Free GARP FRM Part II Market Risk Practice Questions

Market Risk Measurement and Management carries 20% of GARP FRM Part II (GARP). Questions cover non-parametric and extreme value approaches to VaR, backtesting and VaR mapping, validating VaR models, how correlations behave and bottom-up correlation modeling, regression hedging and principal component analysis, term structure models (drift, volatility, Vasicek and Gauss+), volatility smiles and surfaces, and the Fundamental Review of the Trading Book.

233 questions 46 easy 104 medium 83 hard 2026 syllabus

Sample Questions

Question 1 Easy
Which statement about the Vasicek and Cox-Ingersoll-Ross (CIR) short-rate models is correct?
Solution
B is correct. The Vasicek model has constant volatility σ\sigma, so the short rate is Gaussian and places positive probability on negative values. The CIR model has diffusion σr\sigma\sqrt{r}, which collapses to zero as r→0r \to 0; under the Feller condition 2kθ≥σ22k\theta \geq \sigma^{2}, the short rate stays strictly positive.
Question 2 Medium
Since the 1987 equity-market crash, implied volatilities for S&P 500 index options at a given expiration typically exhibit which pattern across strikes?
Solution
C is correct. Post-1987 equity index options display a pronounced negative skew (often called a smirk). Out-of-the-money puts (low strikes) trade at meaningfully higher implied volatilities than at-the-money or out-of-the-money calls (high strikes). This reflects the market's pricing of crash-risk and heavier left-tail risk-neutral density than the lognormal assumed in Black-Scholes.
Question 3 Hard
A portfolio's daily P&L is normally distributed with zero mean. The one-day 95% parametric VaR is $1 million. Using z0.95=1.645z_{0.95} = 1.645, z0.99=2.326z_{0.99} = 2.326, and assuming square-root-of-time scaling, the 10-day 99% expected shortfall is closest to:
Solution
D is correct. First back out the dollar standard deviation from the given 95% VaR: σV=$1,000,000/1.645=$607,903\sigma V = \$1{,}000{,}000 / 1.645 = \$607{,}903. Next, the daily 99% expected shortfall multiplier under normality is ϕ(z0.99)/(1−0.99)=0.02665/0.01≈2.665\phi(z_{0.99})/(1-0.99) = 0.02665/0.01 \approx 2.665, where ϕ\phi is the standard normal density. Then daily 99% ES =2.665×$607,903≈$1,620,000= 2.665 \times \$607{,}903 \approx \$1{,}620{,}000. Scaling to a 10-day horizon by 10≈3.162\sqrt{10} \approx 3.162 gives $1,620,000×3.162≈$5.12\$1{,}620{,}000 \times 3.162 \approx \$5.12 million.
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