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Risk Management puzzles, solved step by step

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All topicsCapital and leverage6Compounding and drawdowns8Correlation and diversification8Counterparty exposure and collateral7Credit risk arithmetic10Duration and rates7Liquidity and balance sheet7Logic, estimation and brainteasers7Operational loss and fraud7Options and Greeks7Probability and base rates8Statistics and estimation10VaR and expected shortfall8
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  1. 016A swap has an expected positive exposure of about Rs 5 crore across its five-year life. The counterparty's credit spread is 200 basis points. Roughly what is the credit valuation adjustment?Counterparty exposure and collateralHardCounterparty riskQuant risk

    Try it first

    Which rough estimate is closest?

    Show the worked solution

    Roughly Rs 45 lakh. The credit spread is what the market charges each year for bearing the counterparty's default risk. Rs 5 crore of expected exposure at 2% is Rs 10 lakh a year. Over five years that is Rs 50 lakh, and discounting at an assumed 4% brings it to about Rs 44.5 lakh. That is the value you give up for trading with a counterparty that can default.

    Why does a spread times an exposure give you CVA?

    Think of a friend who owes you money and a moneylender who would charge that friend 2% more a year than a safe borrower. That 2% is the market's price for the chance your friend does not pay. The credit valuation adjustmentThe reduction in the value of a derivative to reflect the chance that the counterparty defaults while owing you money. is roughly the counterparty's credit spread charged on the amount it is expected to owe you, year by year, discounted to today. The spread already folds together the chance of default and the loss if it happens.

    CVA, roughly: each year's exposure charged at the counterparty's spread0510Rs lakh per year9.62Year 19.25Year 28.89Year 38.55Year 48.22Year 55 crore x 2% = 10 lakh before discountingExposure x spreadx discount factoreach year, at 4%Sum over 5 yearsRs 44.5 lakhabout Rs 45 lakhUndiscounted: Rs 50 lakh. With default timing modelled explicitly at 60% LGD: about Rs 41 lakh.
    Rs 5 crore of expected exposure charged at a 2% credit spread costs Rs 10 lakh a year, and discounting each year at 4% gives Rs 9.62, 9.25, 8.89, 8.55 and 8.22 lakh, a rough CVA of about Rs 44.5 lakh.
    The relationship
    CVA≈s∑t=15EPEt Dt=0.02×5×∑t=151.04−t≈0.445 crore\text{CVA} \approx s \sum_{t=1}^{5} \text{EPE}_t \, D_t = 0.02 \times 5 \times \sum_{t=1}^{5} 1.04^{-t} \approx 0.445 \text{ crore}
    sthe counterparty's credit spread, 2% a year
    EPE_texpected positive exposure in year t, Rs 5 crore
    D_tthe discount factor for year t, at an assumed 4%
    What it says in wordsCharge the spread on each year's expected exposure, discount it to today, and add the years.

    What does the rough version leave out?

    Three things, each worth a sentence. First, the rough formula ignores that a counterparty that defaults in year two cannot default again in year three, so it slightly overstates the charge. Modelled explicitly, with a 60% loss given default and a constant default rate implied by the spread, the CVA comes to about Rs 41 lakh. Second, exposure on a swap is rarely flat; it usually rises and then falls as payments are made. Third, if the counterparty's credit worsens exactly when your exposure grows, the charge is larger, which is wrong-way risk.

    The discount rate here is an assumption for the illustration, and the market spread would come from the counterparty's bonds or credit default swaps where they trade. The structure is what the interviewer wants: exposure profile, times default cost, discounted and summed.

    Where candidates lose it

    The trap is answering Rs 10 lakh, one year of the spread, or treating the spread as a one-off probability of losing the whole exposure. The spread is an annual rate, and the exposure lasts five years.

    The second loss is not naming the assumptions. Say that you used a flat exposure, a 4% discount rate and the spread as a stand-in for default probability times loss given default.

    What the interviewer asks next

    • How does a collateral agreement with a zero threshold change this CVA?
    • The exposure rises from Rs 2 crore in year one to Rs 8 crore in year five, with the same average. Is CVA higher or lower?
    • What is debit valuation adjustment, and why do some people dislike booking it as profit?
  2. 020An auditor samples 60 of 3,000 trades and finds no booking errors. What error rate can you rule out at 95% confidence?Operational loss and fraudHardOperational riskBig Four risk advisory

    Try it first

    What can you conclude from zero errors in 60?

    Show the worked solution

    You can rule out error rates above about 4.9%, and nothing below. If the true error rate were p, the chance of 60 clean trades would be one minus p, to the power 60. That falls to 5% at p of about 4.87%. The rule of three gives the same answer quickly: 3 divided by 60 is 5%. With a real error rate of 1%, a clean sample happens 55% of the time.

    Why does a clean sample not prove a clean book?

    Think of tasting one spoonful from a large pot to check the salt. A good spoonful tells you the pot is not badly over-salted, but a few salty patches could easily be missed. Zero errors in a sample puts an upper bound on the error rate; it never proves the rate is zero. At a true error rate of 1%, one trade in a hundred is wrong, yet 60 random trades would all be clean 55% of the time. The sample simply is not large enough to see errors that rare.

    A clean sample of 60 rules out rates above about 5%, and nothing below0%25%50%75%100%0%2%4%6%8%10%True error rate in the populationChance of seeing zero errors in 605%1% errors: 55% chance of a clean sample4.87%: the 95% boundruled outnot ruled outRule of three3 / 60 = 5%exact: 4.87%
    The chance that a sample of 60 trades shows no errors is 55% when the true error rate is 1% and falls to 5% only at an error rate of 4.87%, so a clean sample rules out rates above about 4.9% but nothing below.
    The relationship
    (1−p)60=0.05  ⇒  p=1−0.051/60≈4.87%rule of three: p≈3n=5%(1-p)^{60} = 0.05 \;\Rightarrow\; p = 1 - 0.05^{1/60} \approx 4.87\% \qquad \text{rule of three: } p \approx \frac{3}{n} = 5\%
    pthe true error rate in the population of trades
    60the sample size
    0.05the chance you accept of being wrong, for 95% confidence
    What it says in wordsFind the error rate at which a clean sample would be a one-in-twenty event; anything higher is ruled out.

    Where does the rule of three come from, and does the 3,000 matter?

    The chance of zero errors is roughly e to the power of minus n times p, and e to the minus 3 is about 5%. So n times p equal to 3 marks the 95% bound, which gives the rule of three: divide 3 by the sample size. The population of 3,000 barely matters here because the sample is only 2% of it. Sampling without replacement tightens the bound slightly: counting exactly, the book could hold at most 144 errors, about 4.80%, instead of the 4.87% the simple formula gives.

    Then give the practical point. If the firm's tolerance for booking errors is 1%, a sample of 60 cannot confirm it; you need about 300 clean trades, 3 divided by 1%, to rule out 1%. The limit is that the sample must be random; a sample of the easiest trades to check says little about the ones that go wrong.

    Where candidates lose it

    The trap is reporting that the error rate is zero, or that it is below one in 60. Neither follows. A clean sample of 60 is quite likely even when one trade in a hundred is wrong.

    The second miss is freezing on the exact formula. Give the rule of three first, then refine it to 4.87% if asked.

    What the interviewer asks next

    • How many trades must you sample, all clean, to rule out a 0.5% error rate?
    • The sample of 60 finds one error. What upper bound can you now give?
    • Why might a random sample still understate errors in complex trades?
  3. 021A desk is short 10,000 put options on a Rs 500 stock. Each put has a delta of minus 0.3 and a gamma of 0.004 per rupee. The stock falls 10%. Compare the delta-only loss estimate with the delta-gamma estimate.Options and GreeksHardBank market riskModel validation

    Try it first

    How much larger is the delta-gamma loss than the delta-only loss?

    Show the worked solution

    Delta alone says a Rs 1.5 lakh loss; delta and gamma say Rs 2.0 lakh, a third more. The stock falls Rs 50. Each put gains 0.3 times 50, Rs 15, from delta, plus half of 0.004 times 50 squared, Rs 5, from gamma. Short 10,000 puts, the desk loses Rs 20 a put. A full revaluation in this example gives about Rs 2.06 lakh, so the gamma term captures most of the gap.

    Why does the delta estimate fall short?

    Think of driving downhill with the brakes a little weak. The first hundred metres feel manageable, but every metre adds speed, so the second hundred covers more ground than the first. A put gains value faster the further the stock falls, because its delta grows more negative as it moves towards being in the money; a short put position therefore loses at an accelerating rate. Delta is the slope at today's price, so a straight line from it misses the curve. GammaHow much an option delta changes for a one rupee move in the underlying price. measures that bend.

    The relationship
    ΔV≈δ ΔS+12 Γ (ΔS)2=(−0.3)(−50)+12(0.004)(2,500)=15+5=20\Delta V \approx \delta\,\Delta S + \tfrac{1}{2}\,\Gamma\,(\Delta S)^2 = (-0.3)(-50) + \tfrac{1}{2}(0.004)(2{,}500) = 15 + 5 = 20
    deltathe put delta, minus 0.3
    Gammathe put gamma, 0.004 per rupee
    Delta Sthe stock move, minus Rs 50
    Delta Vthe change in each put's value, Rs
    What it says in wordsEach put gains Rs 15 from the slope and Rs 5 from the bend; the desk is short, so it loses both.
    Short puts: the loss curves away from the delta line+0.5-1.0-2.0-3.00-10%-5%+5%+10%Move in the stockP&L, Rs lakhdelta only: -1.50full revaluation: -2.06At -10% (-Rs 50)Delta: -1.50Delta-gamma: -2.00Full: -2.06Rs lakhshort gamma: big moves hurtin either direction
    For a 10% fall in the stock, the straight delta line predicts a Rs 1.5 lakh loss on 10,000 short puts, but the true P&L curve bends away to about Rs 2.06 lakh, and the delta-gamma estimate of Rs 2.0 lakh lands close to it.

    Why does this matter for VaR?

    Because a VaR built on delta alone treats every option book as a straight line. For a book that is short options, the straight line understates the loss in exactly the large moves that VaR is meant to capture, and the understatement grows with the square of the move. On a 5% fall the gamma term adds Rs 1.25 a put against Rs 7.50 from delta, one sixth; on a 10% fall it adds a third; on a 20% fall it would add two thirds. A risk team would use delta-gamma at minimum and full revaluation for large scenarios.

    Say the limit of the gamma fix too. Delta-gamma is still an approximation: gamma itself changes as the stock moves, and volatility usually jumps when a stock falls 10%, which adds a vega loss for a short option book. In this example, built with an assumed 30% volatility and a strike of about Rs 463, full revaluation gives Rs 2.06 lakh, a little above the delta-gamma figure, before any volatility move.

    Where candidates lose it

    The trap is stopping at the delta answer, Rs 1.5 lakh, and treating gamma as a rounding detail. On a 10% move it adds a third to the loss, and for a short option book it always adds, never subtracts.

    The second slip is the sign. Gamma is positive for the option holder; the desk is short, so the gamma term increases its loss whichever way the stock moves.

    What the interviewer asks next

    • What would the P&L be if the stock rose 10% instead?
    • How many shares would you trade to delta hedge the book, and does that remove the gamma loss?
    • Implied volatility rises 5 points as the stock falls. What else would you need to estimate the full loss?
  4. 023You estimate a correlation of 0.5 between two assets from 36 monthly observations. Roughly what is its standard error, and is a later estimate of 0.3 from the next 36 months evidence that the relationship has changed?Statistics and estimationHardQuant riskAsset manager risk

    Try it first

    Is the drop from 0.5 to 0.3 statistically meaningful?

    Show the worked solution

    The standard error is about 0.125, and 0.3 is not evidence of a change. A rough rule is one minus the correlation squared, over the square root of n: 0.75 over 6. Using Fisher's transformation, the 95% interval for 0.5 runs from about 0.21 to 0.71, and 0.3 sits inside it. Testing the difference between two 36 month estimates gives z of about 0.97, well below 1.96.

    Why is a correlation from 36 months so imprecise?

    Think of judging whether two friends like the same films after watching three dozen films together. They might agree on 20 by chance even if their tastes barely overlap. A correlation is estimated from how often two series move together, and with 36 points random coincidences move the estimate a lot: the standard error of 0.5 is about 0.125, a quarter of the estimate itself. The rough rule is one minus the correlation squared over the square root of the sample size, 0.75 over 6.

    The relationship
    SE(ρ^)≈1−ρ2n=0.756=0.125z=tanh⁡−1(ρ^),    SE(z)=1n−3\text{SE}(\hat\rho) \approx \frac{1 - \rho^2}{\sqrt{n}} = \frac{0.75}{6} = 0.125 \qquad z = \tanh^{-1}(\hat\rho), \;\; \text{SE}(z) = \frac{1}{\sqrt{n - 3}}
    rhothe estimated correlation, 0.5
    nthe number of observations, 36
    tanh^{-1}Fisher's transformation, which makes the sampling error close to normal
    What it says in wordsThe error of a correlation shrinks only with the square root of the sample size, and Fisher's transformation gives a fair interval around it.
    Two correlations from 36 months each: the intervals overlap-0.20.00.20.40.60.81.0Correlation0.210.71First 36 months: 0.50-0.030.57Next 36 months: 0.30overlap 0.21 to 0.57Rough standard error: (1 - 0.5 squared) / 6 = 0.125. Test of the difference: z = 0.97, well under 1.96.
    The 95% interval for a correlation of 0.5 from 36 months runs from 0.21 to 0.71, and the interval for 0.3 from another 36 months runs from -0.03 to 0.57, so the two overlap heavily and the drop is within sampling noise.

    How do you test whether the relationship changed?

    Transform both estimates with Fisher's z, which turns 0.5 into 0.549 and 0.3 into 0.310. The difference, 0.240, divided by its standard error of the square root of 2 over 33, about 0.246, gives a z of 0.97: well short of the 1.96 needed for 95% confidence. A gap that size would appear by chance about 33% of the time even if nothing had changed. The honest statement is that the data cannot tell 0.5 from 0.3.

    Then say why a risk manager cares. Hedge ratios, portfolio volatility and diversification benefits are all built on correlations like this one, and treating 0.5 as exact makes a model look more certain than its data allow. The practical responses are longer or higher-frequency data, shrinking noisy estimates towards a sensible prior, and stress testing the portfolio at correlations across the whole interval rather than at the point estimate.

    Where candidates lose it

    The trap is reading the drop from 0.5 to 0.3 as a regime change. It is a 40% fall in the number but well inside the noise that three years of monthly data carry.

    The second miss is quoting a standard error without saying it is rough. The simple formula is an approximation; Fisher's transformation is what you use for an interval or a test.

    What the interviewer asks next

    • How many monthly observations would you need to bring the standard error down to 0.05?
    • Would weekly data over the same three years help, and what could go wrong?
    • How would you stress test a portfolio whose risk depends on this correlation?
  5. 027A stock has 2% daily volatility and no drift. What is the probability that it touches a level 10% below today's price at some point in the next 20 trading days, and how does that compare with the probability of ending below that level?Compounding and drawdownsHardBank market riskQuant risk

    Try it first

    Compared with the chance of finishing below the level, the chance of touching it at some point is:

    Show the worked solution

    About 24% to touch, against about 12% to finish below: roughly double. Twenty days at 2% a day is 8.94% of volatility. A 10% fall is ln 0.9, or 10.5% in log terms, which is 1.18 standard deviations, so the chance of ending below is 11.9%. The reflection principle doubles it for a touch: 23.9%.

    Why is touching twice as likely as finishing below?

    Picture a drunk walker on a pavement with a kerb one step to the left. By the end of the evening he may be well to the right, yet at some point he almost certainly stepped off the kerb. Paths wander. Once a driftless path has touched the level, it is equally likely to finish above it or below it, so for every path that finishes below there is a twin that touched and came back. This is the reflection principleFor a random walk with no drift, the path after it first hits a level is equally likely to be its mirror image about that level.: flip the path after the first touch and you swap an above-finisher for a below-finisher.

    Every path that ends below the line has a twin that touched it and came backtoday-10%stop level05101520Trading daynever touchestouch, day 9ends -1.8%mirror ends -19.2%Ends below the line11.9%Touches at any time23.9%twice the first
    A path that touches the 10% stop level on day 9 and recovers has a mirror twin, drawn dashed, that finishes below the line; because the two are equally likely, the chance of touching is twice the chance of finishing below, 23.9% against 11.9%.

    How do you get the numbers in the room?

    Scale volatility by the square root of time: 2% times the square root of 20 is 8.94%. Work in log returns, because a 10% fall is a log move of 10.5%, not 10%. That is 1.18 standard deviations, and the normal table gives 11.9% below it. Double for the touch. If you skip the log step and use 10% flat you get 26.4%, close enough to earn credit if you say it is an approximation.

    The relationship
    P(touch)=2 Φ ⁣(ln⁡0.90.0220)=2 Φ(−1.18)≈2×0.119=0.239P(\text{touch}) = 2\,\Phi\!\left(\frac{\ln 0.9}{0.02\sqrt{20}}\right) = 2\,\Phi(-1.18) \approx 2 \times 0.119 = 0.239
    \Phithe standard normal cumulative probability
    0.02\sqrt{20}20-day volatility, 8.94%
    \ln 0.9the log return of a 10% fall, minus 0.105
    What it says in wordsFind the chance of ending beyond the level, then double it because every finisher beyond has a twin that touched and came back.

    Where does the doubling rule stop being exact?

    Two places. With drift the mirror twins are no longer equally likely, and a positive drift pulls the touch probability below double. And if the stop only triggers on daily closes, a path can dip through during the day and recover by the close, so fewer touches count; a standard correction for daily monitoring gives about 19%. The practical reading is that a stop-loss level is hit far more often than a finishing-price distribution suggests, which is why stops set from end-of-period VaR fire more than the desk expects.

    Where candidates lose it

    Most candidates compute the finishing probability, 11.9%, and offer that as the answer to the touching question. The interviewer asked about touching precisely to see whether you know the two differ, and by how much.

    The other loss is knowing the doubling rule but not why. Draw the mirror path out loud; it takes ten seconds and proves you are reasoning, not reciting.

    What the interviewer asks next

    • What is the probability of touching a level 10% above instead?
    • How does a positive drift change the answer?
    • A barrier option knocks out at this level. Why is it cheaper than the vanilla option by more than the finishing probability suggests?
  6. 040A portfolio holds Rs 60 crore of asset A with 20% volatility and Rs 40 crore of asset B with 30% volatility, and their correlation is 0.5. Compute the portfolio's volatility and each asset's contribution to it.Correlation and diversificationHardBank market riskQuant risk

    Try it first

    Asset A is 60% of the capital. What share of the portfolio's risk does it contribute?

    Show the worked solution

    Portfolio volatility is about Rs 20.78 crore, 20.8% of the Rs 100 crore, and each asset contributes exactly half of it. A carries 60 x 20% = Rs 12 crore of standalone volatility and B carries 40 x 30% = Rs 12 crore. The variance is 144 + 144 + 2 x 0.5 x 12 x 12 = 432, so volatility is 20.78. With equal standalone risk, each contributes Rs 10.39 crore: 60% of the capital, 50% of the risk.

    Why convert to rupee volatility first?

    A household spends 60% of its budget on rent and 40% on a car loan. If the rent is fixed and the car loan's rate floats, most of the budget's uncertainty comes from the smaller item. Risk depends on size times volatility, so the first step is to express each position's volatility in rupees: that puts the two assets on one scale. Here the smaller, more volatile position carries exactly as much standalone risk as the larger, calmer one: Rs 12 crore each.

    The relationship
    σP2=122+122+2(0.5)(12)(12)=432,σP=20.78\sigma_P^2 = 12^2 + 12^2 + 2(0.5)(12)(12) = 432, \qquad \sigma_P = 20.78
    12rupee volatility of each position, Rs crore
    0.5correlation between A and B
    What it says in wordsPortfolio variance is each position's variance plus twice the shared part.

    How do you split the portfolio's risk between the two assets?

    Use the Euler allocationSplitting total risk so each position gets its own variance plus its share of every covariance, divided by total volatility; the pieces add up exactly to the total.: each asset's contribution is its own variance plus its covariance with the other, divided by total volatility. For A that is (144 + 0.5 x 12 x 12) over 20.78, which is 216 over 20.78, Rs 10.39 crore. B's contribution is the same Rs 10.39 crore, and the two add back to exactly Rs 20.78 crore, so the split is 50/50 while the capital split is 60/40.

    Where the rupees sit is not where the risk sitsA 60%Rs 60 croreB 40%Rs 40 croreCapitalA 50%Rs 10.39 croreB 50%Rs 10.39 croreRisk contributionRupee volatilityA: 60 x 20% = 12B: 40 x 30% = 12Standalone sum: 24Portfolio: 20.78Diversification: 3.22Rs crore a year
    Asset A holds 60% of the capital but contributes only 50% of the risk, Rs 10.39 crore of the Rs 20.78 crore portfolio volatility, because asset B's higher volatility makes its Rs 40 crore as risky as A's Rs 60 crore.

    Two points to add. First, the diversification benefitThe amount by which portfolio risk falls short of the sum of the standalone risks, because the positions do not move in perfect lockstep.: standalone risks sum to Rs 24 crore, the portfolio carries Rs 20.78 crore, and Rs 3.22 crore disappears because correlation is below 1. Euler allocation spreads that benefit across both assets rather than handing it to whoever joined last. Second, the limit: contributions depend on a correlation estimate, and correlations rise in stress. At a correlation of 1 there is no benefit left to share.

    Where candidates lose it

    The common error is assuming risk shares match capital shares, 60/40, because A is the bigger position. The whole point of the puzzle is that the smaller, more volatile asset can carry as much risk as the larger one.

    The second error is adding standalone volatilities, Rs 24 crore, as the portfolio risk. That ignores correlation and overstates risk by more than Rs 3 crore; always go through variance.

    What the interviewer asks next

    • How much of B would you sell to make the risk split 60/40?
    • What happens to both contributions if the correlation rises to 0.9?
    • Why do risk-parity funds size positions by risk contribution instead of capital?
  7. 041Bank A holds one Rs 100 crore loan. Bank B holds one hundred loans of Rs 1 crore each. Every loan has a 2% default probability, total loss on default and independent defaults. Both expect to lose Rs 2 crore. What is each bank's chance of losing more than Rs 10 crore in a year?Credit risk arithmeticHardBank credit riskRisk GCC

    Try it first

    Bank A has a 2% chance of losing more than Rs 10 crore. Bank B's chance is:

    Show the worked solution

    Bank A: 2%. Bank B: about 5.6 in a million, roughly 1 in 177,116. Bank A loses Rs 100 crore whenever its single loan defaults. Bank B loses more than Rs 10 crore only if 11 or more of its 100 loans default when 2 are expected, which independent defaults almost never produce. Expected loss is Rs 2 crore for both; the tail differs by a factor of several thousand.

    Why does splitting the same money into small loans change the tail?

    A shopkeeper who sells to one wholesale buyer is either paid in full or not at all; one who sells to a hundred households loses a few payments every month but never the whole book. Granularity leaves expected loss unchanged but pulls the loss distribution in around its average, because independent small defaults rarely pile up. Bank A's outcome is all or nothing: 98% nothing, 2% the full Rs 100 crore. Bank B's losses cluster between Rs 0 and 6 crore.

    Same expected loss, Rs 2 crore; completely different tailsBank A: one Rs 100 crore loanno loss: 98%2%lose 100Rs 10 crore050100P(loss above 10) = 2%Bank B: one hundred Rs 1 crore loansabove 100246810clusters around 2P(loss above 10) = 5.6 in a millionLoss in the year, Rs crore. Bar heights are probabilities on the same scale in both panels.
    Both banks expect to lose Rs 2 crore, but bank A faces a 2% chance of losing its whole Rs 100 crore while bank B's losses cluster around Rs 2 crore and exceed Rs 10 crore only about 5.6 times in a million.

    How do you estimate bank B's number in the room?

    The defaults are a binomial with mean 2 and standard deviation about 1.40. Eleven defaults is more than six standard deviations above the mean, so the answer is tiny; the exact binomial tail is 5.6 in a million. Compare the standard deviations of loss directly: bank A's is Rs 14.0 crore, bank B's is Rs 1.4 crore, ten times smaller, because splitting one exposure into a hundred independent ones divides the spread by the square root of 100.

    The relationship
    σA=1000.02×0.98=14.0,σB=1×100×0.02×0.98=1.4\sigma_A = 100\sqrt{0.02 \times 0.98} = 14.0, \qquad \sigma_B = 1 \times \sqrt{100 \times 0.02 \times 0.98} = 1.4
    \sigmastandard deviation of the annual loss, Rs crore
    0.02default probability of each loan
    What it says in wordsSplitting one exposure into n independent pieces cuts the spread of losses by the square root of n.

    Now the caveat that matters in practice. The 99% loss quantile is Rs 100 crore for A and Rs 6 crore for B, which is why concentration riskExtra risk from large single exposures or from many exposures that fail together, which average-based measures do not show. gets its own capital charge. But granularity only helps if defaults are independent. A hundred small loans to one industry in one city will default together in a local downturn, and bank B's tail then fattens towards bank A's. Diversification across names is not diversification across causes.

    Where candidates lose it

    The first trap is saying both banks are equally risky because both expect to lose Rs 2 crore. Expected loss is exactly the number that cannot tell these two banks apart.

    The second is assuming more loans means more risk because more loans can default. The count of defaults rises, but each is small, and the chance of many at once collapses. Close by naming the independence assumption as the thing that could undo it.

    What the interviewer asks next

    • What is each bank's 99% VaR?
    • If bank B's loans all go to one sector with default correlation, what happens to its tail?
    • How would you set a single-name concentration limit using this logic?
  8. 043A bank holds Rs 5,000 crore of bonds at market value, of which Rs 3,200 crore are pledged in repo and at the clearing house. Bond prices fall 10%, and the pledged value must be restored to Rs 3,200 crore. How much free collateral is left?Liquidity and balance sheetHardTreasury and ALMBank credit risk

    Try it first

    Free collateral started at Rs 1,800 crore. By roughly how much does it fall?

    Show the worked solution

    Rs 1,300 crore, down 27.8% from Rs 1,800 crore. The holding falls 10% to Rs 4,500 crore. The pledged bonds are now worth Rs 2,880 crore, so Rs 320 crore of free bonds must be posted to restore Rs 3,200 crore. Free bonds, themselves down to Rs 1,620 crore, lose that Rs 320 crore too, leaving Rs 1,300 crore.

    Why does free collateral fall faster than prices?

    Someone who has pawned most of their jewellery owes the lender a fixed amount of value. When gold prices fall, the pawnbroker asks for more pieces to keep the loan covered, and those pieces come from the small pile still at home. The pledged claim is fixed in rupees, so the entire fall in value across all Rs 5,000 crore lands on the free slice. That makes free collateral a leveraged position on bond prices, with leverage of 5,000 over 1,800, about 2.78 times.

    Prices fall 10%; free collateral falls 28%pledged 3,200free 1,8005,000Beforepledged 2,880+320 top-upfree 1,3004,500After a 10% fallFree collateral1,800 to 1,300down 27.8%= 10% x 2.78(5,000 / 1,800)All free bonds gone aftera 36% fallRs crore, market value
    After a 10% price fall, Rs 320 crore of free bonds must be posted to keep the pledged value at Rs 3,200 crore, so free collateral drops from Rs 1,800 crore to Rs 1,300 crore, a 27.8% fall.
    The relationship
    free1=total (1−d)−pledged=5,000×0.9−3,200=1,300\text{free}_1 = \text{total}\,(1 - d) - \text{pledged} = 5{,}000 \times 0.9 - 3{,}200 = 1{,}300
    dthe fall in bond prices, 10%
    pledgedthe market value that must stay posted, Rs 3,200 crore
    What it says in wordsFree collateral is whatever is left of the whole holding after the fixed pledged value is carved out.

    At what point does the bank run out, and what would you add?

    Solve for zero: 5,000 x (1 minus d) equals 3,200 when d is 36%. A 36% price fall exhausts every free bond, and any further fall produces a margin call the bank cannot meet from securities. Real life is worse than this sum: lenders raise haircutsThe discount a lender applies to collateral value; a 5% haircut means Rs 100 of bonds secures only Rs 95 of borrowing. in a sell-off, which increases the pledged amount needed at exactly the moment prices fall, and the same stress usually brings deposit outflows that draw on the same free bonds.

    That is why a liquidity risk team tracks the encumbrance ratio, here 64% before the fall and 71% after, and stresses free collateral against price falls and haircut rises together. A buffer reported at Rs 1,800 crore is really a buffer of Rs 1,800 crore only on the day prices stand still.

    Where candidates lose it

    The instinct is to take 10% off the free Rs 1,800 crore and answer Rs 1,620 crore. That misses the top-up: the pledged bonds also lost value and must be replenished from the free pile.

    The overcorrection is to subtract the top-up but forget that the free bonds fell too. Do it in one line from the total: 4,500 less 3,200 is 1,300, and the leverage of 5,000 over 1,800 explains the 27.8% in one sentence.

    What the interviewer asks next

    • The repo lender also raises its haircut from 2% to 5%. How much free collateral is left?
    • Which bonds would you pledge first, and which would you keep free?
    • How would a 10% fall combined with a Rs 400 crore deposit outflow change your answer?
  9. 045A bank records only operational losses above Rs 10 lakh. True losses follow an exponential distribution with a mean of Rs 20 lakh. What is the average recorded loss, and what goes wrong if you fit a severity model to the records as if they were complete?Operational loss and fraudHardOperational riskModel validation

    Try it first

    What is the average of the recorded losses?

    Show the worked solution

    Recorded losses average Rs 30 lakh, 50% above the true Rs 20 lakh. Because the exponential is memoryless, losses above Rs 10 lakh exceed it by an average of Rs 20 lakh, so they average Rs 30 lakh. Fitting the records as if complete overstates the size of a typical loss by half, pushes the 99th percentile from Rs 92 to 138 lakh, and misses the 39% of losses below the threshold.

    Why does a threshold raise the recorded average?

    A school that only records exam scores above 60 will report a class average far above the real one, because the weak scores never enter the register. A reporting threshold removes the small losses from the data, so any average taken from what is left overstates the typical loss. For an exponential, the shift is exact: it is memorylessFor an exponential distribution, knowing a value already exceeds some level tells you nothing new about how much further it goes; the excess has the same distribution as the original., so the excess above Rs 10 lakh again averages Rs 20 lakh, and recorded losses average Rs 30 lakh.

    A reporting threshold hides the small losses and inflates the averagereporting threshold, Rs 10 lakhneverrecorded39.3%true mean 20recorded mean 30020406080100Size of a single loss, Rs lakhFit the records as if complete:mean 30, not 20 (+50%)99th percentile 138, not 92and 39% of events missing
    With a Rs 10 lakh reporting threshold, the 39.3% of losses below it are never recorded, and the recorded losses average Rs 30 lakh against a true mean of Rs 20 lakh.
    The relationship
    E[L∣L>u]=u+μ=10+20=30,P(L>u)=e−u/μ=e−0.5=0.607E[L \mid L > u] = u + \mu = 10 + 20 = 30, \qquad P(L > u) = e^{-u/\mu} = e^{-0.5} = 0.607
    uthe reporting threshold, Rs 10 lakh
    \muthe true mean loss, Rs 20 lakh
    What it says in wordsAbove the threshold the average excess is unchanged, so the recorded mean is the threshold plus the true mean, and about 61% of losses are recorded.

    What exactly goes wrong in the fitted model?

    Two errors that pull in opposite directions. Severity is overstated: fit an exponential to the records and you get a mean of 30, so every quantile is 50% too high, with the 99th percentile of a single loss at Rs 138 lakh instead of Rs 92 lakh. Frequency is understated: only 60.7% of events are recorded, so the true count is 1.65 times the recorded one. A model that fits both naively gets the size of losses and the number of losses wrong at once, and the errors do not cancel in the tail.

    The fix is to fit a truncated distribution: treat the records as losses known to exceed Rs 10 lakh and estimate the parameters of the whole curve from that conditional shape. For an exponential that means fitting the excesses over 10, which recovers the mean of 20. The limit is that heavier-tailed distributions are not memoryless, the truncated fit becomes unstable when the threshold is high relative to the data, and the missing small losses still matter for frequency.

    Where candidates lose it

    The common wrong answer is Rs 20 lakh, assuming the threshold only removes data without changing the average. The next most common is Rs 15 lakh, from averaging the threshold and the mean.

    The deeper miss is answering the number but not the modelling consequence. The interviewer wants to hear that severity is inflated, frequency is understated, and that the fix is a truncated fit, not simply adding the small losses back by guesswork.

    What the interviewer asks next

    • What fraction of the total rupee value of losses falls below the threshold?
    • If true losses were lognormal instead, would the recorded mean rise by more or less?
    • How would you combine internal data with a threshold and external loss data?
  10. 046A trader who has sold a call hedges it by buying the stock whenever it trades above the strike and selling whenever it falls below. The hedge looks costless: you only hold the stock when the option is in the money. Why does it lose money over time?Options and GreeksHardQuant riskBank market risk

    Try it first

    If the trader checks the price more often, what happens to the expected cost of the hedge?

    Show the worked solution

    Because you always buy a little above the strike and sell a little below it, and those gaps add up to the option's time value. Prices do not stop at the strike; by the time you trade they have moved through it. Watching more closely shrinks each gap but multiplies the crossings, so the cost never vanishes. Here, with a Rs 100 strike, 1.5 a day of volatility and 40 days, the average cost is about Rs 3.78 a share at any monitoring frequency.

    Where does the money leak out?

    Think of a thermostat set to switch the heater on at 20 degrees and off at 20 degrees. It cannot switch at exactly 20; it reacts at 20.3 on the way up and 19.7 on the way down, and it keeps flicking all evening. A stop-loss hedge buys when the price is already above the strike and sells when it is already below, so every round trip loses the gap between the two. On the path in the figure the trader traded 5 times and the gaps added to Rs 3.86 a share.

    Every buy lands above the strike and every sell below itK 10095105010203040Trading day, daily closesbuy at 101.7sell at 99.5This path5 trades, cost 3.86Average over 2,000 pathsdaily: 3.844 x a day: 3.7916 x a day: 3.79option time value 3.78
    Each buy in the stop-loss hedge happens above the Rs 100 strike and each sell below it, so this path's 5 trades lose Rs 3.86 a share; averaged over many paths the loss is about Rs 3.78, the option's time value, whether the price is checked daily or sixteen times a day.

    Why can faster monitoring not fix it?

    Because of how random paths behave near a level. Check four times as often and each gap roughly halves, but the path crosses the strike roughly twice as often, so the total barely changes. In a simulation of 2,000 paths, the average cost is 3.84 with daily checks (3.9 crossings), 3.79 at four a day (7.9 crossings) and 3.79 at sixteen a day (16.0 crossings). Checking continuously would mean infinitely many crossings of zero size each, and the cost still would not go away.

    The relationship
    E[crossing costs]=E[(ST−K)+]−(S0−K)+≈σT2π=1.5402.507=3.78E[\text{crossing costs}] = E[(S_T - K)^+] - (S_0 - K)^+ \approx \frac{\sigma\sqrt{T}}{\sqrt{2\pi}} = \frac{1.5\sqrt{40}}{2.507} = 3.78
    \sigma\sqrt{T}price volatility over the option's life, 1.5 a day for 40 days, about 9.5
    (S_T - K)^+the call's payoff at expiry
    What it says in wordsThe expected leakage from crossing the strike equals the part of the option's value that comes from time and volatility.

    That identity is the real lesson. The stop-loss hedge only replicates the option's intrinsic value; the crossing costs are the time valueThe part of an option price above what it would pay if exercised now, which reflects the chance of favourable moves before expiry. the trader collected as premium and is now paying back. Higher volatility means more and bigger crossings, which is why volatile underlyings command bigger premiums. Proper delta hedging spreads the same cost smoothly as gamma losses rather than lumpy crossing losses; neither makes the premium free money.

    Where candidates lose it

    Candidates accept the premise that the hedge is costless and look for a trading cost or bid-offer spread to explain the loss. Transaction costs make it worse, but the loss is there even with none: it is built into how prices cross a level.

    The second trap is saying monitor more often and the problem goes away. It does not, and the reason, crossings multiply as gaps shrink, is what separates a memorised answer from an understood one.

    What the interviewer asks next

    • How does delta hedging spread this same cost differently?
    • What happens to the stop-loss hedge's cost if volatility doubles?
    • The stock starts well above the strike. What does the hedge cost now, and why?
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