Comprehensive Exam Practice

Ultra-Complex Multi-Part Questions - 3 Versions (L02-L06, L09)

Versions: 3
Questions: 6 total (2 per version)
Time: ~90-120 min/version
Difficulty: HARD
Lesson Coverage (All Versions)
L02 Descriptive Stats L03 Set Theory & Axioms L04 Conditional Probability L05 Discrete RVs L06 Continuous RVs L09 Sampling & CLT

Version B: Probability Chain (Application-Heavy)

Focus on real-world applications with bank credit scoring and portfolio management contexts

Q1 The Complete Probability Framework - From Axioms to Distributions ~60 min
A bank uses a credit scoring model to assess loan applicants. Historical data shows:
- 30% of applicants are classified as High Risk (H)
- 70% of applicants are classified as Low Risk (L)
- High Risk applicants default with probability 0.25
- Low Risk applicants default with probability 0.03
Part I: Axiomatic Foundations L03
(a)
Define a probability space $(\Omega, \mathcal{F}, P)$ appropriate for this scenario. Specifically:
(i) Construct the sample space $\Omega$ consisting of all possible (Risk Classification, Default Outcome) pairs
(ii) List all elements of $\Omega$ explicitly
(iii) State the probability of each elementary outcome using the given information
(b)
State Kolmogorov's three axioms of probability. For each axiom, verify that your probability assignment in part (a) satisfies it.
(c)
Using only the three axioms (no additional theorems), prove the following foundational results:
(i) $P(\emptyset) = 0$
(ii) $P(A^c) = 1 - P(A)$ for any event A
(iii) Monotonicity: If $A \subseteq B$, then $P(A) \leq P(B)$
Hint for (i): Consider that $\Omega = \Omega \cup \emptyset \cup \emptyset \cup \cdots$
(d)
Prove the Addition Rule (inclusion-exclusion for two events): $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Your proof must start from only the three axioms and clearly identify each step's justification.
(e)
Using your result from (d), prove the subadditivity inequality: $$P(A \cup B) \leq P(A) + P(B)$$ State precisely when equality holds.
Part II: Conditional Probability and Bayes' Theorem L04
(f)
From the definition of conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$, derive:
(i) The Product Rule: $P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)$
(ii) Bayes' Theorem: $P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$
(g)
State and prove the Law of Total Probability: For a partition $\{B_1, B_2, \ldots, B_k\}$ of $\Omega$: $$P(A) = \sum_{i=1}^{k} P(A|B_i) \cdot P(B_i)$$
(h)
Let $D$ denote the event "applicant defaults" and $H$ denote "applicant is High Risk." Apply the Law of Total Probability to calculate $P(D)$, the probability that a randomly selected applicant defaults. Show your calculation explicitly.
(i)
A loan officer reviews an application and discovers the applicant has defaulted. Calculate the probability this applicant was High Risk, i.e., compute $P(H|D)$. Apply Bayes' theorem and interpret this result.
(j)
Define statistical independence for two events A and B. Determine whether H and D are independent by:
(i) Computing $P(H) \cdot P(D)$
(ii) Computing $P(H \cap D)$
(iii) Comparing and stating your conclusion
(k)
Prove: If A and B are independent, then A and $B^c$ are also independent. Your proof must start from the definition of independence and use only axioms and previously proven results.
Part III: Discrete Random Variables L05
(l)
Let $X$ = number of defaults among $n = 10$ randomly selected Low Risk applicants (assume selections are independent).
(i) Identify the distribution of $X$, stating the exact parameters
(ii) Write the probability mass function (PMF) $P(X = k)$ for $k = 0, 1, \ldots, 10$
(iii) State the three conditions that define a valid PMF and verify your expression satisfies them
(m)
Calculate $P(X \geq 1)$ using the complement rule. Interpret this probability in the context of the bank's risk.
(n)
Derive $E[X]$ for the Binomial$(n, p)$ distribution using the definition: $$E[X] = \sum_{k=0}^{n} k \cdot P(X = k)$$ Your derivation must use the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$ and show all algebraic steps to arrive at $E[X] = np$.
(o)
Derive $\text{Var}(X) = np(1-p)$ for Binomial$(n, p)$. First derive $E[X(X-1)]$ using the identity $k(k-1)\binom{n}{k} = n(n-1)\binom{n-2}{k-2}$, then use: $$\text{Var}(X) = E[X^2] - (E[X])^2 = E[X(X-1)] + E[X] - (E[X])^2$$
(p)
The bank processes approximately 1000 Low Risk applications per month. Explain why the Poisson approximation may be appropriate. State the conditions for Poisson approximation to Binomial and calculate the Poisson parameter $\lambda$.
(q)
Prove that the Poisson distribution is the limit of the Binomial distribution: If $X_n \sim \text{Binomial}(n, \lambda/n)$, then as $n \to \infty$: $$P(X_n = k) \to \frac{\lambda^k e^{-\lambda}}{k!}$$ Use the facts that $\lim_{n \to \infty} (1 - \lambda/n)^n = e^{-\lambda}$ and $\lim_{n \to \infty} \frac{n!}{(n-k)! n^k} = 1$.
Part IV: Continuous Random Variables L06
(r)
Suppose the time $T$ (in months) until a Low Risk applicant defaults (given that they eventually default) follows an Exponential distribution with rate $\lambda = 0.1$.
(i) Write the PDF $f_T(t)$ and verify it integrates to 1
(ii) Derive the CDF $F_T(t) = P(T \leq t)$
(iii) Calculate the probability that a defaulting Low Risk applicant defaults within the first 6 months
(s)
Derive $E[T] = 1/\lambda$ for the Exponential$(\lambda)$ distribution using integration by parts. Show all steps.
(t)
Derive $\text{Var}(T) = 1/\lambda^2$ for the Exponential$(\lambda)$ distribution. First compute $E[T^2]$ using integration by parts twice.
(u)
Prove the memoryless property of the Exponential distribution: $$P(T > s + t \mid T > s) = P(T > t)$$ Explain intuitively why this property is reasonable for modeling "waiting times."
(v)
The bank's risk model assumes that among defaulters, the default amount follows a Normal distribution with $\mu = \$50,000$ and $\sigma = \$15,000$.
(i) State the PDF of $X \sim N(\mu, \sigma^2)$
(ii) Prove that if $Z \sim N(0,1)$ (standard normal), then $X = \mu + \sigma Z \sim N(\mu, \sigma^2)$
(iii) What proportion of defaults exceed \$75,000? Show the standardization and use $\Phi(1.67) \approx 0.9525$
(w)
The bank defines a "severe default" as one exceeding \$80,000.
(i) Calculate $P(X > 80000)$ using standardization
(ii) Using Chebyshev's inequality, find an upper bound on $P(|X - 50000| > 30000)$
(iii) Compare your exact answer from (i) with the Chebyshev bound. What does this comparison reveal?
Solution
(a)
(i) $\Omega = \{(H,D), (H,D^c), (L,D), (L,D^c)\}$ where H=High Risk, L=Low Risk, D=Default (ii) Four outcomes: (H,D), (H,D^c), (L,D), (L,D^c) (iii) P(H,D) = 0.30 x 0.25 = 0.075, P(H,D^c) = 0.30 x 0.75 = 0.225 P(L,D) = 0.70 x 0.03 = 0.021, P(L,D^c) = 0.70 x 0.97 = 0.679 Sum = 1.000 (verified)
(b)
Axiom 1: P(A) >= 0 for all events. Check: All four probabilities are non-negative. Axiom 2: P(Omega) = 1. Check: 0.075 + 0.225 + 0.021 + 0.679 = 1. Axiom 3: For disjoint events, P(union) = sum of P. Check: Our outcomes are disjoint.
(c)
(i) P(empty) = 0: Since Omega = Omega U empty U empty U ..., by Axiom 3: P(Omega) = P(Omega) + P(empty) + P(empty) + ... This requires P(empty) = 0, else infinite sum exceeds 1. (ii) P(A^c) = 1 - P(A): Omega = A U A^c (disjoint). By Axiom 3: P(Omega) = P(A) + P(A^c). By Axiom 2: 1 = P(A) + P(A^c). Therefore P(A^c) = 1 - P(A). (iii) Monotonicity: If A subset of B, write B = A U (B cap A^c) (disjoint). By Axiom 3: P(B) = P(A) + P(B cap A^c). By Axiom 1: P(B cap A^c) >= 0. Therefore P(B) >= P(A). QED
(d)
Write A U B = A U (B cap A^c) (disjoint union). By Axiom 3: P(A U B) = P(A) + P(B cap A^c). Now B = (B cap A) U (B cap A^c) (disjoint). So P(B) = P(B cap A) + P(B cap A^c), giving P(B cap A^c) = P(B) - P(A cap B). Substituting: P(A U B) = P(A) + P(B) - P(A cap B). QED
(e)
From P(A U B) = P(A) + P(B) - P(A cap B) and P(A cap B) >= 0 (Axiom 1): P(A U B) <= P(A) + P(B). Equality holds iff P(A cap B) = 0, i.e., A and B are disjoint.
(f)
(i) Product Rule: From P(A|B) = P(A cap B)/P(B), multiply both sides by P(B): P(A cap B) = P(A|B) * P(B). Similarly, P(A cap B) = P(B|A) * P(A). (ii) Bayes: From P(A|B) = P(A cap B)/P(B) = P(B|A)*P(A)/P(B). QED
(g)
Since {B_i} partitions Omega: A = A cap Omega = A cap (U B_i) = U (A cap B_i) (disjoint). By Axiom 3: P(A) = sum P(A cap B_i) = sum P(A|B_i)*P(B_i). QED
(h)
P(D) = P(D|H)*P(H) + P(D|L)*P(L) = 0.25*0.30 + 0.03*0.70 = 0.075 + 0.021 = 0.096
(i)
P(H|D) = P(D|H)*P(H)/P(D) = (0.25 * 0.30)/0.096 = 0.075/0.096 = 0.781 (78.1%) Interpretation: A defaulter is much more likely High Risk (78%) than the base rate (30%).
(j)
(i) P(H)*P(D) = 0.30 * 0.096 = 0.0288 (ii) P(H cap D) = 0.075 (from part a) (iii) Since 0.075 != 0.0288, H and D are NOT independent. Risk classification and default are dependent.
(k)
Given: P(A cap B) = P(A)*P(B) (independence). Want: P(A cap B^c) = P(A)*P(B^c). Since A = (A cap B) U (A cap B^c) (disjoint): P(A) = P(A cap B) + P(A cap B^c) P(A cap B^c) = P(A) - P(A cap B) = P(A) - P(A)*P(B) = P(A)*(1 - P(B)) = P(A)*P(B^c). QED
(l)
(i) X ~ Binomial(n=10, p=0.03) (ii) P(X=k) = C(10,k) * (0.03)^k * (0.97)^(10-k) for k = 0,1,...,10 (iii) Valid PMF: (1) P(X=k) >= 0, (2) sum over k = 1, (3) defined for k in {0,...,10}. All satisfied.
(m)
P(X >= 1) = 1 - P(X = 0) = 1 - (0.97)^10 = 1 - 0.7374 = 0.2626 (26.3%) Interpretation: ~26% chance of at least one default among 10 Low Risk applicants.
(n)
E[X] = sum_{k=0}^n k * C(n,k) * p^k * (1-p)^(n-k) = sum_{k=1}^n k * C(n,k) * p^k * (1-p)^(n-k) (k=0 term vanishes) Using k*C(n,k) = n*C(n-1,k-1): = sum_{k=1}^n n*C(n-1,k-1) * p^k * (1-p)^(n-k) = np * sum_{j=0}^{n-1} C(n-1,j) * p^j * (1-p)^(n-1-j) (let j=k-1) = np * (p + (1-p))^(n-1) = np * 1 = np. QED
(o)
E[X(X-1)] = sum k(k-1)*C(n,k)*p^k*(1-p)^(n-k) Using k(k-1)*C(n,k) = n(n-1)*C(n-2,k-2): = n(n-1)p^2 * sum C(n-2,j)*p^j*(1-p)^(n-2-j) = n(n-1)p^2 E[X^2] = E[X(X-1)] + E[X] = n(n-1)p^2 + np Var(X) = E[X^2] - (E[X])^2 = n(n-1)p^2 + np - n^2p^2 = np(1-p). QED
(p)
Conditions: n large, p small, lambda = np moderate. Here: n = 1000, p = 0.03, lambda = 30. n is large, p is small, lambda = 30 is moderate. Poisson(30) approximates Binomial(1000, 0.03).
(q)
P(X_n = k) = C(n,k) * (lambda/n)^k * (1 - lambda/n)^(n-k) = [n(n-1)...(n-k+1)/k!] * (lambda^k/n^k) * (1 - lambda/n)^n * (1 - lambda/n)^(-k) As n -> infinity: - n(n-1)...(n-k+1)/n^k -> 1 - (1 - lambda/n)^n -> e^(-lambda) - (1 - lambda/n)^(-k) -> 1 Therefore P(X_n = k) -> lambda^k * e^(-lambda) / k!. QED
(r)
(i) f_T(t) = 0.1 * e^(-0.1t) for t > 0. Integral: int_0^inf 0.1*e^(-0.1t) dt = 1. (ii) F_T(t) = 1 - e^(-0.1t) for t >= 0. (iii) P(T <= 6) = 1 - e^(-0.6) = 1 - 0.549 = 0.451 (45.1%)
(s)
E[T] = int_0^inf t * lambda * e^(-lambda*t) dt Let u = t, dv = lambda*e^(-lambda*t)dt, then du = dt, v = -e^(-lambda*t) = [-t*e^(-lambda*t)]_0^inf + int_0^inf e^(-lambda*t) dt = 0 + 1/lambda = 1/lambda. QED
(t)
E[T^2] = int_0^inf t^2 * lambda * e^(-lambda*t) dt = 2/lambda^2 (by integration by parts twice) Var(T) = E[T^2] - (E[T])^2 = 2/lambda^2 - 1/lambda^2 = 1/lambda^2. QED
(u)
P(T > s+t | T > s) = P(T > s+t) / P(T > s) = e^(-lambda(s+t)) / e^(-lambda*s) = e^(-lambda*t) = P(T > t). Intuition: "Fresh start" property - past waiting provides no information about future waiting.
(v)
(i) f(x) = (1/(sigma*sqrt(2*pi))) * exp(-(x-mu)^2/(2*sigma^2)) (ii) If Z ~ N(0,1), then X = mu + sigma*Z has E[X] = mu, Var(X) = sigma^2, and the linear transformation of a normal is normal, so X ~ N(mu, sigma^2). (iii) P(X > 75000) = P(Z > (75000-50000)/15000) = P(Z > 1.67) = 1 - 0.9525 = 0.0475 (4.75%)
(w)
(i) P(X > 80000) = P(Z > 2) = 1 - 0.9772 = 0.0228 (2.28%) (ii) Chebyshev: P(|X - 50000| > 30000) = P(|X - mu| > 2*sigma) <= 1/4 = 0.25 (25%) (iii) Exact = 2.28%, Chebyshev bound = 25%. Chebyshev is very loose for normal distributions because normals have thin tails. The bound is distribution-free, hence conservative.
Q2 From Samples to the Central Limit Theorem ~60 min
A portfolio manager monitors daily log-returns on a stock. Over a long historical period, the returns have been found to have mean $\mu = 0.0008$ (0.08% daily) and standard deviation $\sigma = 0.018$ (1.8% daily). The manager will analyze samples of returns.
Part I: Descriptive Statistics Foundations L02
(a)
For a sample of n observations $\{x_1, x_2, \ldots, x_n\}$, prove algebraically that: $$\sum_{i=1}^{n} (x_i - \bar{x}) = 0$$ Explain why this property makes it impossible to measure spread using simple deviations and necessitates squaring.
(b)
Prove the computational formula for sample variance: $$\sum_{i=1}^{n} (x_i - \bar{x})^2 = \sum_{i=1}^{n} x_i^2 - n\bar{x}^2$$ Show every algebraic step.
(c)
The sample variance is defined as $s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2$. Explain the concept of degrees of freedom and why we divide by $(n-1)$ rather than $n$.
(d)
Prove that linear transformations affect the mean and variance in specific ways. If $y_i = ax_i + b$ for constants $a, b$, then:
(i) $\bar{y} = a\bar{x} + b$
(ii) $s_y^2 = a^2 s_x^2$
(e)
A sample of 5 daily returns (in percent) is: $\{-2.1, 0.5, 1.2, -0.8, 1.7\}$.
(i) Calculate the sample mean $\bar{x}$
(ii) Calculate the sample variance $s^2$ using the computational formula
(iii) If these returns were converted to decimal form (dividing by 100), what would the new mean and variance be?
Part II: Discrete Random Variables in Sampling L05
(f)
Consider a simplified model: each day, the stock has probability $p = 0.52$ of a positive return (independent across days). Let $Y$ = number of positive return days in a trading month of $n = 22$ days.
(i) Identify the distribution of $Y$
(ii) Calculate $E[Y]$ and $\text{Var}(Y)$
(g)
A "bad month" is defined as having fewer than 9 positive days. Express $P(Y < 9)$ as a sum. Using the Normal approximation with continuity correction, approximate this probability.
(h)
Derive the PMF, mean, and variance of the Geometric distribution: Let $W$ = number of days until the first positive return (including that day).
(i) Derive the PMF: $P(W = k) = (1-p)^{k-1} p$ for $k = 1, 2, 3, \ldots$
(ii) Prove this is a valid PMF (sums to 1) using the geometric series
(iii) Derive $E[W] = 1/p$ using the series $\sum_{k=1}^{\infty} k r^{k-1} = \frac{1}{(1-r)^2}$
(i)
Prove the memoryless property of the Geometric distribution: $$P(W > s + t \mid W > s) = P(W > t) \quad \text{for positive integers } s, t$$
Part III: Continuous Distributions in Finance L06
(j)
The manager models daily log-returns using a Normal distribution with $\mu = 0.0008$ and $\sigma = 0.018$.
(i) State the 68-95-99.7 rule for normal distributions
(ii) Calculate the range of returns that captures the middle 95% of days
(iii) On what proportion of days does the return exceed 3%?
(k)
Value at Risk (VaR): The 5% VaR is the return value such that there's only a 5% chance of a worse return.
(i) Find the z-score such that $P(Z < z) = 0.05$ [use $z_{0.05} \approx -1.645$]
(ii) Calculate the 5% daily VaR for this stock
(iii) If you hold \$1,000,000 of this stock, what is the 5% VaR in dollar terms?
(l)
Derive the PDF of a transformed random variable using the CDF method: Let $X \sim \text{Uniform}(0, 1)$ and define $Y = -\frac{1}{\lambda} \ln(X)$.
(i) Derive the CDF $F_Y(y) = P(Y \leq y)$ for $y > 0$
(ii) Differentiate to find the PDF $f_Y(y)$
(iii) Identify this distribution (this shows how to generate Exponential RVs from Uniform)
(m)
The time between large price movements (>$2\sigma$) is modeled as Exponential with mean 5 days.
(i) Calculate the probability of no large movement in the next 10 days
(ii) Given that 7 days have passed without a large movement, what is the probability of no large movement in the next 10 days? Use the memoryless property.
(iii) Why is the memoryless property controversial when applied to financial markets?
Part IV: The Central Limit Theorem L09
(n)
The manager collects a sample of $n = 64$ daily returns. Prove that the sample mean $\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$ is an unbiased estimator of $\mu$: $$E[\bar{X}] = \mu$$ Use the linearity of expectation and show all steps.
(o)
Prove that for a random sample from a population with variance $\sigma^2$: $$\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$$ Use the properties $\text{Var}(aX) = a^2 \text{Var}(X)$ and for independent RVs: $\text{Var}(\sum X_i) = \sum \text{Var}(X_i)$.
(p)
Calculate the standard error of $\bar{X}$ when $n = 64$, $\sigma = 0.018$. Explain in words what the standard error measures and why larger samples produce smaller standard errors.
(q)
Prove that $S^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})^2$ is an unbiased estimator of $\sigma^2$: $E[S^2] = \sigma^2$
Start by showing: $\sum (X_i - \bar{X})^2 = \sum (X_i - \mu)^2 - n(\bar{X} - \mu)^2$
(r)
State the Central Limit Theorem precisely, including:
(i) The setup (what assumptions on $X_1, \ldots, X_n$)
(ii) The convergence statement (what converges to what)
(iii) The type of convergence (in distribution)
(s)
Explain why each condition in the CLT is necessary:
(i) Why must the $X_i$ be identically distributed?
(ii) Why must the variance be finite? Give the Cauchy distribution as a counterexample.
(iii) What role does independence play?
(t)
Apply the CLT: For the $n = 64$ daily returns with $\mu = 0.0008$ and $\sigma = 0.018$:
(i) State the approximate distribution of $\bar{X}$
(ii) Calculate $P(\bar{X} > 0.003)$ - the probability the sample mean exceeds 0.3%
(iii) Calculate $P(\bar{X} < 0)$ - the probability the sample mean is negative
(u)
The daily returns are not actually Normal - they have heavier tails (leptokurtic). Explain why the CLT still allows us to make valid inferences about the sample mean, even when individual returns are non-normal.
(v)
Critical Thinking: A colleague claims, "By CLT, we can assume stock returns are normally distributed." Identify the error in this reasoning and explain what the CLT does and does not tell us about:
(i) Individual returns
(ii) Sample means
(iii) The relationship between sample size and approximation quality
Solution
(a)
sum(x_i - x_bar) = sum(x_i) - sum(x_bar) = sum(x_i) - n*x_bar = sum(x_i) - n*(sum(x_i)/n) = 0. Since positive and negative deviations always cancel, using raw deviations would always give 0. Squaring ensures all terms are positive and captures spread.
(b)
sum(x_i - x_bar)^2 = sum(x_i^2 - 2*x_i*x_bar + x_bar^2) = sum(x_i^2) - 2*x_bar*sum(x_i) + n*x_bar^2 = sum(x_i^2) - 2*x_bar*n*x_bar + n*x_bar^2 = sum(x_i^2) - 2*n*x_bar^2 + n*x_bar^2 = sum(x_i^2) - n*x_bar^2. QED
(c)
Degrees of freedom = number of independent pieces of information. After computing x_bar, the deviations sum to 0, so only n-1 are free. Dividing by n-1 makes s^2 an unbiased estimator of sigma^2.
(d)
(i) y_bar = (1/n)*sum(a*x_i + b) = a*(1/n)*sum(x_i) + b = a*x_bar + b (ii) s_y^2 = (1/(n-1))*sum(y_i - y_bar)^2 = (1/(n-1))*sum(a*x_i + b - a*x_bar - b)^2 = (1/(n-1))*sum(a*(x_i - x_bar))^2 = a^2 * s_x^2
(e)
(i) x_bar = (-2.1 + 0.5 + 1.2 - 0.8 + 1.7)/5 = 0.5/5 = 0.1% (ii) sum(x_i^2) = 4.41 + 0.25 + 1.44 + 0.64 + 2.89 = 9.63 s^2 = (9.63 - 5*0.01)/4 = (9.63 - 0.05)/4 = 2.395 (iii) Decimal: a = 0.01. Mean = 0.001, Variance = (0.01)^2 * 2.395 = 0.0002395
(f)
(i) Y ~ Binomial(22, 0.52) (ii) E[Y] = 22 * 0.52 = 11.44, Var(Y) = 22 * 0.52 * 0.48 = 5.49
(g)
mu = 11.44, sigma = sqrt(5.49) = 2.34 P(Y < 9) approx P(Y <= 8.5) with continuity correction Z = (8.5 - 11.44)/2.34 = -1.26 P(Z < -1.26) = 0.104 (about 10.4%)
(h)
(i) First success on trial k means k-1 failures then 1 success: P(W=k) = (1-p)^(k-1) * p (ii) sum_{k=1}^inf p*(1-p)^(k-1) = p * sum_{j=0}^inf (1-p)^j = p * 1/(1-(1-p)) = p/p = 1 (iii) E[W] = sum k*p*(1-p)^(k-1) = p * sum k*(1-p)^(k-1) = p * 1/(1-(1-p))^2 = p/p^2 = 1/p
(i)
P(W > k) = (1-p)^k (probability of k failures) P(W > s+t | W > s) = P(W > s+t) / P(W > s) = (1-p)^(s+t) / (1-p)^s = (1-p)^t = P(W > t). QED
(j)
(i) 68-95-99.7: 68% within 1 SD, 95% within 2 SD, 99.7% within 3 SD of mean. (ii) mu +/- 2*sigma = 0.0008 +/- 0.036 = [-0.0352, 0.0368] or (-3.52%, +3.68%) (iii) P(R > 0.03) = P(Z > (0.03-0.0008)/0.018) = P(Z > 1.62) = 1 - 0.9474 = 0.0526 (5.3%)
(k)
(i) z = -1.645 for 5th percentile (ii) VaR = mu + z*sigma = 0.0008 + (-1.645)*0.018 = 0.0008 - 0.0296 = -0.0288 = -2.88% (iii) Dollar VaR = 0.0288 * $1,000,000 = $28,800
(l)
(i) F_Y(y) = P(Y <= y) = P(-ln(X)/lambda <= y) = P(ln(X) >= -lambda*y) = P(X >= e^(-lambda*y)) = 1 - e^(-lambda*y) for y > 0 (ii) f_Y(y) = d/dy[1 - e^(-lambda*y)] = lambda * e^(-lambda*y) (iii) This is Exponential(lambda). Shows how to generate exponential RVs from uniform.
(m)
(i) lambda = 1/5 = 0.2. P(T > 10) = e^(-0.2*10) = e^(-2) = 0.135 (13.5%) (ii) By memoryless: P(T > 17 | T > 7) = P(T > 10) = 0.135 (same answer!) (iii) Markets exhibit clustering - large moves tend to follow large moves. The memoryless property ignores this dependence structure.
(n)
E[X_bar] = E[(1/n) * sum X_i] = (1/n) * sum E[X_i] = (1/n) * n * mu = mu. QED (Unbiased)
(o)
Var(X_bar) = Var((1/n) * sum X_i) = (1/n^2) * Var(sum X_i) = (1/n^2) * sum Var(X_i) = (1/n^2) * n * sigma^2 = sigma^2/n. QED
(p)
SE = sigma/sqrt(n) = 0.018/8 = 0.00225 (0.225%) SE measures the typical deviation of sample means from the true mean. Larger n means more information, hence more precise estimates.
(q)
sum(X_i - X_bar)^2 = sum(X_i - mu)^2 - n(X_bar - mu)^2 (expand and simplify) E[sum(X_i - mu)^2] = n*sigma^2, E[(X_bar - mu)^2] = Var(X_bar) = sigma^2/n E[sum(X_i - X_bar)^2] = n*sigma^2 - sigma^2 = (n-1)*sigma^2 Therefore E[S^2] = E[sum/(n-1)] = (n-1)*sigma^2/(n-1) = sigma^2. QED
(r)
(i) X_1, ..., X_n are iid with mean mu and variance sigma^2 < infinity (ii) (X_bar - mu)/(sigma/sqrt(n)) converges to N(0,1) (iii) Convergence in distribution (CDFs converge pointwise)
(s)
(i) Identical: Different means would shift the sum's distribution unpredictably (ii) Finite variance: Cauchy has no mean or variance. Sample mean does not stabilize; no convergence. (iii) Independence ensures the variance formula Var(sum) = sum(Var) holds.
(t)
(i) X_bar ~ N(0.0008, 0.00225^2) approximately (ii) Z = (0.003 - 0.0008)/0.00225 = 0.98. P(X_bar > 0.003) = 1 - 0.8365 = 0.164 (iii) Z = (0 - 0.0008)/0.00225 = -0.356. P(X_bar < 0) = 0.361
(u)
CLT applies to sums/means regardless of the underlying distribution (given finite variance). The sample mean becomes approximately normal even if individual returns are leptokurtic. n=64 is typically sufficient.
(v)
(i) Individual returns: CLT says NOTHING about individual returns. They remain non-normal. (ii) Sample means: CLT says sample means are approximately normal for large n. (iii) Larger n improves the approximation. The colleague confuses individual observations with sample statistics.

Version D: Finance Integration (Applied + Theory Balance)

Balance of rigorous proofs with realistic finance applications: portfolio risk, VaR, and inference

Q1 Portfolio Risk from Probability Foundations ~60 min
A quantitative analyst is developing a risk model for a trading desk. The model must incorporate market regime changes, individual trade outcomes, and extreme loss scenarios.
Part I: Market Regimes and Probability Axioms L03
The analyst identifies three key market conditions for tomorrow:
- $A$ = "Equity markets rise" with $P(A) = 0.55$
- $B$ = "Bond yields increase" with $P(B) = 0.35$
- $C$ = "Volatility spikes (VIX > 25)" with $P(C) = 0.20$

Historical analysis shows:
- $P(A \cap B) = 0.15$, $P(A \cap C) = 0.08$, $P(B \cap C) = 0.12$, $P(A \cap B \cap C) = 0.03$
(a)
State Kolmogorov's three axioms of probability. Verify that the given probabilities are consistent with these axioms.
(b)
Derive the Inclusion-Exclusion Principle for three events: $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$
(c)
Calculate $P(A \cup B \cup C)$ - the probability that at least one of the three conditions occurs tomorrow.
(d)
Calculate $P(A^c \cap B^c \cap C^c)$ - the probability of a "quiet" day where none of the three conditions occur.
(e)
Calculate $P(A \cap B^c \cap C^c)$ - equities rise but yields don't increase and volatility doesn't spike (a "goldilocks" scenario).
Part II: Conditional Probability for Trading Decisions L04
(f)
Calculate $P(C | A)$ - the probability of a volatility spike given that equities rise. Is a volatility spike more or less likely when equities rise?
(g)
Calculate $P(A | C)$ - the probability equities rise given a volatility spike. Does this support the claim "When vol spikes, equities usually fall"?
(h)
Derive Bayes' Theorem from the definition of conditional probability, then verify your answers to (f) and (g) satisfy $P(A|C) \cdot P(C) = P(C|A) \cdot P(A)$.
(i)
Test whether $A$ and $C$ are independent. State your conclusion and interpret what dependence means for trading.
(j)
Prove that if A and C are independent, then $A^c$ and C must also be independent.
(k)
Let $D$ = "at least two of A, B, C occur." Calculate $P(D)$ and $P(A \cap B \cap C | D)$.
Part III: Trade Outcomes as Discrete Random Variables L05
The trading desk executes $n = 50$ independent trades per day. Each trade has probability $p = 0.58$ of being profitable.
(l)
Let $X$ = number of profitable trades. Identify the distribution, write the PMF, and calculate $E[X]$ and $\text{Var}(X)$.
(m)
Derive the mean of the Binomial distribution from the PMF using the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$.
(n)
A "bad day" is defined as fewer than 24 profitable trades. Use the Normal approximation with continuity correction to find $P(X < 24)$.
(o)
Let $W$ = number of trades until first loss. Identify the distribution, calculate $E[W]$, and $P(W > 5)$.
(p)
Prove the memoryless property of the Geometric distribution. Is this realistic for trading?
Part IV: Continuous Models for Loss Severity L06
When a trade loses money, the loss amount $L$ (in thousands of dollars) follows a distribution the analyst must characterize.
(q)
Model $L \sim \text{Exponential}(\lambda = 0.1)$ (mean loss = \$10,000). Write the PDF/CDF, calculate $P(L > 15)$, and find the median loss.
(r)
Derive $E[L]$ and $\text{Var}(L)$ for Exponential$(\lambda)$ using integration by parts.
(s)
Given that a trade has lost at least \$10,000, what is $P(L > 25 | L > 10)$? Calculate using both the definition and the memoryless property.
(t)
Compare a Normal model $N(10, 25)$ for losses: calculate $P(L > 20)$ and $P(L > 25)$. Which model predicts more extreme losses?
(u)
State Chebyshev's Inequality. For Normal$(10, 25)$: use Chebyshev to bound $P(|L - 10| > 10)$, then calculate the exact probability. Compare.
(v)
Prove Chebyshev's Inequality starting from Markov's inequality applied to $(X-\mu)^2$.
Solution
(a)-(e) Market Regimes
(c) P(A U B U C) = 0.55 + 0.35 + 0.20 - 0.15 - 0.08 - 0.12 + 0.03 = 0.78 (d) P(A^c cap B^c cap C^c) = 1 - 0.78 = 0.22 (22% chance of quiet day) (e) P(A cap B^c cap C^c) = P(A) - P(A cap B) - P(A cap C) + P(A cap B cap C) = 0.55 - 0.15 - 0.08 + 0.03 = 0.35
(f)-(k) Conditional Probability
(f) P(C|A) = P(A cap C)/P(A) = 0.08/0.55 = 0.145 (14.5%). Unconditional P(C) = 20%, so vol spike is LESS likely when equities rise. (g) P(A|C) = P(A cap C)/P(C) = 0.08/0.20 = 0.40 (40%). Since unconditional P(A) = 55%, equities are less likely to rise when vol spikes. Supports the claim. (h) P(A|C)*P(C) = 0.40 * 0.20 = 0.08 = P(C|A)*P(A) = 0.145 * 0.55 = 0.08. Verified. (i) P(A)*P(C) = 0.55 * 0.20 = 0.11. But P(A cap C) = 0.08. Not equal, so NOT independent. (k) D = (A cap B) U (A cap C) U (B cap C). Using inclusion-exclusion: P(D) = 0.15 + 0.08 + 0.12 - 2*0.03 = 0.29 P(A cap B cap C | D) = 0.03/0.29 = 0.103
(l)-(p) Discrete RVs
(l) X ~ Binomial(50, 0.58). E[X] = 29, Var(X) = 50*0.58*0.42 = 12.18 (n) mu = 29, sigma = 3.49. P(X < 24) approx P(X <= 23.5) = P(Z < (23.5-29)/3.49) = P(Z < -1.58) = 0.057 (o) W ~ Geometric(0.42) for first loss. E[W] = 1/0.42 = 2.38. P(W > 5) = (0.58)^5 = 0.066
(q)-(v) Continuous Models
(q) f(l) = 0.1*e^(-0.1l), F(l) = 1 - e^(-0.1l). P(L > 15) = e^(-1.5) = 0.223. Median: 0.5 = 1 - e^(-0.1m), m = 6.93 (r) E[L] = 1/0.1 = 10, Var(L) = 1/0.01 = 100 (s) By memoryless: P(L > 25 | L > 10) = P(L > 15) = e^(-1.5) = 0.223 (t) Normal(10,25): P(L > 20) = P(Z > 2) = 0.023, P(L > 25) = P(Z > 3) = 0.0013 Exponential: P(L > 20) = 0.135, P(L > 25) = 0.082. Exponential has heavier tail. (u) Chebyshev: P(|L-10| > 10) <= Var(L)/100 = 25/100 = 0.25. Exact: P(|L-10| > 10) = P(L < 0 or L > 20) = 0 + 0.023 = 0.023. Much tighter!
Q2 Statistical Inference for Risk Management ~60 min
The risk management team monitors daily portfolio returns over a 3-month period ($n = 63$ trading days). Historical data suggests daily return has mean $\mu = 0.0005$ (0.05%) and standard deviation $\sigma = 0.012$ (1.2%).
Part I: Descriptive Statistics of Returns L02
(a)-(d)
Define sample statistics, prove computational formulas, explain degrees of freedom, and prove linear transformation effects.
Part II: Bernoulli Model for Direction Prediction L05
(e)-(i)
Model positive/negative returns as Bernoulli, derive Binomial variance, test for directional edge using Normal approximation.
Part III: Continuous Return Distribution L06
(j)-(o)
Normal model for returns, standardization proofs, VaR calculations at 95% and 99%, fat tails discussion, derive Normal median = mean.
Part IV: Central Limit Theorem for Inference L09
(p)-(w)
Prove unbiasedness of sample mean/variance, state CLT precisely, explain all conditions including Cauchy counterexample, apply CLT for probability calculations, discuss sample size requirements.
Solution
Key Results
Standard Error: SE = 0.012/sqrt(63) = 0.00151 VaR 95%: mu + z*sigma = 0.0005 - 1.645*0.012 = -0.0192 = -1.92% VaR 99%: mu + z*sigma = 0.0005 - 2.326*0.012 = -0.0274 = -2.74% CLT Application: P(X_bar < 0) = P(Z < -0.33) = 0.37 Sample size to halve SE: Need n = 252 days (one year)

Version E: Maximum Integration (Most Comprehensive)

Complete theoretical construction from axioms through CLT with maximum rigor

Q1 Building Probability Theory from Axioms to Distributions ~60 min
This question constructs the entire probability framework systematically, from foundational axioms through random variable theory.
Part I: The Axiomatic Foundation L03
(a)
Define a probability space $(\Omega, \mathcal{F}, P)$ by specifying what each component represents.
(b)
State Kolmogorov's three axioms of probability precisely.
(c)
From the axioms alone, prove:
Theorem 1: $P(\emptyset) = 0$
Theorem 2: $P(A^c) = 1 - P(A)$
Theorem 3: If $A \subseteq B$, then $P(A) \leq P(B)$ (Monotonicity)
Theorem 4: $0 \leq P(A) \leq 1$ for any event $A$
(d)
Prove the Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
(e)
Prove Boole's Inequality (union bound) by induction: $P(\bigcup_{i=1}^{n} A_i) \leq \sum_{i=1}^{n} P(A_i)$
(f)
Derive the Inclusion-Exclusion Formula for three events.
Part II: Conditional Probability and Independence L04
(g)
Define conditional probability $P(A|B)$ and state the required condition on $B$.
(h)
Derive the Product Rule, Bayes' Theorem, and Chain Rule for three events.
(i)
State and prove the Law of Total Probability.
(j)
Define independence for two events. State three equivalent characterizations.
(k)
Prove: If $A$ and $B$ are independent, then so are: (i) $A$ and $B^c$, (ii) $A^c$ and $B$, (iii) $A^c$ and $B^c$.
(l)
Define mutual independence for three events. Explain why pairwise independence does not imply mutual independence.
Part III: Discrete Random Variables L05
(m)
Define a discrete random variable, its PMF, and CDF. State three properties of a valid PMF.
(n)
For $X \sim \text{Bernoulli}(p)$: Write the PMF, derive $E[X] = p$ and $\text{Var}(X) = p(1-p)$.
(o)
For $X \sim \text{Binomial}(n, p)$: Express as sum of Bernoullis, use linearity to show $E[X] = np$, use independence to show $\text{Var}(X) = np(1-p)$.
(p)
Alternative derivation: Derive $E[X] = np$ directly from the PMF using $k\binom{n}{k} = n\binom{n-1}{k-1}$.
(q)
For $X \sim \text{Poisson}(\lambda)$: Write PMF, verify it sums to 1, derive $E[X] = \lambda$ and $\text{Var}(X) = \lambda$.
(r)
Prove the Poisson Limit Theorem with full limit analysis.
(s)
For $X \sim \text{Geometric}(p)$: Derive PMF, prove validity, prove memoryless property.
Part IV: Continuous Random Variables L06
(t)
Define a continuous random variable, its PDF, and the relationship $P(a < X < b) = \int_a^b f_X(x) dx$. State two properties of a valid PDF.
(u)
For $X \sim \text{Uniform}(a, b)$: Write PDF, verify normalization, derive $E[X]$ and $\text{Var}(X)$.
(v)
For $X \sim \text{Exponential}(\lambda)$: Write PDF/CDF, derive $E[X]$ and $\text{Var}(X)$ via integration by parts, prove memoryless property.
(w)
For $Z \sim N(0,1)$: Write PDF, explain why normalization cannot be proven elementarily, state 68-95-99.7, prove $X = \mu + \sigma Z \sim N(\mu, \sigma^2)$.
(x)
CDF Transformation: If $X \sim \text{Uniform}(0,1)$ and $Y = -\frac{1}{\lambda}\ln(X)$, derive the distribution of $Y$.
Solution
Complete Axiomatic Development
All theorems proven from Kolmogorov's three axioms: 1. Non-negativity: P(A) >= 0 2. Normalization: P(Omega) = 1 3. Countable additivity: P(U A_i) = sum P(A_i) for disjoint A_i Key results: P(empty) = 0, P(A^c) = 1 - P(A), Monotonicity, Addition Rule, Boole's Inequality. All derived using set decompositions and countable additivity.
Mutual Independence Counterexample
Consider two fair coin flips. Let: A = "First flip is H", B = "Second flip is H", C = "Both same or both different" P(A) = P(B) = P(C) = 0.5 P(A cap B) = P(A cap C) = P(B cap C) = 0.25 = P(A)*P(B) etc. (Pairwise independent) But P(A cap B cap C) = P(HH) = 0.25 != 0.125 = P(A)*P(B)*P(C). Not mutually independent.
Q2 From Random Variables to Statistical Inference ~60 min
This question connects descriptive statistics to the Central Limit Theorem, building the complete inference chain.
Part I: Descriptive Statistics as Estimators L02
(a)-(f)
Prove sum of deviations = 0, computational formula, parallel axis theorem, linear transformation effects, Chebyshev inequality with proof and tightness analysis.
Part II: Properties of Expectation and Variance L05, L06
(g)-(k)
Prove linearity of expectation (even for dependent RVs), variance shift formula, Var(aX+b) = a^2*Var(X), Var(X+Y) = Var(X) + Var(Y) for independent RVs, generalization to n variables.
Part III: Sampling Distributions L09
(l)-(q)
Define iid, sample mean as RV, prove E[X_bar] = mu, prove Var(X_bar) = sigma^2/n, define standard error, prove E[S^2] = sigma^2, explain degrees of freedom.
Part IV: The Central Limit Theorem L09
(r)-(y)
State CLT precisely with convergence in distribution, explain each condition (identical, independent, finite variance with Cauchy counterexample), apply CLT to normal and non-normal populations, critical analysis of common misconceptions, practical considerations for sample size.
Solution
Central Limit Theorem
CLT Statement: Let X_1, ..., X_n be iid with E[X_i] = mu and Var(X_i) = sigma^2 < infinity. Then: (X_bar - mu)/(sigma/sqrt(n)) -> N(0,1) in distribution as n -> infinity. Cauchy Counterexample: Cauchy has PDF f(x) = 1/(pi*(1+x^2)). - E[X] does not exist (integral diverges) - Var(X) does not exist - Sample mean of Cauchy RVs is also Cauchy (no concentration!) - CLT fails completely: no convergence to normal. Why CLT is "most important": Enables inference about population mean using sample mean, regardless of underlying distribution. Explains ubiquity of normal distribution in nature (sums of many small effects).