定量方法（Quantitative Methods）— 概率论模块 · 综合练习课
一、本课定位
L115-L122 完成了概率论模块全部核心知识点的学习。本课为综合练习课，通过跨知识点的综合题目帮助巩固理解，查漏补缺。
| 项目 | 说明 |
|---|---|
| 模块 | 2.4 概率论 |
| 覆盖范围 | L115-L122 全部知识点 |
| 后继 | L124 概率周测（10 题） |
| 难度 | ★★★★☆ |
| 考试权重 | 高（概率论约占 Quantitative Methods 的 25-30%） |
| 练习时长 | 约 30 分钟 |
二、知识点速查
核心公式汇总
| 编号 | 知识点 | 核心公式 / 概念 |
|---|---|---|
| 1 | 概率加法法则 | P(A∪B) = P(A) + P(B) − P(A∩B) |
| 2 | 条件概率 | P(A|B) = P(A∩B) / P(B) |
| 3 | 乘法法则 | P(A∩B) = P(A|B) · P(B) |
| 4 | 全概率公式 | P(A) = Σ P(A|Bᵢ) · P(Bᵢ) |
| 5 | 贝叶斯公式 | P(B|A) = [P(A|B)·P(B)] / P(A) |
| 6 | 独立性检验 | P(A∩B) = P(A)·P(B) ⇔ A与B独立 |
| 7 | 排列公式 | nPr = n! / (n−r)! |
| 8 | 组合公式 | nCr = n! / [r!(n−r)!] |
| 9 | 期望值 | E(X) = Σ xᵢ·P(xᵢ) |
| 10 | 方差 | Var(X) = Σ (xᵢ − E(X))²·P(xᵢ) = E(X²)−[E(X)]² |
| 11 | 协方差 | Cov(X,Y) = E[(X−μₓ)(Y−μᵧ)] = E(XY)−E(X)·E(Y) |
| 12 | 相关系数 | ρ = Cov(X,Y) / (σₓ·σᵧ) |
| 13 | 正态分布标准化 | z = (X − μ) / σ |
| 14 | 正态分布概率 | 68-95-99.7 法则 |
| 15 | 对数正态均值 | E(X) = e^(μ + σ²/2) |
| 16 | 对数正态中位数 | Median(X) = e^μ |
| 17 | 连续复利收益 | r = ln(P₁/P₀) |
常考分布对比
| 分布 | 支撑集 | 参数 | 关键特征 |
|---|---|---|---|
| 正态分布 | (−∞, +∞) | μ, σ² | 对称钟形，68-95-99.7法则 |
| 标准正态 | (−∞, +∞) | μ=0, σ=1 | z表查概率 |
| 对数正态 | (0, +∞) | μ, σ²（lnX的参数） | 右偏，中位数<均值 |
| 二项分布 | {0, 1, …, n} | n, p | n次独立伯努利试验成功次数 |
三、综合练习题
第一部分：基础概念（Q1-Q4）
Q1 关于概率的基本概念，以下哪项错误？
A. 互斥事件意味着 P(A∩B) = 0
B. 若 A 与 B 独立，则 P(A|B) = P(A)
C. 若 P(A) = 0.4，P(B) = 0.5，则 P(A∪B) 一定等于 0.9
D. 完备事件组（exhaustive events）的概率之和为 1
Q2 以下关于期望值和方差的说法，正确的是：
A. E(X+Y) = E(X) + E(Y) 仅在 X 与 Y 独立时成立
B. Var(X−Y) = Var(X) − Var(Y) 当 X 与 Y 独立时成立
C. E(aX + b) = aE(X) + b 始终成立
D. Var(aX) = a·Var(X) 始终成立
Q3 两个事件 A 和 B，已知 P(A) = 0.3，P(B) = 0.4，P(A|B) = 0.5。求 P(B|A)：
A. 1/3
B. 2/3
C. 0.375
D. 0.500
Q4 从 10 只股票中选取 4 只等权重构建投资组合，共有多少种不同的组合方式？（不考虑顺序）
A. 10 × 9 × 8 × 7 = 5040
B. 10! / (4! × 6!) = 210
C. 10! / 6! = 5040
D. 4¹⁰
第二部分：概率计算（Q5-Q8）
Q5 某投资策略有 60% 概率盈利 ¥2000，有 40% 概率亏损 ¥1000。该策略的期望收益是：
A. ¥600
B. ¥800
C. ¥1000
D. ¥1200
Q6 接上题，该策略的方差（以千元²为单位）最接近：
A. 1.44
B. 2.16
C. 2.56
D. 3.24
Q7 某基金经理想从 15 只候选股票中选出 3 只，按权重从小到大排序后构建一个递增权重的投资组合。有多少种不同的排序结果？
A. 15³ = 3375
B. C(15,3) = 455
C. P(15,3) = 2730
D. 3 × 15 = 45
Q8 一家公司的股票在任意交易日上涨的概率为 0.55，各日独立。连续观察 5 个交易日，恰好有 3 天上涨的概率最接近：
A. 0.275
B. 0.336
C. 0.500
D. 0.550
第三部分：贝叶斯应用（Q9-Q10）
Q9 某基金公司有两位基金经理：张经理管理 70% 的资金，其跑赢基准的概率为 0.6；李经理管理 30% 的资金，其跑赢基准的概率为 0.8。现随机选一只基金，结果它跑赢了基准。这只基金由张经理管理的概率是：
A. 0.42
B. 0.56
C. 0.636
D. 0.700
Q10（进阶） 使用贝叶斯更新：某检测方法识别财务造假的真阳性率为 90%（造假被检出的概率），假阳性率为 5%（没造假但被误判为造假的概率）。若市场上财务造假的公司占比约为 2%，那么当一家公司被检测判定为造假时，它真正造假的概率最接近：
A. 2%
B. 18%
C. 27%
D. 90%
第四部分：正态分布与 z 分数（Q11-Q13）
Q11 已知基金收益率服从 N(12%, 20%²)。收益率超过 52% 的概率最接近（用 68-95-99.7 法则估算）：
A. 16%
B. 5%
C. 2.5%
D. 0.15%
Q12 接上题，收益率低于 −8% 的概率最接近：
A. 2.5%
B. 5%
C. 16%
D. 32%
Q13 某股票日收益率服从 N(0.1%, 2%²)。标准差（波动率）是多少？
A. 0.04%
B. 0.1%
C. 2%
D. 4%
第五部分：协方差与相关性（Q14-Q15）
Q14 已知股票 A 和 B 的协方差 Cov(A,B) = 0.018。股票 A 的波动率为 15%，股票 B 的波动率为 20%。两股票的相关系数最接近：
A. 0.45
B. 0.60
C. 0.67
D. 0.75
Q15 以下关于协方差和相关系数的说法，错误的是：
A. 协方差的符号表示两个变量线性关系的方向
B. 相关系数取值范围为 [−1, +1]
C. 若 ρ = 0，则两个随机变量一定独立
D. Cov(X,X) = Var(X)
第六部分：对数正态分布（Q16-Q17）
Q16 已知 ln X ~ N(3, 0.4²)。中位数最接近：
A. e³ ≈ 20.09
B. e^(3.08) ≈ 21.76
C. e^(3.16) ≈ 23.57
D. e^(3.2) ≈ 24.53
Q17 接上题，E(X) 最接近：
A. 20.09
B. 21.76
C. 23.57
D. 24.53
第七部分：综合应用（Q18-Q20）
Q18 投资组合由两种资产构成：资产 X 权重 60%，资产 Y 权重 40%。已知 σₓ = 10%，σᵧ = 15%，ρₓᵧ = 0.3。两资产的协方差 Cov(X,Y) 为：
A. 0.0030
B. 0.0045
C. 0.0180
D. 0.0450
Q19 关于正态分布与对数正态分布的关系，以下说法正确的是：
A. 若 X ~ 对数正态，则 e^X ~ 正态
B. 对数正态分布用于建模可以为负的变量
C. 对数正态分布的众数 > 中位数 > 均值
D. 股票价格在 Black-Scholes 框架下假设服从对数正态分布
Q20 某分析师使用贝叶斯公式评估一只股票"被低估"的概率。已知：
- 该行业内股票被低估的基准概率：30%
- 若股票被低估，出现当前估值指标的可能性：80%
- 若股票未被低估，出现当前估值指标的可能性：20%
根据当前的估值指标，股票被低估的后验概率为：
A. 24%
B. 36.8%
C. 54.5%
D. 63.2%
四、答案与解析
【Q1 答案】C
逐项分析：
- A ✅ 互斥事件定义：A∩B = ∅，故 P(A∩B) = 0
- B ✅ 独立事件定义：P(A|B) = P(A)
- C ❌ 若 A 和 B 有交集，P(A∪B) = 0.4+0.5−P(A∩B) < 0.9；只有当 P(A∩B)=0 时才等于 0.9。题干未说明互斥，所以不能断定一定为 0.9
- D ✅ 完备事件组定义
🧠 CFA 常见陷阱：默认假设事件互斥而直接用 P(A)+P(B)。
【Q2 答案】C
逐项分析：
- A ❌ E(X+Y) = E(X) + E(Y) 始终成立，不依赖独立性
- B ❌ Var(X−Y) = Var(X) + Var(Y) − 2Cov(X,Y)，不独立时含协方差项；即使独立也是 Var(X) + Var(Y)，不是减法
- C ✅ 线性变换的期望始终成立
- D ❌ Var(aX) = a²·Var(X)，注意是 a² 而不是 a
🧠 方差运算的核心记忆：方差二次方（Var(aX)=a²Var(X)），协方差一次方。
【Q3 答案】B — 2/3
计算过程：
$$\begin{aligned} P(A \cap B) &= P(A|B) \cdot P(B) = 0.5 \times 0.4 = 0.2 \ P(B|A) &= \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.3} = \frac{2}{3} \approx 0.667 \end{aligned}$$
🧠 条件概率两种方向的转换：通过 P(A∩B) 桥梁。
【Q4 答案】B — 210
等权重组合中股票的选择无顺序差异 → 使用组合公式 C(10,4)：
$$C(10, 4) = \frac{10!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
A 和 C 是排列 P(10,4)=5040，D 无意义。
🧠 判断用排列还是组合：顺序不重要 → 组合 / 顺序重要 → 排列。
【Q5 答案】B — ¥800
$$E(X) = 0.6 \times 2000 + 0.4 \times (-1000) = 1200 - 400 = 800$$
【Q6 答案】B — 2.16
$$\begin{aligned} E(X^2) &= 0.6 \times 2000^2 + 0.4 \times (-1000)^2 \ &= 0.6 \times 4,000,000 + 0.4 \times 1,000,000 \ &= 2,400,000 + 400,000 = 2,800,000 \[8pt] \text{Var}(X) &= E(X^2) - [E(X)]^2 \ &= 2,800,000 - 800^2 \ &= 2,800,000 - 640,000 = 2,160,000 \end{aligned}$$
以千元²为单位：2,160,000 / 1,000,000 = 2.16
🧠 用 E(X²)−[E(X)]² 比直接用 Σ(xᵢ−μ)²P(xᵢ) 更快。
【Q7 答案】C — P(15,3) = 2730
题干要点："按权重从小到大排序" → 顺序重要 → 排列：
$$P(15, 3) = 15 \times 14 \times 13 = 2730$$
🧠 同样选 3 只股票，如果只是"选 3 只不计顺序"→ C(15,3)=455，但此处有排序要求 → 排列。
【Q8 答案】B — 0.336
二项分布：n=5，p=0.55，k=3
$$\begin{aligned} P(X=3) &= C(5,3) \cdot p^3 \cdot (1-p)^{2} \ &= 10 \cdot 0.55^3 \cdot 0.45^2 \ &= 10 \cdot 0.1664 \cdot 0.2025 \ &= 10 \cdot 0.0337 = 0.3369 \end{aligned}$$
🧠 C(5,3)=10 种排列方式 × 每种方式的概率相同。
【Q9 答案】C — 0.636
使用贝叶斯公式：
$$\begin{aligned} P(\text{张} \mid \text{跑赢}) &= \frac{P(\text{跑赢} \mid \text{张}) \cdot P(\text{张})}{P(\text{跑赢})} \[4pt] P(\text{跑赢}) &= 0.7 \times 0.6 + 0.3 \times 0.8 = 0.42 + 0.24 = 0.66 \[4pt] P(\text{张} \mid \text{跑赢}) &= \frac{0.6 \times 0.7}{0.66} = \frac{0.42}{0.66} = 0.6364 \end{aligned}$$
🧠 贝叶斯三步走：① 先验概率 ② 全概率求分母 ③ 贝叶斯求后验。
【Q10 答案】C — 约 27%
贝叶斯公式直接验算：
$$\begin{aligned} P(\text{造假} \mid \text{阳性}) &= \fracP(\text{造假} \mid \text{阳性}) &= \frac{P(\text{阳性} \mid \text{造假}) \cdot P(\text{造假})}{P(\text{阳性})} \[4pt] P(\text{阳性}) &= 0.90 \times 0.02 + 0.05 \times 0.98 = 0.018 + 0.049 = 0.067 \[4pt] P(\text{造假} \mid \text{阳性}) &= \frac{0.90 \times 0.02}{0.067} = \frac{0.018}{0.067} \approx 0.2687 \end{aligned}$$
🧠 这个结果令人惊讶：即使检测"看起来"很准（真阳性 90%，假阳性仅 5%），因为造假实际比例极低（2%），被标记造假的公司真正造假的概率只有约 27%。这是基础率忽略（Base Rate Neglect）的经典案例——CFA 考试中常见的认知偏误考点。
【Q11 答案】C — 2.5%
μ = 12%，σ = 20%。52% = 12% + 2×20% = μ + 2σ。
68-95-99.7 法则：μ±2σ 覆盖约 95%，尾部每侧约 2.5%。
$$P(X > 52\%) = P(Z > 2) \approx 2.5\%$$
🧠 精确值（查 z 表）：P(Z>2) ≈ 2.28%，2.5% 是 68-95-99.7 法则的快速近似。
【Q12 答案】C — 16%
−8% = 12% − 1×20% = μ − σ。
μ±σ 覆盖约 68%，则左侧尾部为 (100%−68%)/2 = 16%。
$$P(X < -8\%) = P(Z < -1) \approx 16\%$$
🧠 对称性：P(X > 32%) 也 ≈ 16%。
【Q13 答案】C — 2%
N(0.1%, 2%²) 中：第二个参数 2%² 是方差 σ²，不是 σ。因此 σ = 2%。
| 术语 | 值 |
|---|---|
| 均值 μ | 0.1% |
| 方差 σ² | 4‱（即 2%²） |
| 标准差 σ | 2% |
⚠️ 考试陷阱：N(μ, σ²) 的第二个参数是方差而不是标准差。CFA 习惯用 N(μ, σ²) 记法，容易与日常用语混淆。
【Q14 答案】B — 0.60
$$\rho = \frac{\text{Cov}(A,B)}{\sigma_A \cdot \sigma_B} = \frac{0.018}{0.15 \times 0.20} = \frac{0.018}{0.030} = 0.60$$
🧠 相关系数的记忆：协方差除以两标准差之积 → 标准化到 [−1, 1]。
【Q15 答案】C
- A ✅ 协方差为正 → 正相关，为负 → 负相关
- B ✅ 相关系数的定义域
- C ❌ ρ=0 不意味着独立。独立 ⇒ ρ=0，但反过来不成立（反例：Y=X²，X~N(0,1)，Cov(X,Y)=0 但很显然不独立）
- D ✅ X 和自己的协方差就是方差
🧠 相关系数只衡量线性关系。两个变量可以存在完美的非线性关系（如 Y = X²）但 ρ=0。
【Q16 答案】A — e³ ≈ 20.09
对数正态分布的中位数 = e^μ：
$$\text{Median}(X) = e^{\mu} = e^{3} \approx 20.09$$
🧠 中位数只取决于 μ，与 σ 无关。这点和均值不同（均值含 σ²/2 修正项）。
【Q17 答案】B — e^(3.08) ≈ 21.76
$$E(X) = e^{\mu + \sigma^2/2} = e^{3 + 0.16/2} = e^{3 + 0.08} = e^{3.08} \approx 21.76$$
🧠 均值 > 中位数（21.76 > 20.09）体现了右偏特征。
【Q18 答案】B — 0.0045
$$\text{Cov}(X,Y) = \rho_{XY} \cdot \sigma_X \cdot \sigma_Y = 0.3 \times 0.10 \times 0.15 = 0.0045$$
注意单位一致性：10% = 0.10，15% = 0.15。
【Q19 答案】D
逐项分析：
- A ❌ 反过来：若 X ~ 对数正态，则 ln X ~ 正态
- B ❌ 对数正态用于建模非负变量
- C ❌ 顺序错了：众数 < 中位数 < 均值（右偏特征）
- D ✅ Black-Scholes 期权定价模型假定股票价格服从对数正态分布
【Q20 答案】D — 63.2%
使用贝叶斯公式：
$$\begin{aligned} P(\text{低估} \mid \text{指标}) &= \frac{P(\text{指标} \mid \text{低估}) \cdot P(\text{低估})}{P(\text{指标})} \[4pt] P(\text{指标}) &= 0.80 \times 0.30 + 0.20 \times 0.70 = 0.24 + 0.14 = 0.38 \[4pt] P(\text{低估} \mid \text{指标}) &= \frac{0.80 \times 0.30}{0.38} = \frac{0.24}{0.38} \approx 0.6316 \end{aligned}$$
🧠 从先验概率 30% 更新到后验概率 63.2%——估值指标提供了有力证据。贝叶斯思维的核心：用新证据更新信念。
五、错题分析与备考建议
高频失分点
| 序号 | 易错点 | 对应题号 | 避坑策略 |
|---|---|---|---|
| 1 | 默认事件互斥直接相加概率 | Q1 | 先判断是否互斥，不确定时使用加法法则 |
| 2 | Var(aX)=a·Var(X) | Q2 | 记住方差运算带平方：Var(aX)=a²Var(X) |
| 3 | 排列 vs 组合混淆 | Q4, Q7 | 顺序重要 → 排列；顺序不重要 → 组合 |
| 4 | N(μ,σ²) 第二参数是方差 | Q13 | CFA 考试统一记法：括号里是方差 |
| 5 | 对数正态均值忘加 σ²/2 | Q17 | E(X)=e^(μ+σ²/2)，不是 e^μ |
| 6 | ρ=0 ⇒ 独立 | Q15 | ρ=0 只说明无线性关系，可能有非线性关系 |
| 7 | 基础率忽略 | Q10 | 低基础率场景下，高准确率检测也可能高误判 |
| 8 | 68-95-99.7 法则区间搞反 | Q11 | 单侧尾部概率 = (100%−覆盖%)/2 |
备考策略
- 贝叶斯公式是高频考点，建议手算 3-5 道不同类型题目强化肌肉记忆
- 正态分布查表在考试中会提供，但 68-95-99.7 法则务必烂熟
- 排列组合题目题干通常有暗示词："排序""顺序"→排列；"选择""组合"→组合
- 对数正态重点记三点：中位数=e^μ、均值=e^(μ+σ²/2)、右偏特征
📚 下一课 L124：概率周测（10 题）——检验本模块掌握程度
Quantitative Methods — Probability Module · Comprehensive Practice
I. Lesson Positioning
L115–L122 covered all core knowledge points in the Probability module. This lesson is a comprehensive practice session designed to reinforce understanding through cross-topic exercises.
| Item | Description |
|---|---|
| Module | 2.4 Probability Theory |
| Coverage | All topics from L115–L122 |
| Next | L124 Probability Weekly Quiz (10 questions) |
| Difficulty | ★★★★☆ |
| Exam Weight | High (Probability ≈ 25–30% of Quantitative Methods) |
| Practice Time | ~30 minutes |
II. Quick Reference — Core Formulas
| # | Topic | Key Formula / Concept |
|---|---|---|
| 1 | Addition Rule | P(A∪B) = P(A) + P(B) − P(A∩B) |
| 2 | Conditional Probability | P(A|B) = P(A∩B) / P(B) |
| 3 | Multiplication Rule | P(A∩B) = P(A|B) · P(B) |
| 4 | Total Probability Theorem | P(A) = Σ P(A|Bᵢ) · P(Bᵢ) |
| 5 | Bayes' Theorem | P(B|A) = [P(A|B)·P(B)] / P(A) |
| 6 | Independence Test | P(A∩B) = P(A)·P(B) ⇔ A and B independent |
| 7 | Permutation Formula | nPr = n! / (n−r)! |
| 8 | Combination Formula | nCr = n! / [r!(n−r)!] |
| 9 | Expected Value | E(X) = Σ xᵢ·P(xᵢ) |
| 10 | Variance | Var(X) = Σ (xᵢ − E(X))²·P(xᵢ) = E(X²)−[E(X)]² |
| 11 | Covariance | Cov(X,Y) = E[(X−μₓ)(Y−μᵧ)] = E(XY)−E(X)·E(Y) |
| 12 | Correlation Coefficient | ρ = Cov(X,Y) / (σₓ·σᵧ) |
| 13 | Normal Distribution Standardization | z = (X − μ) / σ |
| 14 | Normal Distribution Probabilities | 68-95-99.7 Rule |
| 15 | Lognormal Mean | E(X) = e^(μ + σ²/2) |
| 16 | Lognormal Median | Median(X) = e^μ |
| 17 | Continuously Compounded Return | r = ln(P₁/P₀) |
Key Distribution Comparison
| Distribution | Support | Parameters | Key Feature |
|---|---|---|---|
| Normal | (−∞, +∞) | μ, σ² | Symmetric bell curve; 68-95-99.7 rule |
| Standard Normal | (−∞, +∞) | μ=0, σ=1 | z-table lookup for probabilities |
| Lognormal | (0, +∞) | μ, σ² (parameters of ln X) | Right-skewed; median < mean |
| Binomial | {0, 1, …, n} | n, p | Number of successes in n independent Bernoulli trials |
III. Comprehensive Practice Questions
Part 1: Basic Concepts (Q1–Q4)
Q1 Regarding basic probability concepts, which of the following is incorrect?
A. Mutually exclusive events imply P(A∩B) = 0
B. If A and B are independent, then P(A|B) = P(A)
C. If P(A) = 0.4 and P(B) = 0.5, then P(A∪B) must equal 0.9
D. The sum of probabilities for exhaustive events equals 1
Q2 Which of the following statements about expected value and variance is correct?
A. E(X+Y) = E(X) + E(Y) holds only when X and Y are independent
B. Var(X−Y) = Var(X) − Var(Y) holds when X and Y are independent
C. E(aX + b) = aE(X) + b always holds
D. Var(aX) = a·Var(X) always holds
Q3 For two events A and B: P(A) = 0.3, P(B) = 0.4, P(A|B) = 0.5. Find P(B|A):
A. 1/3
B. 2/3
C. 0.375
D. 0.500
Q4 From 10 stocks, select 4 to form an equal-weighted portfolio. How many different combinations are possible? (Order does not matter.)
A. 10 × 9 × 8 × 7 = 5040
B. 10! / (4! × 6!) = 210
C. 10! / 6! = 5040
D. 4¹⁰
Part 2: Probability Calculations (Q5–Q8)
Q5 An investment strategy has a 60% probability of earning ¥2,000 and a 40% probability of losing ¥1,000. The expected return is:
A. ¥600
B. ¥800
C. ¥1,000
D. ¥1,200
Q6 Continuing from Q5, the variance (in thousands² of CNY) is closest to:
A. 1.44
B. 2.16
C. 2.56
D. 3.24
Q7 A fund manager needs to select 3 stocks from 15 candidates, then sort them from smallest to largest weight to build an increasing-weight portfolio. How many different ordering outcomes are possible?
A. 15³ = 3375
B. C(15,3) = 455
C. P(15,3) = 2730
D. 3 × 15 = 45
Q8 A company's stock rises on any given trading day with probability 0.55, with days independent. Over 5 consecutive trading days, the probability of exactly 3 up days is closest to:
A. 0.275
B. 0.336
C. 0.500
D. 0.550
Part 3: Bayes' Applications (Q9–Q10)
Q9 A fund company has two managers: Zhang manages 70% of assets with a 0.6 probability of beating the benchmark; Li manages 30% of assets with a 0.8 probability of beating the benchmark. A fund is randomly selected and it has beaten the benchmark. The probability that it is managed by Zhang is:
A. 0.42
B. 0.56
C. 0.636
D. 0.700
Q10 (Advanced) A fraud detection method has a true positive rate of 90% and a false positive rate of 5%. If the true proportion of fraudulent companies in the market is approximately 2%, when a company is flagged as fraudulent by this method, the probability that it is actually fraudulent is closest to:
A. 2%
B. 18%
C. 27%
D. 90%
Part 4: Normal Distribution & z-Scores (Q11–Q13)
Q11 Fund returns follow N(12%, 20%²). The probability of returns exceeding 52% is closest to (use the 68-95-99.7 rule):
A. 16%
B. 5%
C. 2.5%
D. 0.15%
Q12 Continuing from Q11, the probability of returns falling below −8% is closest to:
A. 2.5%
B. 5%
C. 16%
D. 32%
Q13 A stock's daily return follows N(0.1%, 2%²). What is the standard deviation (volatility)?
A. 0.04%
B. 0.1%
C. 2%
D. 4%
Part 5: Covariance & Correlation (Q14–Q15)
Q14 Cov(A,B) = 0.018. Stock A volatility = 15%, Stock B volatility = 20%. The correlation coefficient is closest to:
A. 0.45
B. 0.60
C. 0.67
D. 0.75
Q15 Regarding covariance and correlation, which statement is incorrect?
A. The sign of covariance indicates the direction of the linear relationship
B. The correlation coefficient ranges from −1 to +1
C. If ρ = 0, then the two random variables must be independent
D. Cov(X,X) = Var(X)
Part 6: Lognormal Distribution (Q16–Q17)
Q16 Given ln X ~ N(3, 0.4²), the median is closest to:
A. e³ ≈ 20.09
B. e^(3.08) ≈ 21.76
C. e^(3.16) ≈ 23.57
D. e^(3.2) ≈ 24.53
Q17 Continuing from Q16, E(X) is closest to:
A. 20.09
B. 21.76
C. 23.57
D. 24.53
Part 7: Integrated Application (Q18–Q20)
Q18 A portfolio consists of two assets: Asset X weight 60%, Asset Y weight 40%. Given σₓ = 10%, σᵧ = 15%, ρₓᵧ = 0.3, the covariance Cov(X,Y) is:
A. 0.0030
B. 0.0045
C. 0.0180
D. 0.0450
Q19 Regarding the relationship between normal and lognormal distributions, which statement is correct?
A. If X ~ lognormal, then e^X ~ normal
B. The lognormal distribution is used to model variables that can be negative
C. For the lognormal distribution, mode > median > mean
D. In the Black-Scholes framework, stock prices are assumed to follow a lognormal distribution
Q20 An analyst uses Bayes' theorem to evaluate the probability that a stock is "undervalued." Given:
- Base rate of undervalued stocks in the industry: 30%
- If a stock is undervalued, probability of observing the current valuation signal: 80%
- If a stock is not undervalued, probability of observing the current valuation signal: 20%
Based on the current valuation signal, the posterior probability that the stock is undervalued is:
A. 24%
B. 36.8%
C. 54.5%
D. 63.2%
IV. Answers & Explanations
【Q1 Answer】C
- A ✅ Mutually exclusive events: A∩B = ∅, so P(A∩B) = 0 (by definition)
- B ✅ Independence definition: P(A|B) = P(A)
- C ❌ If A and B overlap, P(A∪B) = 0.4 + 0.5 − P(A∩B) < 0.9. Only equals 0.9 when P(A∩B) = 0. Not stated in the question.
- D ✅ Exhaustive events definition
🧠 Common CFA trap: assuming events are mutually exclusive by default and directly adding probabilities.
【Q2 Answer】C
- A ❌ E(X+Y) = E(X) + E(Y) always holds, regardless of independence
- B ❌ Var(X−Y) = Var(X) + Var(Y) − 2Cov(X,Y). Even when independent, it equals Var(X) + Var(Y), not Var(X) − Var(Y)
- C ✅ Linearity of expectation always holds
- D ❌ Var(aX) = a²·Var(X), not a·Var(X)
🧠 Key memory aid: variance involves squares: Var(aX) = a²Var(X); covariance is first-order.
【Q3 Answer】B — 2/3
$$\begin{aligned} P(A \cap B) &= P(A|B) \cdot P(B) = 0.5 \times 0.4 = 0.2 \ P(B|A) &= \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.3} = \frac{2}{3} \approx 0.667 \end{aligned}$$
🧠 Converting between two conditional probability directions: use P(A∩B) as the bridge.
【Q4 Answer】B — 210
Equal-weighted portfolio → order does not matter → Combination C(10,4):
$$C(10, 4) = \frac{10!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
A and C compute permutations P(10,4) = 5040; D is meaningless.
🧠 The key question: order matters → permutation / order does not matter → combination.
【Q5 Answer】B — ¥800
$$E(X) = 0.6 \times 2000 + 0.4 \times (-1000) = 1200 - 400 = 800$$
【Q6 Answer】B — 2.16
$$\begin{aligned} E(X^2) &= 0.6 \times 2000^2 + 0.4 \times (-1000)^2 = 2,400,000 + 400,000 = 2,800,000 \[8pt] \text{Var}(X) &= E(X^2) - [E(X)]^2 = 2,800,000 - 800^2 = 2,800,000 - 640,000 = 2,160,000 \end{aligned}$$
In thousands²: 2,160,000 / 1,000,000 = 2.16
🧠 Using E(X²) − [E(X)]² is faster than Σ(xᵢ−μ)²P(xᵢ).
【Q7 Answer】C — P(15,3) = 2730
Key phrase: "sort from smallest to largest weight" → order matters → permutation:
$$P(15, 3) = 15 \times 14 \times 13 = 2730$$
🧠 Same 3 stocks selected, but with a ranking requirement → permutation, not combination.
【Q8 Answer】B — 0.336
Binomial distribution: n = 5, p = 0.55, k = 3
$$\begin{aligned} P(X=3) &= C(5,3) \cdot p^3 \cdot (1-p)^{2} \ &= 10 \cdot 0.55^3 \cdot 0.45^2 \ &= 10 \cdot 0.1664 \cdot 0.2025 = 0.3369 \end{aligned}$$
🧠 C(5,3) = 10 arrangements × same probability for each.
【Q9 Answer】C — 0.636
Bayes' theorem:
$$\begin{aligned} P(\text{Zhang} \mid \text{Beat}) &= \frac{P(\text{Beat} \mid \text{Zhang}) \cdot P(\text{Zhang})}{P(\text{Beat})} \[4pt] P(\text{Beat}) &= 0.7 \times 0.6 + 0.3 \times 0.8 = 0.42 + 0.24 = 0.66 \[4pt] P(\text{Zhang} \mid \text{Beat}) &= \frac{0.6 \times 0.7}{0.66} = \frac{0.42}{0.66} = 0.6364 \end{aligned}$$
🧠 Bayes in 3 steps: ① Prior probability ② Total probability for denominator ③ Bayes for posterior.
【Q10 Answer】C — ~27%
Bayes' theorem:
$$\begin{aligned} P(\text{Fraud} \mid \text{Positive}) &= \frac{P(\text{Positive} \mid \text{Fraud}) \cdot P(\text{Fraud})}{P(\text{Positive})} \[4pt] P(\text{Positive}) &= 0.90 \times 0.02 + 0.05 \times 0.98 = 0.018 + 0.049 = 0.067 \[4pt] P(\text{Fraud} \mid \text{Positive}) &= \frac{0.90 \times 0.02}{0.067} = \frac{0.018}{0.067} \approx 0.2687 \end{aligned}$$
🧠 Surprising result: despite a seemingly accurate test (90% true positive, only 5% false positive), because the true fraud rate is extremely low (2%), the probability that a flagged company is actually fraudulent is only ~27%. This is the classic Base Rate Neglect fallacy — a common behavioral bias tested on the CFA exam.
【Q11 Answer】 C — 2.5%
μ = 12%, σ = 20%. 52% = 12% + 2×20% = μ + 2σ.
68-95-99.7 Rule: μ ± 2σ covers ~95%, leaving ~2.5% in each tail.
$$P(X > 52\%) = P(Z > 2) \approx 2.5\%$$
🧠 Exact value (z-table): P(Z > 2) ≈ 2.28%; 2.5% is the 68-95-99.7 rule approximation.
【Q12 Answer】C — 16%
−8% = 12% − 1×20% = μ − σ.
μ ± σ covers ~68%, so the left tail is (100% − 68%)/2 = 16%.
$$P(X < -8\%) = P(Z < -1) \approx 16\%$$
🧠 By symmetry: P(X > 32%) ≈ 16% as well.
【Q13 Answer】C — 2%
N(0.1%, 2%²): the second parameter 2%² is the variance σ², not σ. Therefore σ = 2%.
| Term | Value |
|---|---|
| Mean μ | 0.1% |
| Variance σ² | 2%² |
| Standard deviation σ | 2% |
⚠️ Exam trap: the second parameter in N(μ, σ²) is variance, not standard deviation. CFA uses this notation and it is a frequent source of confusion.
【Q14 Answer】B — 0.60
$$\rho = \frac{\text{Cov}(A,B)}{\sigma_A \cdot \sigma_B} = \frac{0.018}{0.15 \times 0.20} = \frac{0.018}{0.030} = 0.60$$
🧠 Correlation = covariance divided by the product of the two standard deviations → normalized to [−1, 1].
【Q15 Answer】C
- A ✅ Positive covariance → positive linear relationship; negative → negative
- B ✅ Definition of correlation coefficient
- C ❌ ρ = 0 does not imply independence. Independence ⇒ ρ = 0, but the converse is false. Counterexample: Y = X², X ~ N(0,1) — Cov(X,Y) = 0 but clearly not independent.
- D ✅ Covariance of X with itself equals variance
🧠 Correlation only measures linear relationships. Two variables can have a perfect nonlinear relationship (e.g., Y = X²) yet ρ = 0.
【Q16 Answer】A — e³ ≈ 20.09
For a lognormal distribution, Median = e^μ:
$$\text{Median}(X) = e^{\mu} = e^{3} \approx 20.09$$
🧠 The median depends only on μ, not σ. This differs from the mean, which includes a σ²/2 adjustment.
【Q17 Answer】B — e^(3.08) ≈ 21.76
$$E(X) = e^{\mu + \sigma^2/2} = e^{3 + 0.16/2} = e^{3 + 0.08} = e^{3.08} \approx 21.76$$
🧠 Mean > Median (21.76 > 20.09) reflects the right-skewed characteristic.
【Q18 Answer】B — 0.0045
$$\text{Cov}(X,Y) = \rho_{XY} \cdot \sigma_X \cdot \sigma_Y = 0.3 \times 0.10 \times 0.15 = 0.0045$$
Be consistent with units: 10% = 0.10, 15% = 0.15.
【Q19 Answer】D
- A ❌ The reverse is true: if X ~ lognormal, then ln X ~ normal
- B ❌ Lognormal is used for non-negative variables
- C ❌ The order is wrong: mode < median < mean (right-skewed characteristic)
- D ✅ Black-Scholes option pricing model assumes stock prices follow a lognormal distribution
【Q20 Answer】D — 63.2%
Applying Bayes' theorem:
$$\begin{aligned} P(\text{Undervalued} \mid \text{Signal}) &= \frac{P(\text{Signal} \mid \text{Undervalued}) \cdot P(\text{Undervalued})}{P(\text{Signal})} \[4pt] P(\text{Signal}) &= 0.80 \times 0.30 + 0.20 \times 0.70 = 0.24 + 0.14 = 0.38 \[4pt] P(\text{Undervalued} \mid \text{Signal}) &= \frac{0.80 \times 0.30}{0.38} = \frac{0.24}{0.38} \approx 0.6316 \end{aligned}$$
🧠 Updated from a prior of 30% to a posterior of 63.2% — the valuation signal provides strong evidence. The core of Bayesian thinking: update beliefs with new evidence.
V. Error Analysis & Exam Preparation Tips
Common Pitfalls
| # | Trap | Relevant Q | How to Avoid |
|---|---|---|---|
| 1 | Assuming mutual exclusivity; directly adding probabilities | Q1 | Check for mutual exclusivity first; use the addition rule when uncertain |
| 2 | Var(aX) = a·Var(X) | Q2 | Remember: variance uses squares — Var(aX) = a²Var(X) |
| 3 | Confusing permutation and combination | Q4, Q7 | Order matters → permutation; order does not matter → combination |
| 4 | N(μ,σ²) second param is variance | Q13 | CFA notation: second parameter is variance, not standard deviation |
| 5 | Forgetting σ²/2 in lognormal mean | Q17 | E(X) = e^(μ+σ²/2), not e^μ |
| 6 | ρ = 0 ⇒ independence | Q15 | ρ = 0 only means no linear relationship; nonlinear may exist |
| 7 | Base rate neglect | Q10 | With low base rates, even accurate tests produce many false positives |
| 8 | 68-95-99.7 rule tail confusion | Q11 | One-tail probability = (100% − coverage %)/2 |
Exam Strategy
- Bayes' Theorem is a high-frequency topic — practice 3–5 different problem types to build muscle memory
- Normal distribution tables will be provided on the exam, but the 68-95-99.7 rule must be second nature
- Permutation vs combination — look for cue words: "sort"/"rank"/"order" → permutation; "select"/"choose"/"combination" → combination
- Lognormal distribution — focus on three key points: Median = e^μ, Mean = e^(μ+σ²/2), right-skewed character
📚 Next: L124 Probability Weekly Quiz (10 questions) — test your mastery of this module