，而不用对称区间。

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7. Using the Cumulative Distribution Function | 使用累积分布函数

The cumulative probability formulae are the fastest way to determine critical bounds. For the lower tail we use P(X ≤ k) = 1 – (1 – p)^k. For the upper tail we use P(X ≥ k) = (1 – p)^{k – 1}. Because we are working with discrete values, we often construct a small table to scan for the required c. In an exam, you are expected to show these calculations, not just state the answer. Always state the critical region in set notation or as an inequality.

累积概率公式是确定临界界限的最快途径。下尾使用 P(X ≤ k) = 1 – (1 – p)^k，上尾使用 P(X ≥ k) = (1 – p)^{k – 1}。由于我们处理的是离散值，通常可以建立一个小表格来扫描所需的 c。考试中要求写出这些计算过程，而不能只给出答案。最后要用集合符号或不等式清楚地陈述临界区域。

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8. Worked Example 1: Lower-Tail Test | 例题 1：下尾检验

Scenario: ance is higher. You decide to play until you win once, recording the number of attempts, X ~ Geo(p). Test H₀: p = 0.4 against H₁: p > 0.4 at the 5% significance level. Find the critical region.

情景： 某游戏声称每次获胜概率为 0.4。你怀疑真实胜率更高。你决定一直玩到首次获胜，记录尝试次数 X ~ Geo(p)。在 5% 显著性水平下检验 H₀: p = 0.4，H₁: p > 0.4。求临界区域。

This is a lower‑tail test because H₁ claims a larger p (early success). We need the largest c with P(X ≤ c) ≤ 0.05. Using P(X ≤ c) = 1 – 0.6^c:

• c = 1: 1 – 0.6 = 0.4 > 0.05
• c = 2: 1 – 0.36 = 0.64 > 0.05
• ... In fact 0.6^c must be ≥ 0.95 for the probability to be ≤ 0.05. 0.6¹ = 0.6, 0.6² = 0.36, 0.6³ = 0.216, 0.6⁴ = 0.1296, 0.6⁵ = 0.07776, 0.6⁶ = 0.046656. So for 1 – 0.6^c ≤ 0.05 we need 0.6^c ≥ 0.95, which never happens. So no lower‑tail critical region exists at this α.

p₀ = 0.1, α = 0.05, H₁: p > 0.1. Already saw no region. So instead, we demonstrate a lower‑tail test with p₀ = 0.05, H₁: p > 0.05, α = 0.05? Let's choose a scenario that works: Suppose p₀ = 0.2, α = 0.10, H₁: p > 0.2. Then 1 – 0.8^c ≤ 0.10 → 0.8^c ≥ 0.90. 0.8¹=0.8, 0.8²=0.64, 0.8³=0.512, 0.8⁴=0.4096, 0.8⁵=0.32768, 0.8⁶=0.262144, 0.8⁷=0.2097, 0.8⁸=0.1678 → none ≥0.9. So lower‑tail tests with geometric often need large α or very small p₀? Actually if p₀ is very small, P(X=1)=p₀ is small, so you can get a lower‑tail critical region. For p₀=0.05, P(X=1)=0.05 ≤ 0.05, so c=1 works. Let' p₀=0.05, α=0.05, H₁: p > 0.05. Then P(X ≤ 1) = 0.05 ≤ 0.05, so c=1 is candidate. Could c=2? P(X ≤ 2)=1–75 >0.05, so largest c is 1. Critical region: X ≤ 1. That shows the point. For a more exam‑friendly example, use p₀=0.15? P(X=1)=0.15 >0.05, no region. Let's just state that when p₀ is not very small, lower‑ ₀ (upper tail) works. But for H₁: p > p₀, they often use a larger α or a smaller p₀. In Edexcel papers, you might see a question where they ask "show that the critical region is empty" or "explain why a critical region cannot be found". So it's worth including this possibility.

这是下尾检验，因为 H₁ 声称 p 更大（较早成功）。我们需要最大的 c 满足 P(X ≤ c) ≤ 0.05。使用 P(X ≤ c) = 1 – 0.6^c：可见最早的概率已 > 0.05，因此无临界区域。考试中可能会特意设计存在临界区域的题目，例如取 p₀ = 0.05，α = 0.05，则 c = 1 给出 P(X ≤ 1) = 0.05 ≤ 0.05，且 c = 2 时概率超过 0.05，因此临界区域为 X ≤ 1。另一种可能是题目要求说明临界区域为空。

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9. Worked Example 2: Upper-Tail Test | 例题 2：上尾检验

Question: A manufacturer claims that only 20% defect. Sofia believes the defect rate is higher, meaning the probability of finding a non‑ than 0.8. She defines X as the number of items inspected up to and including the first non‑defective one. So X ~ Geo(p), with p = P(non‑ H₁: p < 0.8 at α = 0.05. Find the critical region and state the actual significance level.

问题： 某制造商声称仅有 20% 的产品存在某种缺陷。Sofia 认为缺陷率更高，即首次找到合格品（成功）的概率 p 低于 0.8。她定义 X 为直至找到第一个合格品时所检查的物品数，则 X ~ Geo(p)，检验 H₀: p = 0.8 与 H₁: p < 0.8，α = 0.05。求临界区域并给出实际显著性水平。

This is an upper‑ because p < 0.8 means longer waits. We need minimum c with P(X ≥ c) ≤ 0.05. P(X ≥ c) = 0.2^{c – 1}.

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