Risk Management puzzles, solved step by step

100 scenarios define the tail, and you need 40,000 paths to halve the error. At 99%, 1% of 10,000 paths is 100 scenarios, and the VaR is read at the edge of that group. Monte Carlo error falls as one over the square root of the number of paths, so halving it means four times the paths. For a book with Rs 10 crore of daily volatility, the error is about Rs 37 lakh at 10,000 paths.

Why does a 99% VaR rest on so few scenarios?

Picture an exit poll that interviews 10,000 voters but reports only on the one voter in a hundred who picked a small party. Your estimate for that party rests on about 100 people, not 10,000. A 99% VaR is read at the boundary of the worst 1% of outcomes, so only about 100 of 10,000 paths carry information about where that boundary is. For a normal book with Rs 10 crore of daily volatility the true VaR is Rs 23.26 crore, and the estimate wobbles around it by about Rs 37 lakh from one run of 10,000 paths to the next.

Why does halving the error cost four times the paths?

Because the paths sit under a square root. To halve the error, the square root of the path count has to double, which means the path count itself must quadruple. That is why precision in Monte Carlo is expensive: going from about 1.6% error to 0.8% of the VaR costs four times the computing time. It also explains why desks use variance reductionSimulation techniques, such as antithetic paths or importance sampling, that give a more precise estimate from the same number of paths., especially importance sampling, which pushes more paths into the tail where the VaR is decided.

Say the limit plainly. More paths shrink sampling error only; they do nothing about model error. If the simulated distribution has the wrong tail, a million paths give a very precise estimate of the wrong number. And expected shortfall at 97.5% averages the worst 250 paths, so it is usually more stable than the 99% VaR read from a single boundary.

For a normal P&L they are almost identical: 2.33 against 2.34 standard deviations. On Rs 10 crore of daily volatility that is Rs 23.3 crore against Rs 23.4 crore. The 97.5% level was chosen so the switch would be neutral for a normal book. For a fat-tailed book with the same volatility, VaR is Rs 26.2 crore and expected shortfall Rs 29.1 crore, about 11% higher.

What does each number actually measure?

Think of a river's flood level. VaR is the line on the wall that the water passes one year in a hundred. Expected shortfallThe average loss on the days when losses exceed the VaR at a chosen confidence level. is how deep the water gets, on average, in the years it passes a lower line. VaR reads a single point in the tail; expected shortfall averages everything beyond its cut-off, so it responds to how far the tail stretches. For a normal distribution the 97.5% cut-off is 1.96 standard deviations, and the average of the losses beyond it is 2.338 standard deviations.

Why does the fat-tailed book pay more under expected shortfall?

Because its extreme losses are larger even though its everyday volatility is the same. Expected shortfall averages the tail, so a book that sells protection against crashes, whose losses are rare but very large, shows a much higher number under ES than under VaR. Using a Student t with three degrees of freedom and the same Rs 10 crore volatility, the 99% VaR rises only to Rs 26.2 crore, but the 97.5% expected shortfall reaches Rs 29.1 crore. That gap is exactly the risk VaR was criticised for ignoring.

Give both sides, because the question asks for advantages and disadvantages. Expected shortfall sees the tail's depth and adds up sensibly across desks, since it is subadditive. But it is harder to backtest, because you are checking an average of rare events rather than a count of breaches, and it needs more data to estimate. VaR is easy to backtest and explain, and blind beyond its own line.

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.

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.

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.

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.

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.

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.

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.

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.

The gain settles at about 0.39, and the filter becomes an exponentially weighted moving average. In steady state the prior variance P solves P squared minus P minus 4 equals zero, so P is about 2.56. The gain is P over P plus 4, about 0.39. Each new estimate is then 0.39 times the new reading plus 0.61 times the old estimate, which is exactly an EWMA.

What is a Kalman filter doing, in one picture?

Think of estimating how many people are in a stadium from a noisy turnstile count that you update every few minutes. Your last estimate is useful but the crowd keeps changing, and each new count is useful but noisy. A Kalman filter blends the old estimate and the new reading, weighting each by how much you trust it; the weight on the new reading is the Kalman gainThe share of the gap between a new measurement and the prior estimate that the filter accepts as news.. Process variance of 1 says the true value drifts by about one unit each step; measurement variance of 4 says each reading is off by about two units.

Why does the gain settle rather than fall to zero?

Because the thing being tracked keeps moving. If the true value were fixed, every reading would add certainty and the gain would shrink towards zero, like a running average; with a random walk, each step adds one unit of variance back, so certainty stops improving at a balance point. Plugging in Q of 1 and R of 4, the prior variance solves P squared minus P minus 4, giving (1 plus the square root of 17) over 2, about 2.56, and a gain of 0.390. Only the ratio of R to Q matters: a noisier measurement lowers the gain, a faster-moving state raises it.

That is the link worth saying in a risk interview. An EWMA volatility or correlation estimate, the kind many risk systems use, is a Kalman filter in steady state with a particular noise ratio, whether or not anyone calls it that. The filter's advantage is that it chooses the weight from stated assumptions about noise and drift, and adapts it early on; the limit is that those assumptions, a linear model with normal noise, have to be right for the weight to be the best one.

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.

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.

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.

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.

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.

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.

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.

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.

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.