Risk Management puzzles, solved step by step

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

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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
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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.
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
$\text{free}_1 = \text{total}\,(1 - d) - \text{pledged} = 5{,}000 \times 0.9 - 3{,}200 = 1{,}300$
d the fall in bond prices, 10%
pledged the 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?
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
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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.
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
$E[L \mid L > u] = u + \mu = 10 + 20 = 30, \qquad P(L > u) = e^{-u/\mu} = e^{-0.5} = 0.607$
u the reporting threshold, Rs 10 lakh
\mu the 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?
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
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If the trader checks the price more often, what happens to the expected cost of the hedge?
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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.
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
$E[\text{crossing costs}] = E[(S_T - K)^+] - (S_0 - K)^+ \approx \frac{\sigma\sqrt{T}}{\sqrt{2\pi}} = \frac{1.5\sqrt{40}}{2.507} = 3.78$
\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?
047A desk makes or loses Rs 1 crore each day, winning with probability 0.52. It stops as soon as it is up Rs 10 crore or down Rs 10 crore. What is the probability it reaches the profit target first?Probability and base ratesHardBank market riskQuant risk
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The daily edge is 52 against 48. What is the chance of hitting plus 10 before minus 10?
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About 69%. This is the gambler's ruin. With r = 0.48 / 0.52, starting 10 steps from each barrier, the chance of hitting the top first is 1 / (1 + r to the power 10). r to the 10 is about 0.449, so the answer is 1 / 1.449, or 69.0%. A 2-point daily edge becomes a 19-point edge because the race takes many days.
Why does a small daily edge grow so much?
A casino wins only a couple of points more than half its bets at roulette, yet over thousands of spins it almost never ends a month behind. An edge that barely shows in one step accumulates over many, because the random part grows with the square root of the number of steps while the edge grows in proportion to it. Reaching plus or minus 10 takes around 95 days on average here, long enough for the edge of 0.04 a day to push the walk decisively one way.
The relationship
$P(\text{+10 first}) = \frac{1 - r^{10}}{1 - r^{20}} = \frac{1}{1 + r^{10}}, \qquad r = \frac{0.48}{0.52} = 0.923, \; r^{10} = 0.449$
r the ratio of losing to winning odds on one day
10 the distance in steps from the start to each barrier
What it says in wordsThe gambler's ruin formula gives the chance of reaching one barrier before the other; starting in the middle it simplifies to one over one plus r to the power of the distance.
The chance of reaching plus Rs 10 crore before minus Rs 10 crore rises steeply with the daily win probability, from 50% for a fair coin to 69.0% at 0.52 and 91.8% at 0.56, far above the dashed line where the race would only match the daily edge.
How do you check the answer, and what does it say about stop-losses?
Two sanity checks. At 0.50, r is 1 and the formula gives 50%, as symmetry demands. Widen both barriers to 20 and the answer at 0.52 rises to 83.2%. The wider the barriers relative to the daily step, the more the edge dominates the noise. That is the risk manager's reading: a tight stop-loss relative to daily volatility turns a trader with a genuine edge into a near coin flip, while a wide one lets the edge show but exposes more capital to a trader without one.
State the limits. The model assumes a constant edge, equal step sizes and independent days. Real P&L has fat tails, the edge decays, and a losing streak may itself be evidence that the edge was never there. The formula tells you how an edge would play out, not whether you have one; with drawdownThe fall from a running peak in cumulative profit, measured in rupees or as a percentage. limits, that second question is usually the harder one.
Where candidates lose it
The most common answer is 52%, treating the race as one big coin toss with the same odds as a single day. The whole puzzle is about how a small edge compounds over many steps.
The second trap is knowing the formula but fumbling it. Say the general form, simplify it for the symmetric start, and check it at 0.50; the check earns as much as the number.
What the interviewer asks next
What is the expected number of days before the desk stops?
The loss limit is Rs 5 crore and the target stays at Rs 10 crore. What is the answer now?
How would you test whether a trader really has a 0.52 edge from their P&L history?
064A bank has CET1 of Rs 100 crore on risk-weighted assets of Rs 1,000 crore, a 10% ratio. A Rs 20 crore loan loss also removes Rs 50 crore of risk-weighted assets as the defaulted loans are written off. How much capital must it raise to get back to 10%?Capital and leverageHardBank credit riskTreasury and ALM
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How much new equity does the bank need?
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Rs 15 crore. The loss takes CET1 from Rs 100 crore to Rs 80 crore, and writing off the defaulted loans takes risk-weighted assets from Rs 1,000 crore to Rs 950 crore, a ratio of 8.42%. To be back at 10% the bank needs 10% of Rs 950 crore, which is Rs 95 crore, so it must raise Rs 15 crore of new equity, assuming the new cash carries no risk weight.
Why is the answer not simply the Rs 20 crore lost?
A cricket team that needs a run rate of 6 an over has lost two overs to rain. It needs fewer runs than before, not the same total, because the target is a rate. A capital ratio is a rate too. The loss cuts the numerator by 20%, but the write-off also cuts the denominator by 5%, and the target is 10% of the new, smaller denominator. That makes the gap Rs 15 crore.
CET1 falls 20%, from Rs 100 crore to Rs 80 crore, while risk-weighted assets fall only 5%, from Rs 1,000 crore to Rs 950 crore, so the ratio drops to 8.42%. Ten per cent of Rs 950 crore is Rs 95 crore, which leaves a Rs 15 crore capital gap.
Where does the Rs 50 crore of RWA come from, and is it realistic?
Think of Rs 50 crore of defaulted loans at a 100% risk weight. The bank loses Rs 20 crore on them, a 40% loss given default, and recovers Rs 30 crore in cash, which carries no risk weight. Both sides of the ratio move on the same event, but a loss always hits capital at full size while the RWA falls only by the risk weight of what left. A write-off never rescues the ratio: 80 over 950 is still well below 10%.
The relationship
$\text{raise} = 0.10 \times (1{,}000 - 50) - (100 - 20) = 95 - 80 = 15$
0.10 the target CET1 ratio
1,000 - 50 risk-weighted assets after the write-off
100 - 20 CET1 after the loss
What it says in wordsThe raise is the target ratio times the new risk-weighted assets, less the capital left after the loss.
The limitations are worth one line each. The new equity is assumed to sit in cash or government bonds at a zero risk weight; if the bank lends it out, RWA grows and it needs more. And a real bank must also rebuild any buffer above the minimum, whose size depends on current rules you would check rather than assume.
Where candidates lose it
The common answer is Rs 20 crore, replacing the loss. It forgets the denominator moved. The opposite slip is to credit the RWA fall at full size and answer Rs 5 crore, as if removing Rs 50 crore of assets were worth Rs 50 crore of capital.
Write the ratio after the loss, 80 over 950, then solve for the numerator that gives 10%. Two lines of arithmetic, said out loud, and the answer defends itself.
What the interviewer asks next
What if the recovered Rs 30 crore is relent at a 100% risk weight?
How much RWA would the bank have to shed instead of raising capital?
Why might a bank prefer to shrink RWA rather than raise equity in a downturn, and what does that do to lending?
065You can bet repeatedly at even odds with a 60% chance of winning each bet. Compare the long-run growth of your capital if you bet 20% of it each time with betting 50% each time.Compounding and drawdownsHardQuant riskAsset manager risk
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Betting 50% a time on a 60% edge: what happens over many bets?
Show the worked solution
Betting 20% grows capital about 2.0% a bet; betting 50% shrinks it about 3.3% a bet. Growth per bet is the average log return: 0.6 ln(1 + f) + 0.4 ln(1 - f). At 20%, the Kelly fraction 2p - 1, that is +2.01%. At 50% it is -3.40%. After 100 bets the typical 20% bettor has about 7.5 times the start; the 50% bettor about 0.03 times.
How can a bet with a positive edge lose money?
A shopkeeper who stakes half his stock on every festival season, and wins more seasons than he loses, can still go under: one bad season halves him, and a good one only adds half back. Losses on a shrunken base are expensive to recover. The typical outcome of repeated betting is set by the average log return, not the average return, and at 50% the losses dominate the logs.
With a 60% chance of winning at even odds, growth per bet peaks at the 20% Kelly fraction, about +2.0% a bet, crosses zero near 39% and is -3.4% a bet at 50%, so oversizing a winning bet turns steady growth into steady decline.
Where does the 20% come from?
Set the slope of the growth curve to zero. For even odds the Kelly fractionThe fraction of capital to bet each time that maximises the long-run compound growth rate of the capital. is the probability of winning minus the probability of losing: 0.6 minus 0.4, or 20%. Betting less than Kelly gives up some growth for a much smoother ride; betting more than about twice Kelly gives up growth altogether. Here the curve crosses zero near 39%, just under twice Kelly.
The relationship
$g(f) = p\ln(1+f) + (1-p)\ln(1-f) \qquad f^* = 2p - 1 = 0.2$
f the fraction of capital bet each time
p the probability of winning a bet, 0.6
g(f) the expected log growth of capital per bet
What it says in wordsLong-run growth is the probability-weighted average of the log of what each outcome does to your capital.
For a risk manager the lesson is position sizing. A desk with a genuine edge can still destroy capital by running too much leverage, and the average P&L will look fine right up until the drawdowns compound. Real edges are also uncertain, which is why practitioners bet a fraction of Kelly.
Where candidates lose it
The trap is answering that 50% grows faster because each bet has a positive expected value and bigger bets earn more of it. That is true of the arithmetic average and false for the capital you actually end up with, which compounds.
The second slip is knowing Kelly but not being able to show why 50% loses. Have the one-line check ready: six wins and four losses at 50% multiply capital by about 0.71.
What the interviewer asks next
What is the Kelly fraction if you win 55% of the time at even odds?
What if the payoff is 2 to 1 with a 40% chance of winning?
Why do traders often bet half Kelly rather than full Kelly?
066A long-short pair holds equal rupee amounts of two stocks, long one and short the other, each with 2% daily volatility. What happens to the pair's daily volatility when the correlation between the stocks falls from 0.9 to 0.6?Correlation and diversificationHardBank market riskAsset manager risk
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The correlation drops by a third. What does the pair's volatility do?
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It doubles, from about 0.89% to 1.79% a day, as a share of one leg. For equal and opposite legs with the same volatility, pair variance is 2 x sigma squared x (1 - rho). The unhedged part, 1 - rho, goes from 0.1 to 0.4, four times as large, so volatility doubles. On Rs 100 crore a leg, daily volatility goes from about Rs 0.89 crore to Rs 1.79 crore.
Why does a modest fall in correlation hit a hedged book so hard?
Two boats tied close together on a river rise and fall almost as one; the rope between them barely strains. Loosen the tie and every wave that hits one boat and not the other shows up as strain on the rope. A long-short pair is the rope. It only carries the part of the two stocks that does not move together, and that part is 1 minus rho, which grows fast as correlation slips from a high level.
With equal legs of 2% daily volatility, the pair's volatility is 0.89% at a correlation of 0.9 and 1.79% at 0.6, double, because the uncorrelated part 1 minus rho quadruples from 0.1 to 0.4.
What does this mean for the desk's risk numbers?
VaR scales with volatility, so the pair's VaR doubles with no change in position. On Rs 100 crore a leg the 99% one-day VaR moves from about Rs 2.1 crore to Rs 4.2 crore, entirely from a correlation the desk does not control. Correlations between related stocks do fall in exactly the markets where one company's news breaks from its peer's, which is when the pair was supposed to be safe.
The relationship
$\sigma_{\text{pair}} = \sqrt{\sigma^2 + \sigma^2 - 2\rho\sigma^2} = \sigma\sqrt{2(1-\rho)}$
\sigma each leg's daily volatility, 2%
\rho the correlation between the two stocks, 0.9 then 0.6
What it says in wordsThe pair's volatility is each leg's volatility times the square root of twice the uncorrelated share.
Say the limitation. The formula assumes equal volatilities and equal notionals; a real pair is often beta-weighted, which changes the coefficients but not the lesson. A risk limit that assumes the historical correlation of 0.9 is a limit on a number that can quietly double.
Where candidates lose it
The trap is reasoning linearly: correlation fell by a third, so risk rises by a third. The risk depends on 1 minus rho, which went up fourfold, and the square root turns that into a doubling.
The second trap is treating the pair as market neutral and therefore low risk. Neutral to the market is not neutral to correlation, and a good answer names correlation as the position's real exposure.
What the interviewer asks next
At what correlation does the pair's volatility equal a single leg's 2%?
How would you hedge the correlation risk itself?
If the legs are beta-weighted instead of equal, how does the formula change?
069Portfolio A holds only 5-year zero-coupon bonds. Portfolio B holds 2-year and 8-year zeros, weighted so its duration is also 5. Which portfolio does better after a large parallel move in yields, and which one loses if the curve steepens?Duration and ratesHardTreasury and ALMBank market risk
Try it first
After a 200 basis point parallel move in either direction, which portfolio is ahead?
Show the worked solution
The barbell wins a large parallel move either way; it loses when the curve steepens. Both portfolios have duration 5, but the barbell's convexity is 34 against the bullet's 25. For a 200 basis point rise the barbell loses 9.35% against 9.52%, and for a fall it gains 10.72% against 10.52%. If the curve steepens 10 basis points per year around 5 years, the bullet is flat and the barbell loses 0.88%.
If the durations match, why do the portfolios behave differently?
Two families each spend an average of Rs 50,000 a month. One spends it evenly; the other spends Rs 20,000 some months and Rs 80,000 others. A small price rise hits them alike, but a big shock hits the second family's heavy months much harder. Averages match; the spread does not. Duration is the average maturity, so it only describes small, parallel moves; the barbell spreads its money further from 5 years, which gives it more convexity and exposes it to changes in the curve's shape.
The bullet sits at 5 years and the barbell splits evenly between 2 and 8 years, both at duration 5. The barbell does better in a 200 basis point parallel move either way because its convexity is 34 against 25, and it loses 0.88% when the curve steepens around the 5-year point while the bullet is unaffected.
Why does the steepener hurt only the barbell?
In this steepener the 5-year yield does not move, so the bullet does not move. The 2-year yield falls 30 basis points and the 8-year yield rises 30. The barbell's long leg carries four times the duration of its short leg, so the loss on the 8-year bond swamps the gain on the 2-year. Roughly: half the money times 8 years times 0.30% is a 1.2% loss, and half times 2 years times 0.30% is a 0.3% gain, a net loss of about 0.9%.
The relationship
$w\cdot 2 + (1-w)\cdot 8 = 5 \Rightarrow w = 0.5 \qquad C_{\text{barbell}} = 0.5(2^2) + 0.5(8^2) = 34 > 25 = 5^2$
w the share of value in the 2-year zero
C convexity; for a zero with continuous compounding it is maturity squared
What it says in wordsMatching duration fixes the weights at half and half, and the barbell ends up with more convexity.
The limitation: the gains from convexity are small for small moves and are usually priced, since a barbell typically yields less than a bullet of the same duration. In a quiet market that yield give-up can outweigh the convexity benefit, so the choice is a view on volatility and curve shape, not a free lunch.
Where candidates lose it
The trap is saying the two are the same risk because duration matches. Duration is one number describing one kind of move, and the interviewer built the question to see if you know the two other risks it misses: convexity and curve shape.
The second slip is getting convexity right and missing the steepener. Give both halves: the barbell wins a big parallel move and loses when the long end sells off relative to the short end.
What the interviewer asks next
Which portfolio is hurt by a flattening instead?
How would you hedge the barbell's steepener exposure?
Why does a barbell usually yield less than a bullet of the same duration?
070A fund holds Rs 1,000 crore of bonds financed by repo, putting up Rs 20 crore of its own capital against a 2% haircut. The lender raises the haircut to 10%. With its capital unchanged, how much must the fund sell?Liquidity and balance sheetHardTreasury and ALMAsset manager risk
Try it first
How much of the Rs 1,000 crore position has to go?
Show the worked solution
It must sell Rs 800 crore, four fifths of the position. The haircut is the share of the position the fund funds itself. At 2%, Rs 20 crore supports Rs 1,000 crore of bonds, 50 times leverage. At 10%, the same Rs 20 crore supports only Rs 200 crore, 10 times leverage. Keeping the full position would need Rs 100 crore of capital, Rs 80 crore more than it has.
Why does a small change in haircut force such a big sale?
Buying a flat with a 5% down payment lets Rs 10 lakh of savings control a Rs 2 crore flat. If the bank suddenly demands 20% down, the same Rs 10 lakh controls only a Rs 50 lakh flat. The haircutThe share of a security value a lender will not lend against, so the borrower must fund that share with its own capital. sets leverage directly, at one over the haircut, so moving it from 2% to 10% cuts the position the capital can support from 50 times to 10 times.
At a 2% haircut, Rs 20 crore of capital and Rs 980 crore of repo carry Rs 1,000 crore of bonds. At 10%, the same Rs 20 crore carries only Rs 200 crore with Rs 180 crore of repo, so Rs 800 crore of bonds must be sold.
What happens next in a real market?
The sale itself moves prices. If many leveraged holders face the same haircut increase, they all sell the same bonds, prices fall, their capital shrinks, and the lower capital supports an even smaller position. That feedback loop between haircuts, prices and forced selling is the liquidity spiral seen in past funding crises. A fund at 50 times leverage loses its entire Rs 20 crore of capital on a 2% price fall.
The relationship
$\text{max position} = \frac{\text{capital}}{\text{haircut}} = \frac{20}{0.10} = 200 \qquad \text{sell} = 1{,}000 - 200 = 800$
capital the fund's own money in the position, Rs 20 crore
haircut the share the lender will not finance, now 10%
What it says in wordsThe largest position the fund can hold is its capital divided by the haircut.
The limitation: the answer assumes the fund cannot raise capital or find another lender at the old haircut. In practice it would try both, and a risk manager's job is to know in advance how many days that would take and how much the forced sale would cost at stressed prices.
Where candidates lose it
The trap is answering Rs 80 crore: 8% more haircut on Rs 1,000 crore. That is the extra capital needed to keep the position, not the amount the fund must sell when it has no extra capital. The question asks for the sale.
The second miss is stopping at the arithmetic. Say what the sale does to prices and to the fund's remaining capital; the interviewer wants to hear the word spiral.
What the interviewer asks next
What price fall on the remaining position would wipe out the fund's capital at 10 times leverage?
The fund can raise Rs 30 crore more. How much must it sell now?
Why do lenders raise haircuts precisely when markets are falling?
072A 99% one-day VaR model is exceeded independently with probability 1% each trading day. On average, how many trading days pass before you first see exceptions on two consecutive days?Probability and base ratesHardBank market riskQuant risk
Try it first
Before setting up equations: which is it?
Show the worked solution
10,100 trading days on average. Let E0 be the expected wait from no current run and E1 the wait after one exception. Then E0 = 1 + 0.99 E0 + 0.01 E1 and E1 = 1 + 0.99 E0. Solving gives E0 = (1 + p) / p squared = 1.01 / 0.0001 = 10,100, about 40 years of trading days if exceptions really are independent.
Why is the answer not simply one over p squared?
Waiting for two heads in a row with a fair coin takes six tosses on average, not four, because each time you get one head and then a tail, you are back to the start and the first head was wasted. A run has memory: after one exception you are one step from the finish, and a miss throws you back, so the waiting time needs one equation per state, not one probability. For VaR exceptions at 1% the extra is 100 days, the time spent waiting for each first exception.
From no run, an exception moves you to state 1 with probability 0.01; from state 1 another exception ends the wait, but a quiet day with probability 0.99 sends you back to the start. Solving the two waiting-time equations gives 10,100 days, not 10,000.
Why would a risk manager care about a 40-year wait?
Because it turns an observation into a test. If the model is right and exceptions are independent, two in a row should take about forty years of trading to appear; seeing them twice in one year says the exceptions cluster, and clustering means the model misses changes in volatility. Backtesting frameworks test independence for exactly this reason, alongside the plain count of exceptions.
The relationship
$E_0 = 1 + qE_0 + pE_1,\quad E_1 = 1 + qE_0 \;\Rightarrow\; E_0 = \frac{1+p}{p^2} = \frac{1.01}{0.0001} = 10{,}100$
E_0 expected days to finish from no current run
E_1 expected days to finish just after one exception
p, q the daily exception probability 0.01 and its complement 0.99
What it says in wordsWrite the wait from each state in terms of where the next day takes you, then solve the two equations.
The limitation is the independence assumption, which is the thing being tested. Real exceptions tend to arrive in bursts when volatility jumps, so the observed wait is usually far shorter, and that gap is evidence against the model rather than bad luck.
Where candidates lose it
The common wrong answer is 10,000, one over p squared. It treats every pair of days as a fresh independent trial, ignoring that the pairs overlap and that a failed attempt costs the wait for a new first exception.
The other trap is 200 days, the wait for any two exceptions. Consecutive is the whole point. Set up the states before you calculate, and the 100 extra days explain themselves.
What the interviewer asks next
How many days on average before three exceptions in a row?
How many exceptions would you expect in 250 days, and what is the chance of seeing 5 or more?
What does it tell you if exceptions cluster in the same weeks?

d	the fall in bond prices, 10%
pledged	the market value that must stay posted, Rs 3,200 crore

u	the reporting threshold, Rs 10 lakh
\mu	the true mean loss, Rs 20 lakh

\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

r	the ratio of losing to winning odds on one day
10	the distance in steps from the start to each barrier

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Why does free collateral fall faster than prices?

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

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Why does a threshold raise the recorded average?

What exactly goes wrong in the fitted model?

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Where does the money leak out?

Why can faster monitoring not fix it?

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Why does a small daily edge grow so much?

How do you check the answer, and what does it say about stop-losses?

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Why is the answer not simply the Rs 20 crore lost?

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How can a bet with a positive edge lose money?

Where does the 20% come from?

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Why does a modest fall in correlation hit a hedged book so hard?

What does this mean for the desk's risk numbers?

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If the durations match, why do the portfolios behave differently?

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Why does a small change in haircut force such a big sale?

What happens next in a real market?

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Why is the answer not simply one over p squared?

Why would a risk manager care about a 40-year wait?

Where candidates lose it

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