Quantitative Analysis

• Topic 01

  Probability Distributions

  63 questions
• Topic 02

  Statistical Inference

  51 questions
• Topic 03

  Linear Regression

  30 questions
• Topic 04

  Time-Series Analysis

  30 questions
• Topic 05

  Monte Carlo Simulation

  30 questions
• Topic 06

  Volatility Estimation

  30 questions
• Topic 07

  Correlation and Copulas

  30 questions
• Topic 08

  Extreme Value Theory

  30 questions
• Topic 09

  Bayesian Analysis

  78 questions
• Topic 10

  Machine Learning in Risk Management

  60 questions

A few questions from this unit, with the answer and a full explanation. The complete bank is available when you start practising.

1. A risk manager simulates two correlated equity returns for a portfolio VaR calculation. The correlation matrix is $\Sigma = \begin{pmatrix} 1 & 0.6 \\ 0.6 & 1 \end{pmatrix}$. She performs Cholesky decomposition $\Sigma = LL^T$ and generates correlated draws $X = LZ$ where $Z = (Z_1, Z_2)^T$ with $Z_1, Z_2 \sim i.i.d.\ N(0,1)$. What is $L$?

   • A
     $L = \begin{pmatrix} 1 & 0.6 \\ 0.6 & 1 \end{pmatrix}$
   • B
     $L = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
   • C
     $L = \begin{pmatrix} 1 & 0.6 \\ 0 & 0.8 \end{pmatrix}$
   • D
     $L = \begin{pmatrix} 1 & 0 \\ 0.6 & 0.8 \end{pmatrix}$
     Correct answer
   Explanation

   For a 2×2 correlation matrix $\Sigma = \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix}$, the lower-triangular Cholesky factor is $L = \begin{pmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{pmatrix}$. With $\rho = 0.6$: $\sqrt{1 - 0.36} = \sqrt{0.64} = 0.8$. So $L = \begin{pmatrix} 1 & 0 \\ 0.6 & 0.8 \end{pmatrix}$. Verify: $LL^T = \begin{pmatrix} 1 & 0.6 \\ 0.6 & 0.36+0.64 \end{pmatrix} = \begin{pmatrix} 1 & 0.6 \\ 0.6 & 1 \end{pmatrix} = \Sigma$. The correlated draws are $X_1 = Z_1$ and $X_2 = 0.6 Z_1 + 0.8 Z_2$.

2. An EGARCH model is estimated as $\ln(\sigma^2_t) = -0.10 + 0.85\ln(\sigma^2_{t-1}) + 0.15(|z_{t-1}| - E|z_{t-1}|) - 0.08 z_{t-1}$, where $z_{t-1} = r_{t-1}/\sigma_{t-1}$. Yesterday's standardised return was $z_{t-1} = -1.5$. Which of the following statements correctly describes the asymmetric impact?

   • A
     EGARCH cannot capture the leverage effect because it models log-variance; the leverage effect requires modelling variance directly (as in GARCH or GJR-GARCH).
   • B
     For $z_{t-1} = -1.5$: the asymmetric term is $0.15(|-1.5| - E|z|) - 0.08 \times (-1.5) = 0.15(1.5 - 0.798) + 0.12 \approx 0.105 + 0.12 = 0.225$(using$E|z| \approx \sqrt{2/\pi} \approx 0.798$for standard normal). For$z_{t-1} = +1.5: the term is \0.15(1.5 - 0.798) - 0.08 \times 1.5 \approx 0.105 - 0.12 = -0.015$. The negative return contributes much more to log-variance than the positive return of the same magnitude.
     Correct answer
   • C
     The $\gamma = -0.08$ parameter in EGARCH captures the leverage effect only when $z_{t-1} > 0$; for negative standardised returns, the $\alpha$ parameter governs the response.
   • D
     For both $z_{t-1} = +1.5$ and $z_{t-1} = -1.5$, the log-variance impact is identical because $|z| = 1.5$ in both cases and the magnitude term dominates; the $\gamma$ parameter only affects the constant.
   Explanation

   EGARCH includes two asymmetric terms: $\alpha|z_{t-1}|$ (size effect) and $\gamma z_{t-1}$ (sign effect). For a negative shock $z = -1.5$: size contribution = $0.15(1.5 - 0.798) \approx 0.105$; sign contribution =$-0.08 \times (-1.5) = +0.12$; total asymmetric addition$\approx 0.225$. For a positive shock$z = +1.5$: size contribution$\approx 0.105$; sign contribution =$-0.08 \times 1.5 = -0.12$; total asymmetric addition$\approx -0.015$. So the negative shock raises log-variance by approximately 0.24 more than the positive shock of the same magnitude.$\gamma < 0$ captures the leverage effect in EGARCH notation. EGARCH operates in log-variance space, guaranteeing positive variance without non-negativity constraints.

3. In Bayes' theorem, $P(H | D) = \frac{P(D | H) \cdot P(H)}{P(D)}$, the term $P(D)$ is called the marginal likelihood (or evidence). What is its primary role in the equation?

   • A
     It adjusts the likelihood for the sample size of the data.
   • B
     It is the probability of the complement hypothesis $P(\neg H)$.
   • C
     It measures how well the hypothesis predicts the observed data.
   • D
     It acts as a normalising constant that ensures the posterior probabilities sum to 1.
     Correct answer
   Explanation

   The marginal likelihood $P(D)$ ensures the posterior is a proper probability distribution (i.e., that posteriors across all hypotheses sum or integrate to 1). It is computed via the law of total probability: $P(D) = \sum_i P(D | H_i) P(H_i)$. While it is a crucial normalising constant, it does not directly encode the analyst's prior belief (that is $P(H)$) nor measure predictive accuracy (that is the likelihood $P(D | H)$). In practice, computing $P(D)$ analytically can be intractable for complex models — motivating MCMC methods.