Risk Management puzzles, solved step by step

Puzzles: 100
Traced to a firm: 17
Topics: 13
Hard: 30

Showing 21–30 of 100

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
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How much larger is the delta-gamma loss than the delta-only loss?
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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
$\Delta V \approx \delta\,\Delta S + \tfrac{1}{2}\,\Gamma\,(\Delta S)^2 = (-0.3)(-50) + \tfrac{1}{2}(0.004)(2{,}500) = 15 + 5 = 20$
delta the put delta, minus 0.3
Gamma the put gamma, 0.004 per rupee
Delta S the stock move, minus Rs 50
Delta V the 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.
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?
022An institutional investor asks for an 8% expected annual return with 10% volatility. Assuming returns are normal, what is the chance of a losing year, and what volatility would keep that chance below 10%?Probability and base ratesCoreMSCIAnonymous interview candidate in · 2013
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Roughly how often does this portfolio lose money in a year?
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About 21%, and volatility would need to fall to about 6.2%. A loss means a return below zero, which is 8 points, or 0.8 standard deviations, under the mean. About 21.2% of a normal distribution lies below that, roughly one year in five. For a 10% chance, zero must sit 1.28 standard deviations below the mean, so volatility must be 8 divided by 1.28, about 6.2%.
How do a return target and a volatility target fix the chance of loss?
Think of a commute that takes 40 minutes on average but varies from day to day. Whether you are ever late for a 50 minute deadline depends on how much it varies, not only on the average. The chance of a losing year depends on how many standard deviations the expected return sits above zero: here 8 divided by 10, which is 0.8. Look up 0.8 in the normal table and about 21.2% of years fall below zero. The investor who hears 8% and thinks losses are rare is wrong one year in five.
With an 8% expected return and 10% volatility, 21.2% of the return distribution falls below zero, while cutting volatility to 6.2% narrows the curve until exactly 10% of years show a loss.
The relationship
$P(R < 0) = N\!\left(-\frac{\mu}{\sigma}\right) = N(-0.8) \approx 21.2\% \qquad \sigma_{\max} = \frac{\mu}{1.2816} \approx 6.2\%$
mu the expected annual return, 8%
sigma the annual volatility
N the standard normal cumulative distribution
1.2816 the number of standard deviations that leaves 10% in the lower tail
What it says in wordsDivide the expected return by the volatility, and the normal table tells you how often returns fall below zero.
What would you actually set as targets, and what is wrong with this model?
Set the targets as a pair, and state the trade-off. If the investor cannot tolerate losing more than one year in ten, then either volatility must come down to about 6.2%, which usually lowers the expected return too, or the loss tolerance must be stated over a longer horizon. Over five years the mean grows five times but the volatility only by the square root of five, so the chance of a losing five-year stretch is much lower. Asking about the horizon is the question a good risk manager raises first.
Then name the model's limits. Real returns have fatter left tails than a normal curve, so the chance of a large loss is understated; returns are not independent from year to year; and the 8% expected return is an assumption, not a promise. A drawdown limit, such as no more than a 15% fall from peak, is often more useful to an institution than a probability of a losing year.
Where candidates lose it
The trap is assuming that a positive expected return makes losing years rare. At 0.8 standard deviations above zero, they happen about one year in five.
The second miss is solving for volatility with the wrong number from the normal table. For a 10% tail you need 1.28 standard deviations, not 1.645, which is the 5% tail.
What the interviewer asks next
What is the chance of a negative return over five years with the same targets, assuming independent years?
The investor adds a limit of no more than a 15% loss in any year. What volatility does that imply at 99% confidence?
Why might a pension fund care more about a drawdown limit than a volatility target?
Asked at MSCI, Risk Management, Anonymous interview candidate in, 2013 (Wall Street Oasis): What risk-return targets would you set for an institutional investor?
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
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Is the drop from 0.5 to 0.3 statistically meaningful?
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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
$\text{SE}(\hat\rho) \approx \frac{1 - \rho^2}{\sqrt{n}} = \frac{0.75}{6} = 0.125 \qquad z = \tanh^{-1}(\hat\rho), \;\; \text{SE}(z) = \frac{1}{\sqrt{n - 3}}$
rho the estimated correlation, 0.5
n the 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.
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?
024A desk's daily P&L has a standard deviation of Rs 2 crore. The head of the desk wants a loss number that is exceeded about one day a month. Which confidence level is that, and what is the VaR?VaR and expected shortfallCoreBank market riskTreasury and ALM
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Which confidence level matches one bad day a month?
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About 95% confidence, and a VaR of about Rs 3.3 crore. A month has about 21 trading days, so one breach a month is a tail of 1 in 21, about 4.8%, a confidence level of about 95.2%. For a normal P&L that cut sits 1.67 standard deviations out, so the VaR is 1.67 times Rs 2 crore, about Rs 3.34 crore. At a round 95% it is Rs 3.29 crore.
How do you turn a frequency into a confidence level?
Think of a weather forecast that says a heavy rain day comes about once a month. That is the same statement as saying about 1 day in 30 is that wet. A VaR confidence level is just a frequency of bad days written as a percentage: one breach in 21 trading days is a 4.8% tail, which is a 95.2% confidence level. Counting trading days matters; using 30 calendar days would make the tail 3.3% and the level about 96.7%.
For daily P&L with a Rs 2 crore standard deviation, one loss day in 21 sits beyond Rs 3.34 crore, 1.67 standard deviations out, which is close to the Rs 3.29 crore 95% VaR.
The relationship
$\text{VaR} = z_{c}\,\sigma = 1.668 \times 2 \approx 3.34 \text{ crore} \qquad z_{95\%} = 1.645 \Rightarrow 3.29 \text{ crore}$
z_c the number of standard deviations that leaves the chosen tail below it
sigma the daily P&L standard deviation, Rs 2 crore
What it says in wordsMultiply the daily standard deviation by the number of standard deviations that matches the chosen frequency of bad days.
Why is this translation useful on a desk?
Because a desk head can act on a frequency and cannot act on a percentage. Saying we expect to lose more than Rs 3.3 crore about once a month, roughly 12 days a year, tells the head of the desk what to expect and when to worry. It also sets up the backtest: if losses beyond Rs 3.3 crore start happening three or four times a month, the model is understating risk, and that is visible within a quarter rather than a year.
Add the limits. The VaR says nothing about how bad the bad day is; a loss of Rs 3.4 crore and one of Rs 10 crore both count as one breach. The normal assumption also understates fat tails, so the real one-in-21 loss may sit further out than Rs 3.3 crore. And the Rs 2 crore standard deviation is itself an estimate that moves with market conditions.
Where candidates lose it
The trap is picking 99% because it is the regulatory number people remember. A 99% VaR is breached about once in 100 trading days, closer to once every five months than once a month.
The second miss is counting 30 calendar days. Losses happen on trading days, so the tail is 1 in 21, not 1 in 30.
What the interviewer asks next
What loss would the desk expect to exceed once a year?
The desk had five breaches last month. What do you conclude, and what would you check first?
How would the answer change if the P&L had fat tails with the same standard deviation?
025A Kalman filter tracks a random walk with process variance 1 and measurement variance 4. What gain does it settle at, and which simple smoother is it then equivalent to?Statistics and estimationHardUBSLondon · 2022
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Where does the gain settle?
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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.
The relationship
$P^- = P + Q, \quad K = \frac{P^-}{P^- + R}, \quad P = (1-K)P^- \;\Rightarrow\; (P^-)^2 - Q P^- - QR = 0$
Q the process variance, 1: how much the true value moves each step
R the measurement variance, 4: how noisy each reading is
P^- the variance of the estimate just before a reading arrives
K the Kalman gain
What it says in wordsIn steady state the uncertainty added by the drift each step exactly balances the uncertainty removed by each reading.
Starting from a vague prior the Kalman gain begins at 0.96, drops to 0.55 and 0.44, and settles at 0.390 by about the sixth update, after which the filter is an exponentially weighted moving average with weight 0.39 on each new reading.
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.
Where candidates lose it
The trap is describing a Kalman filter in general terms and never producing a number. The interviewer wants to see you set up the variance recursion and solve the steady state.
The second miss is saying the gain goes to zero. That is true only for a constant state; for a random walk it settles at a positive value, and that is why the filter reduces to an EWMA.
What the interviewer asks next
What steady-state gain do you get with measurement variance 1 instead of 4?
What EWMA decay factor corresponds to this filter, and what is its half-life in updates?
How would you estimate Q and R from data?
Asked at UBS, Risk, London, 2022 (Wall Street Oasis): Explain what a Kalman filter is
026A Rs 100 crore loan earns Rs 3 crore of net interest income a year, costs Rs 0.8 crore to run, carries an expected loss of Rs 1 crore and needs Rs 8 crore of economic capital. What is its risk-adjusted return on capital, and does it clear a 14% hurdle?Capital and leverageCoreBank credit riskRisk GCC
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Where does the Rs 1 crore of expected loss go in the calculation?
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RAROC is 15.0%, so the loan clears a 14% hurdle, but only just. Income of Rs 3 crore less Rs 0.8 crore of cost and Rs 1 crore of expected loss leaves Rs 1.2 crore of risk-adjusted profit. Divided by Rs 8 crore of economic capital that is 15.0%. After a 14% charge on the capital, Rs 1.12 crore, the loan adds only Rs 0.08 crore a year.
Why does expected loss sit in the numerator and unexpected loss in the denominator?
Think of a restaurant that knows about two plates in a hundred get dropped. That breakage is priced into the menu; it is a running cost. What the owner keeps cash aside for is the night a whole shelf comes down. Expected loss is the average, so it is a cost charged against income; unexpected loss is the bad year above the average, so it is covered by capital. RAROCRisk-adjusted return on capital: profit after costs and expected loss, divided by the capital held against unexpected loss. puts each in its proper place, which is why a bank can compare a thin-margin safe loan with a fat-margin risky one on the same scale.
Net interest income of Rs 3.0 crore, less Rs 0.8 crore of cost and Rs 1.0 crore of expected loss, leaves Rs 1.2 crore of risk-adjusted profit; over Rs 8 crore of economic capital that is a 15.0% RAROC, one point above the 14% hurdle.
The relationship
$\text{RAROC} = \frac{\text{income} - \text{cost} - \text{EL}}{\text{economic capital}} = \frac{3.0 - 0.8 - 1.0}{8} = 15.0\%$
EL expected loss, the average annual credit loss, Rs 1 crore
economic capital the buffer held against losses worse than average, Rs 8 crore
What it says in wordsTake the average loss off the income, then ask how hard the remaining profit works the capital held against a bad year.
How much room is there before the loan fails the hurdle?
Turn the hurdle into rupees. A 14% return on Rs 8 crore of capital is Rs 1.12 crore a year, and the loan earns Rs 1.2 crore, so it clears by Rs 0.08 crore. A margin that thin means one small change flips the answer. If expected loss rises from Rs 1.0 crore to Rs 1.08 crore, or economic capital rises from Rs 8 crore to Rs 8.57 crore, RAROC falls to exactly 14%. Saying those two breakeven numbers is what turns a ratio into a credit judgement.
State the limits too. This is a pre-tax, single-year figure; many banks compute RAROC after tax and add the return earned on the capital itself, which would lift the number. And the Rs 8 crore depends on the bank's own capital model, so the answer is only as good as that model's view of this borrower.
Where candidates lose it
The common slip is to subtract expected loss and then also add it to the capital, or to leave it out of the numerator on the grounds that capital covers losses. Either way the ratio comes out wrong by several points, and the interviewer hears that you do not know what capital is for.
The second slip is stopping at 15% and saying yes. A one-point margin deserves the breakeven sentence: how much worse can the loss or the capital get before the loan stops paying for itself.
What the interviewer asks next
The bank prices the loan 20 basis points lower to win the deal. Does it still clear the hurdle?
How would you estimate the Rs 8 crore of economic capital in the first place?
Why might a bank still make a loan with a RAROC below its hurdle?
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
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Compared with the chance of finishing below the level, the chance of touching it at some point is:
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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.
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
$P(\text{touch}) = 2\,\Phi\!\left(\frac{\ln 0.9}{0.02\sqrt{20}}\right) = 2\,\Phi(-1.18) \approx 2 \times 0.119 = 0.239$
\Phi the standard normal cumulative probability
0.02\sqrt{20} 20-day volatility, 8.94%
\ln 0.9 the 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?
028A fund's annual volatility is 18%, its benchmark's is 16%, and the correlation between their returns is 0.95. What is the fund's tracking error?Correlation and diversificationCoreMSCIMonterrey · 2013
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Quick instinct: roughly how big is the tracking error?
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About 5.73%. Tracking error is the volatility of the fund's return minus the benchmark's. Its variance is 18 squared plus 16 squared minus 2 x 0.95 x 18 x 16, which is 324 plus 256 minus 547.2, or 32.8. The square root is 5.73%, nearly three times the 2-point gap in volatilities.
What exactly is tracking error measuring?
Two friends walk to the same office. How far apart they are at any moment depends less on how fast each walks than on whether they take the same streets. Tracking error is the volatility of the return difference, fund minus benchmark, so it depends on how much the two move apart, not on how much each moves. The fund and index can both swing wildly and still track closely if they swing together. That is why the formula needs the correlationA number from minus 1 to 1 describing how closely two returns move together; 1 means perfect lockstep., not just the two volatilities.
The relationship
$TE = \sqrt{\sigma_F^2 + \sigma_B^2 - 2\rho\,\sigma_F\sigma_B} = \sqrt{324 + 256 - 547.2} = \sqrt{32.8} = 5.73\%$
\sigma_F, \sigma_B fund and benchmark volatility, 18% and 16%
\rho correlation of their returns, 0.95
What it says in wordsThe variance of a difference is the two variances added, less twice the part they share.
Drawing the two volatilities as sides 18 and 16 at the angle whose cosine is 0.95 makes tracking error the short third side, 5.73%; nudging correlation from 0.95 to 0.99 cuts it to 3.12%, and dropping it to 0.90 raises it to 7.85%.
Why does the correlation matter more than the volatilities?
Look at the table in the figure. Keeping 18 and 16 fixed, moving correlation from 0.99 to 0.90 takes tracking error from 3.12% to 7.85%, more than doubling it. At high correlations each hundredth of correlation moves tracking error a lot, because the large shared term 2 x rho x 18 x 16 almost cancels the two variances. Now hold correlation at 0.95 and give both sides 16% volatility: tracking error is still 5.06%. The volatility gap contributes a little; the imperfect correlation contributes most.
Close with the limit. The formula uses a correlation estimated from history, and correlations drift, often falling in stressed markets. A fund reporting 5.7% tracking error in calm years can run well above it in a sell-off, so a risk team watches realised tracking error alongside the model figure.
Where candidates lose it
The fast wrong answer is 2%, subtracting the volatilities. It silently assumes correlation of exactly 1, which the question has just told you is false. Candidates who say it have treated volatility as if it were a return.
The second trap is fumbling the formula under pressure. Anchor it to one line you already know: the variance of A minus B is var A plus var B minus twice the covariance. Everything else follows.
What the interviewer asks next
What correlation would give a tracking error of exactly 2%?
The fund's beta to the benchmark is 1.07. Split the tracking error into a beta part and a residual part.
Why might a fund with low tracking error still underperform its benchmark every year?
Asked at MSCI, Financial Tools, Monterrey, 2013 (Wall Street Oasis): What's the tracking error formula?
029A fully collateralised netting set's value moves with a daily volatility of Rs 2 crore. Assuming a 10-day margin period of risk and normally distributed moves, what is the 99% potential future exposure?Counterparty exposure and collateralCoreCounterparty riskBank market risk
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Which number do you multiply the daily Rs 2 crore by to reach ten days?
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About Rs 14.7 crore. Collateral covers the value up to the last margin call, so exposure is the move during the 10 days it takes to notice the default and close out. Ten days of Rs 2 crore daily volatility is 2 x root 10, Rs 6.32 crore. At 99%, 2.33 standard deviations, potential future exposure is Rs 14.71 crore.
If the trades are fully collateralised, where does any exposure come from?
A landlord holding a month's deposit is covered for last month's rent. If the tenant stops paying and takes three months to evict, the deposit covers one of them. Collateral protects you up to the last margin you actually received; the margin period of riskThe time from the last successful collateral exchange with a defaulting counterparty until its trades are closed out or re-hedged. is the gap after that, while the defaulter pays nothing and the market keeps moving. Ten days covers the missed call, the dispute, the formal default and the close-out.
From the last margin received, the 99% edge of the exposure fan opens with the square root of time, reaching Rs 4.65 crore after one day and Rs 14.7 crore after the 10-day margin period of risk, which is the exposure the collateral never covered.
Why the square root of ten and not ten?
Each day's move is independent, so some days undo others. Variances add across independent days, which makes volatility grow with the square root of time. Ten days is 2 x 3.16, Rs 6.32 crore, and 99% is 2.33 of those. Scaling linearly would give Rs 46.5 crore, more than three times too big, and a counterparty limit set on it would turn away business the bank could safely do.
The relationship
$\text{PFE}_{99\%} = z_{0.99}\,\sigma_{1d}\sqrt{\text{MPOR}} = 2.33 \times 2 \times \sqrt{10} \approx 14.7$
z_{0.99} the 99% one-sided normal quantile, 2.33
\sigma_{1d} daily volatility of the netting set value, Rs 2 crore
MPOR margin period of risk, 10 days
What it says in wordsExposure at a confidence level is that many standard deviations of the move over the days you are unprotected.
Name what the simple version leaves out. Only moves in your favour are exposure, so the lower half of the fan does not count. Collateral thresholds or a minimum transfer amount add a fixed layer on top. And a counterparty that defaults in a crisis defaults when volatility is high, so Rs 2 crore a day is likely too calm. Double the margin period to 20 days, as can happen with disputes or illiquid trades, and the figure rises to Rs 20.8 crore, not Rs 29.4 crore.
Where candidates lose it
Two ways to lose it. Some candidates say zero, because the set is fully collateralised, and miss that collateral only covers values already agreed. Others multiply by 10 instead of the square root of 10 and land on Rs 46.5 crore.
Say the timeline before the formula: last margin in, default, close-out. Once the interviewer hears that you know why the ten days exist, the arithmetic is a formality.
What the interviewer asks next
The counterparty has a Rs 5 crore threshold before it posts collateral. What is the PFE now?
Why might a regulator require a longer margin period of risk for some netting sets?
What is wrong-way risk, and how would it change this number?
030A distressed one-year bond costs 80. It pays 100 at maturity with probability 85%, and if the issuer defaults, holders recover 40. What is the expected return, and at what default probability does the trade break even?Credit risk arithmeticCoreBank credit riskAsset manager risk
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Before you calculate: at what default probability does paying 80 stop making sense?
Show the worked solution
The expected return is 13.75%, and the trade breaks even at a default probability of 33.3%. The expected payoff is 0.85 x 100 plus 0.15 x 40, which is 91, against a price of 80. Breakeven solves 80 = 100 minus 60p: p is 20 over 60, one third. The real question is whether default odds could be that high.
Why is the breakeven more useful than the expected return?
The 15% is somebody's estimate, and estimates of default for a stressed issuer are soft. Expected return inherits every error in the 15%; the breakeven tells you how wrong the estimate can be before you lose money. It is like buying a second-hand car that needs repairs: rather than guess the repair bill, you ask how large the bill could be before the deal stops being worth it. Here default odds could more than double, from 15% to one in three, before the trade loses on average.
Paying 80 for a bond that returns 100 with 85% probability and 40 on default gives an expected payoff of 91, a 13.75% expected return, and the trade only loses on average if the default probability exceeds 33.3%.
The relationship
$80 = (1-p)\,100 + p\,40 \;\Rightarrow\; p^{*} = \frac{100 - 80}{100 - 40} = \frac{1}{3}$
p one-year default probability
100 - 80 the gain if the bond pays at par
100 - 40 the gap between par and recovery
What it says in wordsBreakeven default probability is the discount to par divided by the loss given default measured from par.
What would a credit risk manager add before approving the trade?
Three things. First, the recovery is also a guess, and it moves the breakeven: if holders recover 20 rather than 40, breakeven default odds fall from one third to 25%, because each default now costs 60 from the purchase price instead of 40. Second, money has a time value. If you require an assumed 7% return from cash for the year, breakeven falls from 33.3% to 24.0%, because the bond must beat cash, not zero. Third, the payoff is lopsided, +25% or -50%, so position size matters more than the average: a book of such bonds survives, a single large position may not.
Where candidates lose it
Candidates compute 13.75% and stop, as if the 15% were a fact. The interviewer's follow-up is always what if the probability is wrong, and the candidate who has the one-third breakeven ready answers it before it is asked.
The second trap is using the 20-point discount to par as the breakeven default rate. The loss on default is 40 from where you bought, not 20, and mixing the two gives nonsense.
What the interviewer asks next
What default probability does the market price imply if investors demand a 7% return?
Recovery is uncertain, between 25 and 55. How does that change your view?
Why do distressed investors often care more about recovery analysis than default probability?

delta	the put delta, minus 0.3
Gamma	the put gamma, 0.004 per rupee
Delta S	the stock move, minus Rs 50
Delta V	the change in each put's value, Rs

mu	the expected annual return, 8%
sigma	the annual volatility
N	the standard normal cumulative distribution
1.2816	the number of standard deviations that leaves 10% in the lower tail

rho	the estimated correlation, 0.5
n	the number of observations, 36
tanh^{-1}	Fisher's transformation, which makes the sampling error close to normal

z_c	the number of standard deviations that leaves the chosen tail below it
sigma	the daily P&L standard deviation, Rs 2 crore

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