--- title: "Risk, Validation and Regulation" book: "Interviews II: Markets, Trading Games and the Final Round" subject: quant language: en chapter: 13 exercises: 0 source: https://one-course.com/books/quant/19/en/chapter/13-risk-validation-and-regulation license: CC-BY-NC-SA-4.0 credit: "One Quant Book, One Course (one-course.com)" --- # Chapter 13 — Risk, Validation and Regulation Value at risk and expected shortfall by hand and when they break, backtests and how much they can really tell you, stress tests, factor risk, the capital rules that a risk answer has to survive, and how to validate a model. Every portfolio here is invented. Quantiles of the normal: 2.326 at 99% and 1.96 at 97.5%. **Question 13.1 ★ Risk, QR, Trader • any • junior.** You are long $100 million of an equity index with a daily volatility of 1%. What is the 99% one-day value at risk, or VaR, under a normal model with zero mean? What does the square-root-of-time rule give for ten days? **Answer to Question 13.1.** It is about $2.33 million for one day, and about $7.36 million for ten days. The dollar volatility is $1 million, and the 1% quantile is 2.326 standard deviations. For ten days, multiply by $\sqrt{10} = 3.162$. **Classic wrong answer.** Saying “we can’t lose more than $2.33 million.” VaR is the point the loss passes on one day in a hundred. It says nothing about how far the loss goes on that day. **Question 13.2 ★ Risk, QR, Analyst • any • junior.** You hold $50 million of one index with a daily volatility of 1% and $30 million of another with a daily volatility of 1.5%. Their correlation is 0.8. What is the one-day 99% VaR? How much does diversification save? **Answer to Question 13.2.** It is about $2.10 million, and diversification saves about $0.11 million. The dollar volatilities are 0.5 and 0.45. Then $$ \sigma = \sqrt{0.25 + 0.2025 + 2 \times 0.8 \times 0.5 \times 0.45} = 0.901. $$ And $0.901 \times 2.326 = 2.10$. Adding the standalone numbers gives $0.95 \times 2.326 = 2.21$. **Classic wrong answer.** Multiplying the correlation into the volatilities and forgetting the factor 2 on the cross term. Weight it by both dollar volatilities and count it twice. **Question 13.3 ★ Risk, QR • any • junior.** Define expected shortfall. For a normal distribution with mean 0 and standard deviation 1, what is it at 97.5% and at 99%? How does the 97.5% figure compare with the 99% VaR? **Answer to Question 13.3.** Expected shortfall is the average loss given that the loss is beyond VaR. For a standard normal, it is 2.338 at 97.5% and 2.665 at 99%. So 97.5% shortfall is almost the same as the 99% VaR of 2.326. The formula is $\mathrm{ES}_{\alpha} = \varphi(z)/(1 - \alpha)$, with $z$ the quantile. So $\varphi(1.96)/0.025 = 2.338$. That is why the regulators’ switch to 97.5% shortfall holds capital roughly level for a normal book and asks for more only if the tails are fatter. **Classic wrong answer.** Thinking shortfall is a much higher number at any level. At 97.5%, on a normal, it only barely exceeds the old 99% VaR. **Question 13.4 ★ Risk, Analyst, QR • any • junior.** Over 250 days, your five worst daily losses in millions were 8, 6, 5, 4 and 3.5. Estimate the 99% historical VaR and the 99% expected shortfall. Say which convention you use. **Answer to Question 13.4.** The 99% VaR is about 5, with shortfall about 6.6. One percent of 250 is 2.5 observations. Taking the third-worst as VaR is the conservative convention, and it gives 5. Interpolating between the second and third gives 5.5. The shortfall averages the worst 2.5 observations: $(8 + 6 + 0.5 \times 5)/2.5 = 6.6$. **Classic wrong answer.** Treating a single number as the answer without naming the convention. With so few points, the convention alone moves VaR by 10%. **Question 13.5 ★ Risk, QR • bank • junior.** A bank’s 99% one-day VaR had 7 exceptions in the past 250 days. Under the Basel traffic light, which zone is that, and what does it do to the multiplier of 3? **Answer to Question 13.5.** It is in the yellow zone, and the multiplier goes from 3 to 3.65. Green is 0 to 4 exceptions, yellow is 5 to 9 with add-ons of 0.40, 0.50, 0.65, 0.75 and 0.85, and red is 10 or more, with 1.00. **Classic wrong answer.** Saying that with 2.5 expected, the model has failed. Yellow means “could go either way”, and supervisors look at why the exceptions came. **Question 13.6 ★ Risk, QR, QT • any • junior.** An exponentially weighted volatility uses $\lambda = 0.94$. What is the half-life of a return’s weight, in days? Yesterday’s daily volatility estimate was 1%, and today’s return is 2%. What is the new estimate? **Answer to Question 13.6.** The half-life is about 11.2 days, and the new estimate is about 1.086%. The half-life solves $0.94^k = 0.5$, so $k = \ln 0.5/\ln 0.94 = 11.2$. The variance is $0.94 \times 0.0001 + 0.06 \times 0.0004 = 0.000118$, and its root is 1.086%. **Classic wrong answer.** Updating the volatility rather than the variance: $0.94 \times 1\% + 0.06 \times 2\% = 1.06\%$. That is a different number, and it is not the model. **Question 13.7 ★ Risk, Analyst, QR • any • junior.** A stock has a beta of 1.2 to a market with 15% volatility, and 20% idiosyncratic volatility. What are its total volatility and its $R^2$? What is the volatility of an equal-weighted portfolio of 100 such stocks, with independent idiosyncratic parts? **Answer to Question 13.7.** Total volatility is about 26.9%, $R^2$ is about 45%, and the portfolio of 100 has about 18.1%. Systematic volatility is $1.2 \times 15\% = 18\%$. Total is $\sqrt{0.18^2 + 0.20^2} = 0.269$. The $R^2$ is $0.0324/0.0724 = 44.8\%$. In the portfolio, idiosyncratic volatility falls to $20\%/\sqrt{100} = 2\%$. So the total is $\sqrt{0.18^2 + 0.02^2} = 18.1\%$. Beta stays. **Classic wrong answer.** Thinking diversification lowers it towards zero. It removes the idiosyncratic part only. The market risk does not diversify. **Question 13.8 ★ Risk, Dev, QD • any • junior.** An account’s value goes 100, 120, 90, 130, 65, 80. What is the maximum drawdown? How do you compute it in one pass? **Answer to Question 13.8.** It is 50%, from 130 to 65. Keep the running peak. At each point, compute the fall from the peak, and keep the largest. That is one pass with two variables. **Classic wrong answer.** Using the highest and lowest points overall. The low has to come after the peak, and here that happens to be the same pair, but a low before the high does not count. **Question 13.9 ★ Risk, Trader, QT • any • junior.** You are long a call with delta 0.5 and gamma 0.02 on a $100 stock with a daily volatility of 2%. What is the delta-normal one-day 99% VaR per option? Is that too high or too low once you add gamma? **Answer to Question 13.9.** It is about $2.33 per option. That is too high, as gamma helps a long option. The move is $2.326 \times \$2 = \$4.65$, and half of that is the delta loss. With gamma, $-0.5 \times 4.65 + 0.5 \times 0.02 \times 4.65^2 = -2.11$. For a short option, the error runs the other way, which is worse: delta-normal then understates the loss. **Classic wrong answer.** Dropping gamma on the view that it is small. On a move of 2.3 standard deviations it is 10% of the number here, and for short options near expiry it can be most of the risk. **Question 13.10 ★ Risk, Trader, Analyst • bank, hedge fund • junior.** A bond book has a DV01 of $50,000, and the yield has a daily volatility of 6 basis points. What is the one-day 99% VaR? **Answer to Question 13.10.** It is about $698,000. $\$50{,}000 \times 6 \times 2.326 = \$697{,}800$. **Classic wrong answer.** Mixing up a yield’s volatility in basis points with its volatility as a percentage of the level. A yield volatility quoted as a percentage needs the level to convert to basis points. **Question 13.11 ★★ Risk, QR • any • junior, experienced.** Two independent bonds each lose 100 with probability 4% and nothing otherwise. What is the 95% VaR of each one and of the pair? Compare the expected shortfall at 95%. What property does this show that VaR lacks? **Answer to Question 13.11.** Each bond’s 95% VaR is 0. The pair’s is 100. Shortfall at 95% is 80 each and 103.2 for the pair. So VaR is not subadditive: combining two bonds raised it. Each bond’s loss has probability 4%, which is within the 5% tail, so its 5% quantile is still zero. For the pair, the chance of at least one default is $1 - 0.96^2 = 7.84\%$, which is above 5%, so VaR is 100. For the shortfall, each bond’s tail of 5% holds 4% at 100 and 1% at 0, so $0.04 \times 100/0.05 = 80$. The pair’s tail holds 0.16% at 200 and 4.84% at 100, so it is $(0.32 + 4.84)/0.05 = 103.2$, below $160$. **Classic wrong answer.** Treating subadditivity as a mere technical concern. Here a VaR limit rewards concentration: you pick up rare-loss risks that sit just beyond the quantile, and the number stays at 0. **Question 13.12 ★★ Risk, QR • bank • junior, experienced.** Your 99% VaR has 7 exceptions in 250 days. Run the Kupiec test. Do you reject at 5%? What are the exact binomial tail probability and the test’s verdict at 4 exceptions and at 0? **Answer to Question 13.12.** Yes, you reject: the likelihood ratio is 5.50, above the 3.84 cut-off, with a $p$-value of 1.9%. The exact tail is 1.37%. At 4 exceptions, the ratio is 0.77 and you do not reject. At 0, it is 5.03, and you reject the other way. $$ \mathrm{LR} = -2\ln\frac{(1 - p)^{n - x}p^{x}}{(1 - \hat p)^{n - x}\hat p^{x}}, $$ with $\hat p = x/n$, compared with a $\chi^2_1$. The exact figure, $P(X \ge 7)$ for a binomial with $n = 250$ and $p = 1\%$, is a check on the asymptotic test. The zero case surprises people. With no exceptions in 250 days, $\mathrm{LR} = -2 \times 250\ln(0.99) = 5.03$. The VaR is too large. That costs capital, even if it is a safer error. **Classic wrong answer.** Using only one side. The test is two-sided, and an overly cautious model has failed too. Yet the traffic light punishes only the high side. **Question 13.13 ★★ Risk, QR • bank • junior, experienced.** Across 250 day-pairs, a 99% VaR has these counts: no exception then no exception, 238; no exception then exception, 5; exception then no exception, 4; exception then exception, 3. Compute the transition probabilities and run Christoffersen’s independence test. **Answer to Question 13.13.** The transition probabilities are 2.06% after a quiet day and 42.9% after an exception, the likelihood ratio is 12.5, and independence is firmly rejected, with $p \approx 0.0004$. $\pi_{01} = 5/243$, $\pi_{11} = 3/7$, and the pooled rate is $8/250 = 3.2\%$. The ratio is $$ -2\bigl[\ln L(\text{pooled}) - \ln L(\pi_{01}, \pi_{11})\bigr] = 12.5, $$ against a $\chi^2_1$ cut-off of 3.84. Exceptions in clusters mean that VaR does not react to changes in volatility. So when a bad day comes, more follow, right when you need the number to be right. **Classic wrong answer.** Passing the model because it looks fine in total. The count test does not care when they happen, so a model can pass on count with every exception in one week. **Follow-up.** *What does the conditional coverage test do?* It adds the Kupiec and independence statistics, and compares the sum with a $\chi^2_2$. **Question 13.14 ★★ Risk, QR • any • junior, experienced.** Daily returns have a first-order autocorrelation of 0.2, following an AR(1). By how much does the square-root-of-time rule understate ten-day volatility? What happens with a negative autocorrelation? **Answer to Question 13.14.** By about 20%: the ten-day volatility is 3.79 daily volatilities, not 3.16. For an AR(1), $$ \mathrm{Var}\Bigl(\sum_{i=1}^{m} r_i\Bigr) = \sigma^2\Bigl(m + 2\sum_{k=1}^{m-1}(m - k)\rho^k\Bigr). $$ At $m = 10$ and $\rho = 0.2$, the bracket is 14.375, and the root is 3.79. The ratio is $3.79/3.16 = 1.20$. At a negative autocorrelation, the rule overstates: mean-reverting returns accumulate less. And the formula gives $\sqrt{m}$ again at $\rho = 0$. Also, volatility itself is not constant, and the rule ignores that too. **Classic wrong answer.** Applying $\sqrt{10}$ to everything. Illiquid assets with smoothed prices tend to show positive autocorrelation, and for those it is exactly the wrong direction. **Question 13.15 ★★ Risk, QR • any • junior, experienced.** Returns follow a Student-$t$ with 4 degrees of freedom, rescaled to a variance of 1. The 1% quantile of the unscaled $t_4$ is $-3.747$. What are the 99% VaR and expected shortfall, against the normal numbers? **Answer to Question 13.15.** VaR is about 2.65, and shortfall about 3.69, against 2.33 and 2.67 for the normal. A $t_\nu$ has variance $\nu/(\nu - 2) = 2$. So the unit-variance scale is $\sqrt{1/2}$, and $3.747 \times 0.7071 = 2.65$. For shortfall, the closed form is $$ \mathrm{ES} = \frac{f_\nu(q)}{0.01}\cdot\frac{\nu + q^2}{\nu - 1}\cdot\sqrt{\frac{\nu - 2}{\nu}} = 3.69. $$ So at the same variance, VaR is 14% higher, and shortfall 39% higher. Fat tails hurt shortfall more. **Classic wrong answer.** Forgetting to scale to unit variance. With the raw 3.747, you would compare a distribution of variance 2 with one of variance 1. **Question 13.16 ★★ Risk, QR, QD • any • junior, experienced.** You estimate a 99% VaR by Monte Carlo on a normal P&L with 10,000 paths. What is the standard error of that quantile, in standard deviations and as a share of VaR? How many paths would bring it to 0.5%? **Answer to Question 13.16.** It is about 0.037 standard deviations, or about 1.6% of VaR. For 0.5%, you need about 103,000 paths. The asymptotic standard error of a quantile is $\sqrt{p(1 - p)/n}/f(q)$. Here $\sqrt{0.0099/10{,}000} = 0.000995$, and $f(2.326) = 0.0267$. So $0.0373$, which is $0.0373/2.326 = 1.6\%$. Since it falls as $\sqrt{n}$, going from 1.6% to 0.5% needs $(1.6/0.5)^2 \approx 10.3$ times more paths. And a shortfall has a different standard error, and fatter tails need more paths for the same precision. **Classic wrong answer.** Using the standard error of a mean, $1/\sqrt{n}$. The quantile estimator depends on the density at that point, and the density is low out in the tail. **Question 13.17 ★★ Risk, QR • any • junior, experienced.** For two positions with dollar volatilities of 0.5 and 0.45 and a correlation of 0.8, split the one-day 99% VaR into contributions that add up to the total. Why do they add up? How do you confirm it by a bump? **Answer to Question 13.17.** The contributions are 1.110 and 0.987, adding up to 2.097. The covariance of each with the total, divided by the total’s volatility, gives $(0.25 + 0.18)/0.901 = 0.477$ and $(0.2025 + 0.18)/0.901 = 0.424$. Multiply by 2.326. They add up by Euler’s theorem. Volatility and normal VaR are homogeneous of degree 1 in position size, so the sum of size times marginal risk equals the total. To confirm: bump the first position by 1%, recompute VaR, and the change should be 1% of its contribution, 0.0111. **Classic wrong answer.** Allocating by standalone VaR. Those sum to 2.21, not to the total, and they give no sign of who adds risk at the margin. A hedge would get a positive allocation. **Question 13.18 ★★ Risk, Trader, Analyst • bank, hedge fund • junior, experienced.** A book is long $50 million of equities with beta 1, has a DV01 of $40,000 long duration, and has a CS01 of $30,000 (it loses that when spreads widen 1 basis point). Apply equities $-20\%$, rates $-100$ basis points and spreads $+150$. What is the P&L? How much further must equities fall for a loss of $15 million? **Answer to Question 13.18.** The loss is $10.5 million, and equities need to fall a further 9% for a loss of $15 million. Equities: $50 \times -20\% = -\$10$ million. Rates: $+\$40{,}000 \times 100 = +\$4$ million. Spreads: $-\$30{,}000 \times 150 = -\$4.5$ million. The remaining $4.5 million on $50 million of equities is 9%, so $-29\%$. **Classic wrong answer.** Assuming rates always help. Whether a rate fall comes with an equity fall is exactly the assumption being stressed. In an inflation scare, both legs lose together, and the stress should have that scenario too. **Question 13.19 ★★ Risk, QR, Analyst • hedge fund, bank • junior, experienced.** Three stocks have weights 0.5, 0.3 and 0.2, and betas on two factors of (1.2, 0.3), (0.8, $-0.2$) and (1.0, 0.5). The factor covariance has variances 0.04 and 0.01 and covariance 0.006. Idiosyncratic variances are 0.03, 0.02 and 0.04. What are the portfolio’s volatility and the share of its variance that comes from factors? **Answer to Question 13.19.** The volatility is about 23.9%, and factors explain about 81% of the variance. Portfolio factor exposures are $b = B^\top w = (1.04, 0.19)$. The factor variance is $$ 1.04^2 \times 0.04 + 2 \times 1.04 \times 0.19 \times 0.006 + 0.19^2 \times 0.01 = 0.0460. $$ Idiosyncratic variance is $0.25 \times 0.03 + 0.09 \times 0.02 + 0.04 \times 0.04 = 0.0109$. The total is 0.0569. **Classic wrong answer.** Using weights in place of squared weights for idiosyncratic variance. Those parts are independent, so they add with squared weights, which is why diversification works. **Question 13.20 ★★ Risk, Trader, QT • hedge fund • junior, experienced.** You are long $10 million of one stock and short $10 million of a peer, each with a daily volatility of 2%. What is the one-day 99% VaR at a correlation of 0.9? What is it if the correlation falls to 0.5? What does that say about using recent correlation? **Answer to Question 13.20.** It is about $208,000 at 0.9, and about $465,000 at 0.5. So a correlation change raises VaR by 2.2 times with no change in positions. The spread’s volatility is $\$200{,}000\sqrt{2 - 2\rho}$, and multiply by 2.326. So a low-VaR pair is long correlation: it relies on the correlation staying high. That correlation is lowest when you need it most: a crisis can pull a pair apart, or a takeover can break it in an afternoon. Recent correlation makes a position look safest just before it blows up, and so sizing by that number builds a large position at the wrong time. **Classic wrong answer.** Treating gross of $20 million and a VaR of $208,000 as a small risk. Look at gross and stress the correlation, as gross is what you have to get out of. **Question 13.21 ★★ Risk, QR • any • junior, experienced.** Distribution A is a standard normal. Distribution B is a standard normal with probability 99.5%, and a loss of 50 with probability 0.5%. Compare the 99% VaR and the 99% expected shortfall. Why have regulators moved towards shortfall? What do you lose by moving? **Answer to Question 13.21.** Both have about the same VaR: 2.33 against 2.57. Their shortfalls differ by a factor of ten: 2.67 against 26.4. For B, the loss of 50 is half of the 1% tail. So VaR is the point where the normal part supplies the other 0.5%: $0.995\,\Phi(-q) = 0.005$, so $q = 2.57$. The shortfall is $$ \frac{0.005 \times 50 + 0.995\,\varphi(q)}{0.01} = 26.4. $$ So shortfall sees the size of the tail, and is subadditive, which is why regulators moved to it. **What you lose.** Shortfall is harder to backtest, as a single number doesn’t pin down what you are checking. And it depends on few data points in the deep tail, so its estimate is less stable. It isn’t elicitable alone: no scoring function has it as its best forecast, though it is jointly with VaR. **Classic wrong answer.** Saying shortfall solves tail risk. It still relies on the same data, and if the 0.5% event isn’t in your sample, the number is 2.67. **Question 13.22 ★★ Risk, Trader, QT • hedge fund, bank • junior, experienced.** You hold $10 million of a stock with normal returns, a daily volatility of 2% and a half-spread of 20 basis points, so a sale costs 20 basis points of the amount sold. You must sell in five equal parts, one at each day’s close, so the whole position is at risk on the first day. What is the 99% VaR of the sale, with the spread added as a sure cost? What is it if you can sell everything at the first close? **Answer to Question 13.22.** It is about $710,000 with five equal parts, and about $485,000 if you can sell in a day. One-day VaR is $\$10 \text{ million} \times 2\% \times 2.326 = \$465{,}000$. Selling in equal parts, the position held across day $k$ is $(6 - k)/5$. So the total variance is $\sum_{k=1}^{5}((6 - k)/5)^2 = 2.2$ daily variances, a factor of $\sqrt{2.2} = 1.483$. That gives $690,000. The general factor is $\sqrt{(n + 1)(2n + 1)/(6n)}$. The half-spread of 20 basis points on the $10 million adds $20,000. **Classic wrong answer.** Using $\sqrt{5}$. That holds the entire position for five days, but you are selling on the way, so the factor is 1.48 and not 2.24. And spread is a sure cost, so it adds in directly. **Follow-up.** *What is missing?* Market impact, which grows with the trading rate. Selling quickly reduces this risk but costs more impact, and that trade-off is the optimal-execution problem. **Question 13.23 ★★ Risk, QR, QD • bank • junior, experienced.** A desk asks you to validate its new pricing model before it goes live. What do you do? Give the steps in order, and what would make you refuse it. **Answer to Question 13.23.** Start by defining what it is for, then check the theory, rebuild it independently, test its limits, compare with outcomes, and approve only within a stated range of use. I would refuse if I could not reproduce its numbers or if it is being used outside its assumptions. 1. **Scope.** What it prices, its users, how much P&L and capital depend on it. That sets the depth of review. 2. **Theory.** Are its assumptions right for the product? A one-factor model on a product paying on the correlation of two rates fails here, whatever the code does. 3. **Independent reimplementation.** Build it from the documentation and compare. This finds what the developer can’t see. 4. **Benchmarks and limits.** Closed-form cases, parity, zero and huge volatility, and very short maturity. Check convergence. 5. **Inputs.** Where market data comes from, how it is calibrated, how stable the calibration is day to day. 6. **Outcomes.** P&L explain, hedging simulation, and how far the model’s price is from traded prices and consensus services. 7. **Conditions.** Limits on use, reserves for the uncertainty, a review date and monitoring. **As of November 2026 — The guidance.** In the US, SR 26-2 of April 2026 superseded the earlier model-risk guidance, SR 11-7, and it keeps effective challenge by objective experts, with outcomes analysis and benchmarking in validation. In the UK, the PRA’s SS1/23 has been in effect since May 2024, with five principles. Check the current text when you cite one. **Classic wrong answer.** Treating validation as a re-run of the developer’s tests. Those show the code does what the developer meant. Validation asks whether that was the right thing to mean. **Follow-up.** *The desk says the trade must go today, and validation takes weeks. What do you do?* Offer a conditional approval with a position limit, a reserve, and a date, and write it down. The worst outcome is a model that goes live with no one saying what it may do. **Question 13.24 ★★ Risk, QR • bank • junior, experienced.** A 99% VaR goes through a 250-day traffic light. If the model is right, what is the chance that it stays green? If the true exception rate is 2% or 3%, what is the chance that it stays green, and that it goes red? **Answer to Question 13.24.** If the model is right, it stays green with probability 89.2%. At 2%, it stays green 43.9% of the time, and goes red 3.0%. At 3%, green 12.8%, and red 22.1%. Green is at most 4 exceptions, and red is at least 10. Binomial with $n = 250$. So $P(X \le 4 \mid p = 1\%) = 0.892$. At $p = 2\%$, $P(X \le 4) = 0.439$, and $P(X \ge 10) = 0.030$. So a model that is wrong by a factor of two passes green nearly half the time. And a right model goes yellow 11% of the time. That is why a one-year count is a weak test. To separate 1% from 2% cleanly, you need much more data or a test using more than counts. **Classic wrong answer.** Treating green as a proof the model is right. A green zone is compatible with a doubled exception rate. **Question 13.25 ★★ Risk, Analyst, QR • bank • junior, experienced.** Capital is $\max(\mathrm{VaR}_{t-1}, m\,\overline{\mathrm{VaR}}) + \max(\mathrm{sVaR}_{t-1}, m\,\overline{\mathrm{sVaR}})$. VaR was 12 yesterday and averages 10 over 60 days. Stressed VaR was 18 and averages 20, and $m = 3.65$. What is the capital? How high would yesterday’s VaR need to be to matter? What does stressed VaR add? **Answer to Question 13.25.** The capital is 109.5. Yesterday’s VaR would have to be above 36.5 to matter. Stressed VaR adds 73, twice the 36.5 from the rest. $\max(12, 3.65 \times 10) = 36.5$. $\max(18, 3.65 \times 20) = 73$. So the sum is 109.5. Since the multiplier is above 3, the average nearly always wins. For the first term, a jump to more than 3.65 times the recent average would be needed. Stressed VaR uses the current portfolio on inputs from a period of financial stress over 12 months. It doesn’t fall in calm periods, and so capital is less procyclical than with VaR alone. **As of November 2026 — Where the rules are.** The January 2019 market-risk standard replaces this with 97.5% expected shortfall. Adoption varies by place and date. As of mid-2026, the European Commission had adopted time-limited amendments applying from 1 January 2027, and the UK’s PRA set internal-model changes for 1 January 2028. Check the current position before you cite it. **Classic wrong answer.** Thinking capital tracks yesterday’s VaR. A sudden one-day jump has almost no effect, so it is the average that a desk must manage. **Question 13.26 ★★★ Risk, QR, Trader • any • experienced.** A delta-hedged book is short options with gamma $-500$ per dollar squared, and the stock’s daily move is normal with a standard deviation of $2. Ignore theta. What does delta-normal VaR say? What is the exact 99% VaR? What does a mean-and-variance normal approximation give? **Answer to Question 13.26.** Delta-normal VaR is zero. The exact 99% VaR is about $6,635. The mean-and-variance normal gives about $4,290, about 35% too low. The loss is $0.5 \times 500 \times x^2$ with $x = 2z$, so $1{,}000 z^2$. Its 99% point comes from $z^2$ as a $\chi^2_1$, whose 99% quantile is $6.635$. So the VaR is $6,635. Both tails of the stock count, and $z = 2.576$ is the two-sided 1% point. The approximation has mean 1,000 and standard deviation $1{,}000\sqrt{2} = 1{,}414$. So $1{,}000 + 2.326 \times 1{,}414 = 4{,}290$. The P&L is strongly skewed, so a normal shape misses the tail. **Classic wrong answer.** Reporting no risk since delta is zero. This book has no first-order risk at all, and its risk is all in the second order, which is what the method ignores. **Question 13.27 ★★★ Risk, QR • any • experienced.** Your returns have skewness $-0.5$ and excess kurtosis 3. Give the Cornish-Fisher 1% quantile. Now use the formula on a Student-$t$ with 6 degrees of freedom, scaled to unit variance, whose 1% quantile of the unscaled $t_6$ is $-3.143$. What do you learn? **Answer to Question 13.27.** The Cornish-Fisher quantile is about $-3.30$. On the $t_6$, the formula gives about $-3.03$, where the truth is $-2.57$. So it has overshot by 18%. It is a series, and not a safe answer in fat tails. $$ z_{\mathrm{CF}} = z + \frac{(z^2 - 1)S}{6} + \frac{(z^3 - 3z)K}{24} - \frac{(2z^3 - 5z)S^2}{36}. $$ With $z = -2.326$, $S = -0.5$ and $K = 3$, that gives $-3.30$. The $t_6$ has variance $6/4$, so its unit-variance quantile is $-3.143 \times 0.8165 = -2.57$. Its excess kurtosis is 3, and $S = 0$, so the formula gives $-2.326 - 0.70 = -3.03$. So the same kurtosis can mean different tails. The expansion is exact only for small departures from normal, and it can even produce non-monotone quantiles at higher skew and kurtosis. It is a useful correction for small $S$ and $K$. With fat tails, fit a distribution or use the empirical tail. **Classic wrong answer.** Assuming a higher-moment correction is always closer to the truth. Here it is further from the truth than the plain normal in size: 0.46 too far against 0.24 too near. And sample skew and kurtosis are themselves noisy. **Question 13.28 ★★★ Risk, QR • bank • experienced.** A desk reports 97.5% expected shortfall under a normal model. Over $T = 250$ days, a test statistic is $Z = 1 + \sum_t X_t I_t/(T \times 0.025 \times \mathrm{ES})$, with $X_t$ the P&L and $I_t$ marking days beyond VaR. What is its mean if the model is right? If the true volatility is 1.2 times the model’s? Where would you set the 5% cut-off, and what power does it have? **Answer to Question 13.28.** The mean is 0 if the model is right, and about $-1.16$ if the true volatility is 1.2 times the model’s. A 5% cut-off is about $-0.70$, by simulation, and that detects the 1.2 case about 77% of the time over 250 days. Under the model, each day’s expectation is $E[X I] = -\alpha\,\mathrm{ES}$, so $Z$ has mean $1 - 1 = 0$. If the true volatility is $s$, $E[X I] = -s\,\varphi(v/s)$, with $v = 1.96$. So the mean of $Z$ is $1 - s\,\varphi(v/s)/(0.025\,\mathrm{ES})$, and at $s = 1.2$ that gives $-1.158$. The statistic has one nice property for a bank: since it uses sizes as well as counts, a too-small shortfall shows up even if VaR is right. The cut-off has to come from simulation, and it depends on the assumed model, though it is stable for a range of tails. **Classic wrong answer.** Thinking you can’t backtest shortfall at all. Its lack of elicitability rules out one type of comparison, not all tests. What it needs is the model’s own forecast, day by day. **Question 13.29 ★★★ Risk, QR • any • junior, experienced.** A bank uses a 250-day historical simulation for a 99% VaR. Volatility doubles overnight and stays doubled for 20 days. What is the exception rate on those days, and the expected count? How does an exponentially weighted model with $\lambda = 0.94$ do on the tenth day? **Answer to Question 13.29.** It is about 12% a day at first, so about 2.2 exceptions over 20 days, against 0.2 expected. The exponentially weighted model is at about 3.6% on the tenth day. At first, the threshold is still 2.326 old volatilities, which is 1.163 new ones, and $\Phi(-1.163) = 12.2\%$. Held fixed for 20 days, that would give 2.45 exceptions. But the new days enter the window as the worst ones, and each lifts the third-worst a little. In a simulation, the rate falls from about 12.7% to about 9% over the 20 days, and the total is about 2.2. For the exponential model, after 10 days a share $1 - 0.94^{10} = 46\%$ of the weight is on the new regime. So variance is $0.54 + 0.46 \times 4 = 2.38$ old units, and volatility is 1.54 old units. The threshold is $2.326 \times 1.544/2 = 1.80$ new volatilities, so $\Phi(-1.80) = 3.6\%$. So historical VaR’s exceptions cluster, and they come right after volatility jumps. An exponential model catches up, though it still starts behind. And the historical model’s VaR keeps the high number for a year after the storm ends. **Classic wrong answer.** Raising the confidence level to solve it. A higher quantile is a higher level, but it adapts no faster. **Question 13.30 ★★★ Risk, QR • any • experienced.** Losses above a threshold of 2 follow a generalised Pareto distribution with shape 0.2 and scale 1, and 5% of days exceed the threshold. What are the 99% VaR and the 99% expected shortfall? What happens as the shape reaches 1? **Answer to Question 13.30.** The VaR is about 3.90, and the shortfall about 5.62. As the shape reaches 1, shortfall becomes infinite. With $u = 2$, $\beta = 1$, $\xi = 0.2$, and exceedance rate 5%, $$ \mathrm{VaR}_p = u + \frac{\beta}{\xi}\Bigl(\Bigl(\frac{1 - p}{0.05}\Bigr)^{-\xi} - 1\Bigr) = 2 + 5(0.2^{-0.2} - 1) = 3.90. $$ And $$ \mathrm{ES}_p = \frac{\mathrm{VaR}_p + \beta - \xi u}{1 - \xi} = \frac{3.90 + 1 - 0.4}{0.8} = 5.62. $$ The $1 - \xi$ in the denominator is why. A shape of 1 or more means the tail has no finite mean, so no average loss exists to report. At a shape of 0, the tail is exponential: VaR is $2 + \ln 5 = 3.61$, and shortfall exceeds it by exactly $\beta$. **Classic wrong answer.** Choosing a threshold so high that few points remain, or so low that the Pareto shape doesn’t hold. Plot the estimate of $\xi$ against the threshold, and take a stable stretch. **Question 13.31 ★★★ Risk, QR • any • experienced.** Two assets have a correlation of 0.5. What is the chance that both fall below their own 1% quantile, under a Gaussian copula and under a Student-$t$ copula with 4 degrees of freedom? What is the limit of the tail dependence in each? **Answer to Question 13.31.** It is about 0.13% under the Gaussian copula, and about 0.29% under the $t_4$, against 0.01% for independence. The limit of tail dependence is 0 for the Gaussian and about 0.25 for the $t_4$. The Gaussian figure is the bivariate normal at $(c, c)$, with $c = -2.326$, and the $t$ figure is by simulation. So given that one falls past its 1% point, the other follows with probability about 13% or 29%. The tail dependence coefficient for a $t$ copula is $$ \lambda = 2\,t_{\nu + 1}\Bigl(-\sqrt{\frac{(\nu + 1)(1 - \rho)}{1 + \rho}}\Bigr) = 0.253. $$ For the Gaussian, $\lambda = 0$ whenever $|\rho| < 1$: in the limit, the extremes are independent. So the same correlation can mean twice the joint crash risk, and a Gaussian copula with fat marginals still misses it. **Classic wrong answer.** Thinking fat-tailed marginals fix joint tails. They change each asset’s own tail, but it is the copula that decides whether they crash together. **Question 13.32 ★★★ Risk, QR • bank • experienced.** Under the trading-book rules, a 97.5% shortfall is computed at a 10-day base horizon and then cascaded by liquidity horizons of 10, 20, 40, 60 and 120 days. The 10-day shortfall on all factors is 10, and on the factors with horizon at least 20, 40, 60 and 120 days it is 8, 5, 3 and 2. What is the capital measure? Why is it not $10\sqrt{12}$? **Answer to Question 13.32.** The capital measure is 16. It is not $10\sqrt{12} = 34.6$, because each horizon’s scaling applies only to the factors with that horizon, and factors are shocked over the extra time alone. $$ \mathrm{ES} = \sqrt{\mathrm{ES}_T(P)^2 + \sum_{j \ge 2}\Bigl(\mathrm{ES}_T(P, j)\sqrt{\frac{\mathrm{LH}_j - \mathrm{LH}_{j-1}}{T}}\Bigr)^{2}}. $$ The terms are $10^2$, $(8\sqrt{1})^2 = 64$, $(5\sqrt{2})^2 = 50$, $(3\sqrt{2})^2 = 18$ and $(2\sqrt{6})^2 = 24$. The total is 256, so 16. If only 10-day factors exist, the later terms vanish, and it is just 10. If every factor is 120-day, all are 10, and the sum is $100(1 + 1 + 2 + 2 + 6) = 1{,}200$, so $10\sqrt{12} = 34.6$. So that naive number is the upper limit, and the formula passes this check. **Classic wrong answer.** Scaling the whole portfolio to its longest horizon. That is the top limit, and it charges liquid factors for an illiquid horizon. **Question 13.33 ★★★ Risk, QR, Dev • any • junior, experienced.** A VaR spreadsheet computes a rate change as $(x_t - x_{t-1})/(x_t + x_{t-1})$ when it was meant to divide by the average of the two. What happens to the volatility and the VaR? How many exceptions a year would you expect in a backtest, for normal returns? How would a validator catch this before any backtest? **Answer to Question 13.33.** Volatility and VaR are halved. You would then expect about 31 exceptions a year in a 99% test, against 2.5. A validator catches it by reperforming one small calculation by hand before any data run. For small changes, $x_t + x_{t-1} \approx 2x_{t-1}$, so the ratio is half the intended one, and every return is halved. So every volatility is halved, and VaR with it. It is exact to first order, whatever the volatility. For normal returns, the threshold is $2.326/2 = 1.163$ true standard deviations. So the rate is $\Phi(-1.163) = 12.2\%$, or 31 days in 250. **How to catch it early.** Hand-compute one number. Compare the input volatility with an independent source. Put in a series with a known volatility and check that it comes back. Don’t let the production model live in an unreviewed spreadsheet with manual steps. **Classic wrong answer.** Trusting the backtest to catch it. If a change of model comes with a quiet market, or hedged positions, the exceptions can take months to show. And a model that lowers VaR is rarely questioned by the people it helps. **Question 13.34 ★★★ Risk, QR • bank • experienced.** Exposure at default is $1.4\,(\mathrm{RC} + \mathrm{mult} \times \mathrm{AddOn})$, with $\mathrm{mult} = \min\bigl(1,\ 0.05 + 0.95\exp((V - C)/(2 \times 0.95 \times \mathrm{AddOn}))\bigr)$ and $\mathrm{RC} = \max(V - C, 0)$. Take an add-on of 10. What is it at $V - C = 5$? At $-20$? What is the floor? **Answer to Question 13.34.** It is 21 at $V - C = 5$, and about 5.34 at $-20$. The floor on the multiplier is 5%, which gives 0.7 when RC is zero. At $V - C = 5$, $\mathrm{RC} = 5$, and the multiplier caps at 1: $1.4 \times (5 + 10) = 21$. At $-20$, $\mathrm{RC} = 0$, and the multiplier is $0.05 + 0.95\,e^{-20/19} = 0.382$. So $1.4 \times 3.82 = 5.34$. Being out of the money or over-collateralised reduces future exposure as well, as it would take a move to get back to positive. As $V - C \to -\infty$, the multiplier is $0.05$, and $1.4 \times 0.5 = 0.7$. The floor exists so that excess collateral never takes exposure to zero. **Classic wrong answer.** Treating RC as the whole exposure. Exposure is also the potential change over the rest of the trade, and that’s what the add-on stands for. **Follow-up.** *Why the 1.4?* It is the supervisory scaling factor, $\alpha$, a fixed number in the rule, and it covers things like wrong-way risk and model error. **Question 13.35 ★★★ Risk, QR • bank, hedge fund • experienced.** A portfolio’s P&L is $2x_1 - x_2$ in millions, where the factors have mean 0, variances 4 and 1, and covariance 1. What is the most likely scenario that loses exactly 10? How many standard deviations away is it? Why does the second factor move the way it does? **Answer to Question 13.35.** The most likely such scenario is $x_1 = -5.38$, $x_2 = -0.77$, and it is about 2.77 standard deviations from the origin. Under normal factors, maximise the density given the loss, which means minimising $x^\top\Sigma^{-1}x$ given $b^\top x = -10$. With a Lagrangian, $$ x^* = -\frac{10\,\Sigma b}{b^\top\Sigma b}, $$ with $b = (2, -1)$, $\Sigma b = (7, 1)$ and $b^\top\Sigma b = 13$. So $x^* = -(10/13)(7, 1)$. The distance is $10/\sqrt{13} = 2.77$. The P&L has a standard deviation of 3.61, so 10 is the same 2.77 of them. **The second factor.** You are short it, so you’d gain from a fall, and a fall in $x_2$ gives you $+0.77$. It falls anyway, since it is positively correlated with the first factor, and a large drop in that one drags it along. The scenario is what the correlation says is likely, and not what hurts most per unit. Compare a stress on the first factor alone, $(-5, 0)$, which also loses 10. Its distance is 2.89, so it is less likely than $x^*$. **Classic wrong answer.** Stressing one factor at a time and calling the worst a reverse stress. The point is to find the plausible combination, and it is often not the one you’d have guessed. **Question 13.36 ★★★ Risk, QR • any • junior, experienced.** A risk manager writes a stressed correlation matrix for three assets: correlations of 0.9 for pairs (1, 2) and (2, 3), and $-0.5$ for pair (1, 3). What is wrong with it? What range can the (1, 3) correlation take? What can go wrong if you use the matrix anyway, and how do you repair it? **Answer to Question 13.36.** That matrix isn’t a correlation matrix. Its smallest eigenvalue is $-0.547$. With the two pairs at 0.9, the (1, 3) correlation must lie in $[0.62, 1]$. The range comes from a determinant condition, or from partial correlation: $$ \rho_{13} \in \rho_{12}\rho_{23} \pm \sqrt{(1 - \rho_{12}^2)(1 - \rho_{23}^2)} = 0.81 \pm 0.19. $$ If you use it anyway, there’s a portfolio with negative variance: weight along the eigenvector with the negative eigenvalue, and the formula takes a square root of a negative number, or returns a risk of zero, so an optimiser will pile into it. To repair it, change as little as possible: set negative eigenvalues to a small positive and rescale to a unit diagonal, or solve a nearest-correlation problem. Then re-read the result, since the repair may have moved the numbers you meant to stress. **Classic wrong answer.** Thinking any number between $-1$ and 1 will do. Pairwise bounds aren’t enough, as correlations have to be mutually consistent. Stress tests that set a few pairs by hand break this all the time. **Question 13.37 ★★★ Risk, QR • bank • experienced.** You backtest a ten-day 99% VaR on overlapping ten-day windows over 250 days. For independent daily normal returns, what are the exception count’s mean and variance, against a binomial with independent days? What does this do to the usual test, and what do you do about it? **Answer to Question 13.37.** The mean stays near 2.5, but the variance is about 10.6 against 2.475. So it’s about 4.3 times too large, and the count behaves as if you had only 58 independent observations. The windows share nine of ten days, so an exception today tends to stay one for several days. The mean of a count is unaffected, as each window has a 1% marginal chance. But exceptions come in clumps of up to about ten, which raises the variance. **What it does.** Kupiec assumes a binomial. With the real, wider distribution, you reject a correct model far too often at the usual cut-off. And the independence test fails by construction. **What to do.** Backtest at one day, which the rules do: one-day VaR for backtesting, ten-day for capital. Or use non-overlapping ten-day windows, though that leaves only 25 observations a year. Or simulate the null distribution of the overlapping count, and use its cut-off. **Classic wrong answer.** Counting overlapping windows as 250 observations of ten-day risk. You have only 25 independent ones. Any statement about tail frequency from that has to be weak. **Question 13.38 ★★★★ Risk, QR • bank • experienced.** You are validating a desk’s 99% one-day VaR. In 500 days it had 9 exceptions. The transitions are: 485 no-no, 5 no-yes, 5 yes-no, 4 yes-yes. Run both tests. Then go through what you would look at next, and what you would tell the desk and the regulator. **Answer to Question 13.38.** The Kupiec test passes, with a ratio of 2.61 and a $p$-value of 10.6%. The independence test fails badly, with a ratio of 21.9 and a $p$-value of about 0.0003%. So the count is acceptable and the timing is not. The model is too slow. **Numbers.** With $\hat p = 9/500 = 1.8\%$, Kupiec gives 2.61 against a cut-off of 3.84, and the exact tail $P(X \ge 9) = 6.7\%$. The transitions give $\pi_{01} = 5/490 = 1.02\%$ and $\pi_{11} = 4/9 = 44.4\%$, a ratio of 21.9. An exception makes the next one more than 40 times likelier. **What I would look at, in order.** 1. **Which P&L?** Backtest on hypothetical P&L with positions frozen, without fees and intraday trading. If exceptions come only on actual P&L, the issue may be new trades and not the model. 2. **When?** Line up the clusters with market events. If they fit volatility jumps, the window is too slow: test a volatility-scaled historical simulation or an exponential weighting. 3. **Which positions?** Attribute each exception by factor. A single desk or a factor missing from the model (a basis, a vol skew, a new product) points to a coverage issue, not a slow model. The fix differs. 4. **Data.** Stale proxies and missing days smooth volatility and create breaches when the actual prices catch up. 5. **Mis-sized positions.** Check for gamma and jump risk near expiry, since a linear approximation can look right until a large move. 6. **Comparison.** Run the candidate model on the same history: the new one has to fix the clustering without just inflating the level. **What I would say.** To the desk: the level is fine, but the model is slow, and limits based on it are too loose just after a shock. To the regulator: exactly that, with the evidence, the remediation plan and date, and what we are doing in the meantime, such as a temporary add-on. On a 250-day clock, a cluster can also move the bank into the yellow zone, so raise it early rather than wait for the multiplier to change. **Classic wrong answer.** Reporting the 2.61 and stopping there. On a count alone, this model passes. Risk models fail in the timing of the exceptions more often than in the total number. **Follow-up.** *Why not raise VaR by a flat factor until the count is right?* That raises the level all the time, costing capital, and leaves the clustering in place. The exceptions still come together, just slightly less often. **Question 13.39 ★★★★ Risk, Trader, QR • hedge fund, bank • experienced.** An options book is delta-hedged, with gamma $-20{,}000$ shares per dollar, vega $-\$40{,}000$ per volatility point, and theta $+\$22{,}500$ a day. The stock’s daily move is normal with a standard deviation of $1.50. Implied volatility moves by a normal of 1 point a day, with correlation $-0.7$ to the stock. Estimate the 99% VaR and shortfall: delta-normal, gamma alone, gamma and vega together. What happens with Student-$t$ moves of 4 degrees of freedom? What about a stress of $-\$10$ and $+5$ points? What limit do you set? **Answer to Question 13.39.** Delta-normal says 0. Gamma alone gives a 99% VaR of about $127,000. Gamma and vega together give about $174,000, with shortfall about $223,000. Under Student-$t$ moves, the VaR is about $252,000, and shortfall about $525,000. The stress loses about $1.18 million. I would set the limit on the stress, not the VaR. **The P&L.** With $x$ the stock’s move and $\Delta v$ the volatility change, $$ \mathrm{P\&L} = 22{,}500 - 10{,}000\,x^2 - 40{,}000\,\Delta v. $$ The theta is fair: $\tfrac12 \times 20{,}000 \times 1.5^2 = 22{,}500$, so the expected P&L is zero. **Gamma alone.** $x^2 = 2.25 z^2$, whose 99% point is $2.25 \times 6.635$. So the loss is $149{,}285 - 22{,}500 = \$126{,}800$. **Vega alone.** $40{,}000 \times 2.326 = \$93{,}000$. The two added would be $220,000. They are not perfectly linked, and vega hurts in only one tail, so the combined figure is $174,000 by simulation of $\Delta v = -0.7 z + \sqrt{0.51}\,\varepsilon$. A falling stock brings rising vol, so down moves are hit twice and up moves partly offset. **Fat tails.** Scale both shocks to a unit-variance $t_4$ with one shared scale. VaR rises by 45%, and shortfall more than doubles. Gamma is a square, so it makes fat tails much worse. **The stress.** $-\$10$ and $+5$ points: $10{,}000 \times 100 - 22{,}500 + 200{,}000 = \$1{,}177{,}500$. That is nearly seven times the normal VaR. A move of $-\$10$ is 6.7 standard deviations. On a normal, that never happens. In a single-stock book, an earnings surprise or takeover can do it in one jump. **The limit.** Set it on scenarios: a grid of stock moves and vol moves, with a maximum loss in the grid, such as $\pm 10\%$ with $\pm 5$ to $10$ points. VaR doesn’t see the gap, as it’s a quadratic that grows with the square of the move. Then keep VaR as a daily measure, and add limits on gamma and vega by expiry. **Classic wrong answer.** Reporting delta-normal for a delta-hedged book. The number that says “no risk” is the first one every risk manager should distrust. Quote what breaks it. **Follow-up.** *The stock jumps $+\$10$ and vol falls 3 points. What then?* The loss is $1{,}000{,}000 - 22{,}500 - 120{,}000 = \$857{,}500$. An up gap still costs a lot, so the book is hurt both ways. **Question 13.40 ★★★★ Risk, QR, QD • bank • experienced.** A colleague’s binomial tree for a European call gives these, with $S = K = 100$, $r = 5\%$, $\sigma = 20\%$ and $T = 1$. The closed form is 10.4506. At $N = 50$, 10.4107. At 51, 10.4850. At 100, 10.4306. At 101, 10.4680. At 200, 10.4406. At 201, 10.4593. At a strike of 180, $N = 50$ gives 0.0249 and $N = 200$ gives 0.0276, and the closed form is 0.0286. At $\sigma = 0.1\%$ and $N = 500$, it returns $2.7 \times 10^{159}$. What do you conclude? What is the fix for each? What else do you test before you sign? **Answer to Question 13.40.** The model is correct at the money and in the normal range, is first-order accurate with an even-odd oscillation, is badly off in relative terms far out of the money, and has a real defect at low volatility. I would approve it with conditions once that defect is fixed. **Oscillation.** The errors are $-0.0399$, $+0.0344$, $-0.0200$, $+0.0174$, $-0.0100$, $+0.0087$. The sign follows parity of $N$, as the final node either sits on the strike or straddles it. The size halves as $N$ doubles, so the order is $1/N$. *Fix:* average two adjacent steps. For $N = 100, 101$ the average is 10.4493, an error of $-0.0013$, which beats $N = 1{,}000$ alone at $-0.0020$. Or smooth the last step by using the closed form there. **Far out of the money.** At a strike of 180, $N = 50$ is $12.9\%$ low in relative terms and $N = 200$ is $3.6\%$ low. The absolute error is tiny, but a desk selling wings at a low price cares about the ratio. *Fix:* a relative tolerance by strike, more steps there, or a closed form where it exists. State the range of use. **Low volatility.** The up probability is $p = (e^{r\Delta t} - d)/(u - d)$, and it needs $\sigma > r\sqrt{\Delta t}$. At $\sigma = 0.1\%$, $N = 500$, $\Delta t = 0.002$: $r\sqrt{\Delta t} = 0.00224$, well above $\sigma$. So $p > 1$, and the weights are not probabilities. Backward induction then multiplies errors at every step until the result blows up. The limit should be $S - Ke^{-rT} = 4.88$. *Fix:* check $0 \le p \le 1$ and refuse or re-grid, or use steps above 2,500 here. A trinomial or a drift-matched lattice avoids this. **What else before signing.** - Parity: call minus put against $S - Ke^{-rT}$ across the grid. At these inputs, it agrees to 12 digits. - Greeks: check bumped deltas and gammas for noise. A tree’s gamma can oscillate even when the price looks fine. - Limits: zero rate, very short and very long expiry, huge volatility, dividends, deep in the money. - Exercise: if it is for American options, compare with a separate fine grid, as there’s no closed form. Check that an American call with no dividends equals the European. - Discounting: a tree with one rate error is still close to the right answer. Test a case that singles out the discount, such as the no-arbitrage forward. - Speed and memory, and what happens on bad input. **Classic wrong answer.** Comparing one price at one step count, seeing a match to two decimals and signing. A limit case is what catches this defect, since no typical input shows it. The worst failure is a large wrong number with no warning raised.