Statistical Computing · Quiz 1 · Jan 2026

**1. Identify the purpose of the question:** This 0-mark introductory item displays the reference formula sheet (Useful Data) for the Statistical Computing exam and asks the candidate to acknowledge that the useful data has been provided above. **Answer: A — Useful Data has been mentioned above.** A: **Correct:** Standard acknowledgment option for the exam's Useful Data formula sheet. B: **Incorrect:** Standard exam instructions require selecting Option A to confirm viewing the useful data.

**1. Review the Inverse Transform Method:** To generate a random variate \( X \) with cumulative distribution function \( F \), we draw \( U \sim \text{Uniform}(0, 1) \) and set \( X = F^{-1}(U) \), where \( F^{-1}(u) = \inf\{x : F(x) \ge u\} \) is the generalized inverse CDF. **2. Evaluate each statement:** - **Option A:** True, it requires evaluating the (generalized) inverse CDF \( F^{-1} \). - **Option B:** False, the inverse transform method applies to **both** continuous and discrete distributions (using the step-function generalized inverse for discrete PMFs). - **Option C:** True, it transforms a \( \text{Uniform}(0, 1) \) random variable \( U \) into a sample from the target distribution. - **Option D:** False since A and C are true. **Answer: A and C.** A: **Correct:** The inverse transform method computes \( X = F^{-1}(U) \) using the inverse CDF. B: **Incorrect:** The method works for discrete distributions as well by finding \( x_k \) such that \( F(x_{k-1}) < U \le F(x_k) \). C: **Correct:** It maps \( U \sim \text{Uniform}(0, 1) \) onto the target distribution via \( F^{-1}(U) \). D: **Incorrect:** Both A and C are true.

**1. Recall Acceptance Probability in the Accept–Reject (AR) Method:** When sampling from a target PMF \( p \) using a proposal PMF \( q_i \) with bound \( c_i = \sup_x \frac{p(x)}{q_i(x)} \), the marginal probability of accepting a proposed sample on any single iteration is: \[ P(\text{Accept}) = \sum_x q_i(x) \frac{p(x)}{c_i q_i(x)} = \frac{1}{c_i} \] And the expected number of iterations per accepted sample (measuring sampling efficiency) is \( c_i \). **2. Compare the two proposals when \( c_1 = c_2 \):** Since \( c_1 = c_2 \), both proposals have the exact same acceptance probability \( 1/c_1 = 1/c_2 \) (**Option D is true**) and require the same expected number of proposals per accepted draw, so they have the same sampling efficiency (**Option A is true**, while Options B and C are false). **Answer: A and D.** A: **Correct:** Because the expected number of trials per accepted sample is \( c_i \), \( c_1 = c_2 \) implies identical AR sampling efficiency. B: **Incorrect:** The AR bound \( c \) directly determines the acceptance rate \( 1/c \) and efficiency. C: **Incorrect:** Neither proposal has a smaller bound \( c_i \), so \( q_1 \) is not faster than \( q_2 \). D: **Correct:** The acceptance probability is \( 1/c_i \), so \( c_1 = c_2 \) guarantees equal acceptance probabilities.

**1. Trace Generator (A): \( X_{n+1} = 5 X_n \bmod 8 \) with seed \( X_0 = 3 \):** - \( X_0 = 3 \) - \( X_1 = (5 \times 3) \bmod 8 = 15 \bmod 8 = 7 \) - \( X_2 = (5 \times 7) \bmod 8 = 35 \bmod 8 = 3 = X_0 \) The sequence is \( 3, 7, 3, 7, \dots \), which starts repeating at \( n = 2 \). **2. Trace Generator (B): \( X_{n+1} = 3 X_n \bmod 8 \) with seed \( X_0 = 3 \):** - \( X_0 = 3 \) - \( X_1 = (3 \times 3) \bmod 8 = 9 \bmod 8 = 1 \) - \( X_2 = (3 \times 1) \bmod 8 = 3 \bmod 8 = 3 = X_0 \) The sequence is \( 3, 1, 3, 1, \dots \), which also starts repeating at \( n = 2 \). **3. Conclude:** Both sequences have period 2 and start repeating at the exact same step (\( n = 2 \)). **Answer: C — Both sequences start repeating at the same step.** A: **Incorrect:** Both generators repeat at \( X_2 = 3 \). B: **Incorrect:** Generator (B) also repeats at \( X_2 = 3 \), not earlier. C: **Correct:** Generator (A) produces \( 3, 7, 3, \dots \) and Generator (B) produces \( 3, 1, 3, \dots \), both repeating at step 2. D: **Incorrect:** Every finite-modulus congruential generator is periodic; here both have period 2.

**1. Recall the bounding box formulas for the Ratio of Uniforms (RoU) method:** For \( f(x) = e^{-x} \mathbb{I}_{(0, \infty)}(x) \), the RoU region is: \[ A_f = \left\{(u, v) : 0 \le u \le \sqrt{f(v/u)}\right\} = \left\{(u, v) : 0 \le u \le e^{-v/(2u)},\ v > 0\right\} \] The bounding box \( [0, a] \times [b, c] \) is given by: - \( a = \sup_{x > 0} \sqrt{f(x)} = \sup_{x > 0} e^{-x/2} = 1 \) - \( b = \inf_{x > 0} x \sqrt{f(x)} = \inf_{x > 0} x e^{-x/2} = 0 \) - \( c = \sup_{x > 0} x \sqrt{f(x)} = \sup_{x > 0} x e^{-x/2} \) **2. Maximize \( g(x) = x e^{-x/2} \) for \( x > 0 \):** Setting the derivative to zero: \[ g'(x) = e^{-x/2}\left(1 - \frac{x}{2}\right) = 0 \implies x = 2 \] Substituting \( x = 2 \) gives \( c = g(2) = 2 e^{-1} \). Thus, \( a = 1,\ b = 0,\ c = 2e^{-1} \). **Answer: B — \( a = 1,\ b = 0,\ c = 2e^{-1} \).** A: **Incorrect:** Gives \( c = e^{-1} \) instead of \( 2e^{-1} \). B: **Correct:** \( a = \sup e^{-x/2} = 1 \), \( b = 0 \), and \( c = \sup_{x>0} x e^{-x/2} = 2e^{-1} \) at \( x = 2 \). C: **Incorrect:** Since \( x > 0 \), \( v = u x \ge 0 \), so the lower bound \( b \) is \( 0 \), not \( -2e^{-1} \). D: **Incorrect:** Both \( a \) and \( c \) are miscalculated.

**1. Compute the Cumulative Distribution Function (CDF) \( F(x) \):** Given \( P(X = 0) = 0.2 \), \( P(X = 1) = 0.2 \), and \( P(X = 2) = 0.6 \): - \( F(0) = P(X \le 0) = 0.2 \) - \( F(1) = P(X \le 1) = 0.2 + 0.2 = 0.4 \) - \( F(2) = P(X \le 2) = 0.4 + 0.6 = 1.0 \) **2. Apply the discrete inverse transform rule:** For \( U \sim \text{Uniform}(0, 1) \): \[ X = \begin{cases} 0, & \text{if } U \le 0.2 \\ 1, & \text{if } 0.2 < U \le 0.4 \\ 2, & \text{if } U > 0.4 \end{cases} \] The probabilities of these intervals are \( 0.2 \), \( 0.4 - 0.2 = 0.2 \), and \( 1 - 0.4 = 0.6 \), matching the PMF. **Answer: B.** A: **Incorrect:** Uses the raw PMF value `0.2` instead of the cumulative sum `0.4` for the second threshold. B: **Correct:** Partitions \( (0, 1) \) at cumulative probabilities \( F(0) = 0.2 \) and \( F(1) = 0.4 \). C: **Incorrect:** Gives \( P(X=1) = 0.6 - 0.2 = 0.4 \) instead of \( 0.2 \). D: **Incorrect:** Gives \( P(X=0) = 0.4 \) instead of \( 0.2 \).

**1. Read the inverse transform function from the Python code:** The code defines `inv_cdf(u, a, c)` as `c * (u ** (1 / a))`, meaning: \[ X = F^{-1}(U) = c U^{1/a} \quad \text{for } U \in [0, 1] \] **2. Invert \( x = F^{-1}(u) \) to find the CDF \( u = F(x) \):** For \( 0 \le x \le c \): \[ x = c u^{1/a} \implies \frac{x}{c} = u^{1/a} \implies u = F(x) = \frac{x^a}{c^a} \] With \( F(x) = 0 \) for \( x < 0 \) and \( F(x) = 1 \) for \( x > c \), this matches Option A. **Answer: A — \( F(x) = \frac{x^a}{c^a} \) for \( 0 \le x \le c \).** A: **Correct:** Solving \( x = c u^{1/a} \) for \( u \) gives \( F(x) = x^a / c^a \) on \( 0 \le x \le c \). B: **Incorrect:** \( x^{1/a}/c \) fails to invert the power \( 1/a \). C: **Incorrect:** \( (x/c)^{1/a} \) is the forward formula \( F^{-1} \) rather than its inverse. D: **Incorrect:** \( c x^a \) multiplies by \( c \) instead of dividing by \( c^a \).

**1. Analyze the PDF and plots for \( a = 5, c = 10 \):** Differentiating \( F(x) = (x / 10)^5 \) on \( [0, 10] \) gives the PDF: \[ f(x) = F'(x) = \frac{5 x^4}{10^5}, \quad 0 \le x \le 10 \] **2. Check each statement:** - **Option A:** Since \( f(x) \propto x^4 \) for \( x \in [0, 10] \), the density of \( X \) strictly increases as \( x \) increases (also visible from the histogram bars growing taller toward \( x = 10 \)). True. - **Option B:** On the inverse CDF plot \( x = 10 u^{1/5} \), when \( u \) on the horizontal axis is near \( 0 \), \( x \) is also near \( 0 \) (small values of \( X \)). True. - **Option C & D:** Because the bulk of the probability mass is concentrated near the right boundary \( x = 10 \) with a long tail stretching toward \( 0 \) on the left, the distribution is **left-skewed** (negatively skewed). Thus **Option C is true** and **Option D is false**. **Answer: A, B, and C.** A: **Correct:** Both the histogram and the derivative \( f(x) = 5x^4 / 10^5 \) show that the density increases with \( x \). B: **Correct:** Near \( u = 0 \) on the horizontal axis, \( F^{-1}(u) = 10 u^{1/5} \) is near \( 0 \). C: **Correct:** The histogram has a long left tail toward 0 and peaks at 10, which is a left-skewed distribution. D: **Incorrect:** The tail is on the left, not the right.

**1. Set up the inverse CDF formula from the code:** From the code snippet, \( a = 5 \), \( c = 10 \), and: \[ X = c \cdot U^{1/a} = 10 \cdot U^{1/5} \] **2. Substitute \( U = 0.243 \):** Notice that \( 0.243 = \frac{243}{1000} = \frac{3^5}{10^3} \), or directly: \[ (0.243)^{0.2} \approx 0.75355 \] Multiplying by \( c = 10 \): \[ X = 10 \times 0.75355 = 7.5355 \approx 7.5 \] (Note: if \( U = 0.243 \) was intended alongside \( 3/4 = 0.75 \) from \( 0.243 \approx (0.75)^5 = 0.2373 \), both give \( 7.5 \) to one decimal place!) **Answer: 7.5.**

**1. Analyze the mixture simulation procedure:** The algorithm draws \( U \sim \text{Uniform}(0, 1) \) and sets: - \( X = X_1 \) if \( U < 0.6 \) (which occurs with probability \( 0.6 \)) - \( X = X_2 \) otherwise (which occurs with probability \( 1 - 0.6 = 0.4 \)) **2. Apply the Law of Total Probability:** \[ P(X = k) = P(U < 0.6) P(X_1 = k) + P(U \ge 0.6) P(X_2 = k) = 0.6 p_k^{(1)} + 0.4 p_k^{(2)} \] **Answer: D — \( P(X = k) = 0.6 p_k^{(1)} + 0.4 p_k^{(2)} \).** A: **Incorrect:** Swaps the mixture weights \( 0.6 \) and \( 0.4 \). B: **Incorrect:** Multiplies the PMFs instead of taking their convex combination. C: **Incorrect:** Sums the unweighted PMFs (which would sum to 2 instead of 1). D: **Correct:** By the law of total probability, \( P(X = k) = 0.6 p_k^{(1)} + 0.4 p_k^{(2)} \).

**1. Read \( p_4^{(1)} \) and \( p_4^{(2)} \) from the given PMFs:** - From the distribution of \( X_1 \) on \( \{1, 2, 3, 4\} \): \( p_4^{(1)} = P(X_1 = 4) = 0.3 \) - From the distribution of \( X_2 \) on \( \{2, 3, 4, 5\} \): \( p_4^{(2)} = P(X_2 = 4) = 0.2 \) **2. Substitute into the mixture PMF formula:** \[ P(X = 4) = 0.6 p_4^{(1)} + 0.4 p_4^{(2)} = 0.6(0.3) + 0.4(0.2) = 0.18 + 0.08 = 0.26 \] **Answer: 0.26.**

**1. Check the support of \( X_1 \) and \( X_2 \):** - \( X_1 \) takes values only in \( \{1, 2, 3, 4\} \), so \( P(X_1 = 6) = 0 \). - \( X_2 \) takes values only in \( \{2, 3, 4, 5\} \), so \( P(X_2 = 6) = 0 \). **2. Compute \( P(X = 6) \):** \[ P(X = 6) = 0.6 P(X_1 = 6) + 0.4 P(X_2 = 6) = 0.6(0) + 0.4(0) = 0 \] Therefore, \( P(X = 6) > 0 \) is false (**No**). **Answer: B — No.** A: **Incorrect:** Neither \( X_1 \) nor \( X_2 \) can take the value 6, so \( P(X = 6) = 0 \). B: **Correct:** Since \( 6 \) is outside the support \( \{1, 2, 3, 4, 5\} \) of both component distributions, \( P(X = 6) = 0 \).

**1. Set up the Accept-Reject bound \( M \):** The target density is \( f(x) = 12 x^2 (1 - x) \) for \( x \in (0, 1) \), and the proposal density is \( g(x) = 1 \) (\( \text{Uniform}(0, 1) \)). The optimal envelope constant \( M \) is: \[ M = \sup_{x \in (0, 1)} \frac{f(x)}{g(x)} = \sup_{x \in (0, 1)} 12(x^2 - x^3) \] **2. Maximize \( f(x) \) on \( (0, 1) \):** Set the derivative \( f'(x) = 0 \): \[ f'(x) = 12(2x - 3x^2) = 12x(2 - 3x) = 0 \implies x = \frac{2}{3} \] Evaluate \( M = f(2/3) \): \[ M = 12 \left(\frac{2}{3}\right)^2 \left(1 - \frac{2}{3}\right) = 12 \times \frac{4}{9} \times \frac{1}{3} = \frac{16}{9} \] **3. Compute the probability of acceptance:** \[ P(\text{Acceptance}) = \frac{1}{M} = \frac{9}{16} = 0.5625 \approx 0.56 \] **Answer: 0.56.**