Risk Management puzzles, solved step by step

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

Showing 41–50 of 100

041Bank A holds one Rs 100 crore loan. Bank B holds one hundred loans of Rs 1 crore each. Every loan has a 2% default probability, total loss on default and independent defaults. Both expect to lose Rs 2 crore. What is each bank's chance of losing more than Rs 10 crore in a year?Credit risk arithmeticHardBank credit riskRisk GCC
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Bank A has a 2% chance of losing more than Rs 10 crore. Bank B's chance is:
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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.
Both banks expect to lose Rs 2 crore, but bank A faces a 2% chance of losing its whole Rs 100 crore while bank B's losses cluster around Rs 2 crore and exceed Rs 10 crore only about 5.6 times in a million.
How do you estimate bank B's number in the room?
The defaults are a binomial with mean 2 and standard deviation about 1.40. Eleven defaults is more than six standard deviations above the mean, so the answer is tiny; the exact binomial tail is 5.6 in a million. Compare the standard deviations of loss directly: bank A's is Rs 14.0 crore, bank B's is Rs 1.4 crore, ten times smaller, because splitting one exposure into a hundred independent ones divides the spread by the square root of 100.
The relationship
$\sigma_A = 100\sqrt{0.02 \times 0.98} = 14.0, \qquad \sigma_B = 1 \times \sqrt{100 \times 0.02 \times 0.98} = 1.4$
\sigma standard deviation of the annual loss, Rs crore
0.02 default probability of each loan
What it says in wordsSplitting one exposure into n independent pieces cuts the spread of losses by the square root of n.
Now the caveat that matters in practice. The 99% loss quantile is Rs 100 crore for A and Rs 6 crore for B, which is why concentration riskExtra risk from large single exposures or from many exposures that fail together, which average-based measures do not show. gets its own capital charge. But granularity only helps if defaults are independent. A hundred small loans to one industry in one city will default together in a local downturn, and bank B's tail then fattens towards bank A's. Diversification across names is not diversification across causes.
Where candidates lose it
The first trap is saying both banks are equally risky because both expect to lose Rs 2 crore. Expected loss is exactly the number that cannot tell these two banks apart.
The second is assuming more loans means more risk because more loans can default. The count of defaults rises, but each is small, and the chance of many at once collapses. Close by naming the independence assumption as the thing that could undo it.
What the interviewer asks next
What is each bank's 99% VaR?
If bank B's loans all go to one sector with default correlation, what happens to its tail?
How would you set a single-name concentration limit using this logic?
042You want a curve steepener that is neutral to parallel moves: buy the 2-year bond and short the 10-year. The 10-year has a DV01 of 0.09 per 100 of face and the 2-year has 0.019. How much 2-year do you buy for each Rs 100 crore of 10-year you short?Duration and ratesCoreTreasury and ALMBank market risk
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Roughly how much 2-year face value balances Rs 100 crore of 10-year?
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About Rs 474 crore of 2-year for each Rs 100 crore of 10-year. A parallel-neutral trade matches rupees per basis point, not face value. Rs 100 crore of 10-year moves Rs 9 lakh per basis point. The 2-year moves 0.019 per 100, so matching Rs 9 lakh needs 100 x 0.09 / 0.019, Rs 473.7 crore. A parallel shift then leaves the book flat; only a change in the gap between the two yields makes or loses money.
Why size by DV01 rather than by face value?
Balancing a see-saw is about weight times distance from the pivot, not the number of children on each end. DV01 is each leg's weight: the rupees it gains or loses when its yield moves one basis point. A 10-year bond is about 4.7 times as sensitive per rupee of face as a 2-year, so a parallel-neutral trade needs about 4.7 times as much 2-year face. Equal face value would leave the book mostly a bet on the 10-year, and a parallel move would swamp the curve view.
Rs 473.7 crore of 2-year bonds bought and Rs 100 crore of 10-year bonds sold each carry Rs 9 lakh of DV01, so the trade is level under a parallel move even though the 2-year face is almost five times larger.
The relationship
$N_2 = N_{10} \times \frac{DV01_{10}}{DV01_{2}} = 100 \times \frac{0.09}{0.019} = 473.7$
N face value of each leg, Rs crore
DV01 price change per 100 of face for a one basis point yield move
What it says in wordsSet the two legs' rupees per basis point equal, then solve for the face value of the hedge leg.
What does the trade make, and what can still go wrong?
If the 10-year yield rises 10 basis points while the 2-year stays put, the curve steepens and the short 10-year gains Rs 90 lakh. If both yields rise 10 basis points together, the 2-year loses Rs 90 lakh and the 10-year short gains Rs 90 lakh: flat, which is the design. Had you used equal face, that same parallel rise would have made about Rs 71 lakh, a large outright bet you did not mean to place.
Name the limits. DV01 is a local measure and drifts as yields move and time passes, so the ratio must be rebalanced. CarryThe income a position earns or pays while it is held unchanged: coupons received less the cost of funding and of shorting. differs across the two legs and can quietly dominate a slow trade. And short and long yields rarely move by the same amount; some desks weight the 2-year leg by its historical beta to the 10-year instead of one for one, which gives a different ratio.
Where candidates lose it
The fast wrong answer is equal face value, Rs 100 crore against Rs 100 crore. It leaves the trade roughly four-fifths an outright short of the 10-year, and the first parallel sell-off or rally decides the P&L, not the curve.
Candidates also invert the ratio and answer about Rs 21 crore, dividing 0.019 by 0.09. Check the direction: the less sensitive bond always needs the bigger face.
What the interviewer asks next
The trade should also be neutral to a 1-for-0.8 move between 2-year and 10-year yields. How does the ratio change?
How much does the position make if the curve steepens by 15 basis points?
Why might a treasury desk prefer futures to cash bonds for this trade?
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?
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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?
044Estimate the cash a city's ATMs must dispense each weekday, given a population of 40 lakh, 75% of them adults, 30% of adults withdrawing once a week, an average withdrawal of Rs 3,000, and withdrawals spread evenly over five weekdays.Logic, estimation and brainteasersCoreTreasury and ALMRisk GCC
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Which order of magnitude is right for cash dispensed per weekday?
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About Rs 54 crore a weekday. 40 lakh people x 75% adults is 30 lakh adults. 30% of them withdraw weekly, 9 lakh withdrawals. At Rs 3,000 each that is Rs 270 crore a week, and spread over five weekdays, Rs 54 crore a day. Moving the weekly share between 20% and 40% puts the answer between Rs 36 and 72 crore.
How do you structure the estimate so the interviewer can follow it?
Planning food for a wedding, you do not guess the total rice; you count guests, portions per guest and grams per portion. A sizing answer is a chain of multipliers, each one a stated assumption, so the interviewer can challenge any link without the whole answer collapsing. Say the chain before you multiply: people, adults, weekly users, ticket size, days. Then the arithmetic is almost an afterthought.
Multiplying 40 lakh people by 75% adults, 30% weekly users and Rs 3,000 a withdrawal gives Rs 270 crore a week, about Rs 54 crore a weekday, and the answer stays between Rs 36 and 72 crore while the weekly share sits between 20% and 40%.
How do you sanity check Rs 54 crore?
Test it from a second angle. Rs 270 crore a week across 30 lakh adults is Rs 900 per adult per week, a plausible cash habit in a city where digital payments carry much of the load. It also means about 1.8 lakh transactions a weekday; if you assumed a number of ATMs, you could check whether each would handle a sensible number of withdrawals a day. Then name the weakest link: the 30% weekly share. Doubling it doubles the answer; the population and adult share are far less uncertain.
The relationship
$40\text{ lakh} \times 0.75 \times 0.30 \times \text{Rs } 3{,}000 \div 5 = \text{Rs } 54 \text{ crore}$
0.75 the share of the population who are adults
0.30 the share of adults who withdraw once a week
5 weekdays the withdrawals are spread across
What it says in wordsMultiply the chain of assumptions, then divide by the days the total is spread over.
Finish with why a treasury desk cares. Cash is a liquidityThe ability to meet a payment obligation, here physical cash in machines, on time and without loss. obligation that cannot be deferred, and the average is not the peak. Salary days, festivals and long weekends bunch demand, so the cash plan is sized to a peak day plus a buffer, and an empty machine is an operational failure as much as a customer one. Rs 54 crore is the average the peaks are built on.
Where candidates lose it
The common loss is racing to a number without stating the chain, then being unable to defend it when the interviewer challenges one input. The interviewer cares more about the structure than whether you land on 54.
The other slip is forgetting the last step and quoting the weekly Rs 270 crore as the daily figure. Say the units at every link: people, withdrawals a week, rupees a week, rupees a day.
What the interviewer asks next
How would you size the peak day, around the first of the month?
If 10% of the city's ATMs are down at any time, how does the cash plan change?
How would you estimate the number of ATMs the city needs?
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?
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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?
048You estimate a desk's daily P&L variance from five observations, once dividing the sum of squared deviations by 5 and once by 4. Which estimator is unbiased, which is consistent, and how large is the bias?Statistics and estimationCoreUBSZurich · 2021
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Which statement is true?
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Dividing by 4 is unbiased; both estimators are consistent; dividing by 5 is low by one fifth. The sample mean is estimated from the same five points, which uses up one degree of freedom, so the divide-by-n estimator averages (n - 1)/n of the true variance, 80% here. If the true variance is 4, it centres on 3.2, a bias of -0.8. As n grows that factor tends to 1, so both estimators converge on the truth.
Why does dividing by n come out too low?
Measure how spread out five friends' heights are by comparing each to the group's own average, and you will understate the spread, because that average was pulled towards those five people. Deviations measured from the sample mean are smaller on average than deviations from the true mean, so their sum of squares understates the spread by exactly one observation's worth. Dividing by n minus 1, the degrees of freedomThe number of independent pieces of information left after estimating something from the same data; estimating the mean uses up one., corrects it exactly.
With five observations and a true variance of 4, the divide-by-4 estimator is centred on 4 while the divide-by-5 estimator is centred on 3.2, 80% of the truth, yet the biased version is narrower and has a lower mean squared error, 5.76 against 8.00.
What is the difference between unbiased and consistent?
Unbiased is about the average over many repeated samples of the same size; consistent is about what happens to one estimate as the sample grows. Divide-by-5 fails the first: repeat the five-day exercise many times and the estimates average 3.2, not 4. It passes the second: with 250 days the bias is only -0.016, and it shrinks to zero with more data. An estimator can be unbiased but inconsistent too, such as using only the first observation to estimate a mean: right on average, never improving.
The relationship
$E\!\left[\tfrac{1}{n}\textstyle\sum (x_i - \bar{x})^2\right] = \frac{n-1}{n}\,\sigma^2 = 0.8 \times 4 = 3.2$
n the number of observations, 5
\bar{x} the sample mean, estimated from the same five points
\sigma^2 the true variance, 4 in the illustration
What it says in wordsDividing by n recovers only (n minus 1) over n of the true variance on average.
Now the twist a model validator should add. For normal data the unbiased estimator has variance 8.00 here, while the divide-by-5 version has 5.12 plus a squared bias of 0.64, a mean squared error of 5.76. The biased estimator is closer to the truth on a typical sample. Which you prefer depends on the use: unbiasedness matters when estimates are averaged across many desks; a smaller typical error matters for a single desk's limit. With five data points, neither is reliable, and that is the more important thing to say.
Where candidates lose it
The usual slip is to treat unbiased and consistent as the same thing, and so to call the divide-by-5 estimator inconsistent. The interviewer asked both words together precisely to hear you separate them.
The second miss is answering from memory without the reason. One sentence on the sample mean using up a degree of freedom shows you know why n minus 1 exists, not just that it does.
What the interviewer asks next
Give an example of an estimator that is unbiased but not consistent.
Is the sample standard deviation, the square root of the unbiased variance, itself unbiased?
With 250 days of P&L, does the choice between n and n - 1 matter for VaR?
Asked at UBS, Risk Management, Zurich, 2021 (Wall Street Oasis): And several other questions on econometrics - what is an unbiased estimator vs consistent estimator?
049A fund has an expected annual return of 12% and annual volatility of 20%, on Rs 100 crore. What is its one-year 95% VaR with and without the expected return, and when does the mean matter?VaR and expected shortfallCoreAsset manager risk
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Including the 12% expected return, what is the one-year 95% VaR?
Show the worked solution
Rs 32.9 crore ignoring the mean and Rs 20.9 crore including it. The 95% cut is 1.645 standard deviations below the mean: 1.645 x 20% is 32.9%, and a 12% expected return lifts the cut to minus 20.9%. At one year the mean cuts VaR by more than a third. At one day it barely matters: Rs 2.07 crore against Rs 2.02 crore.
Why does the mean matter at one year but not at one day?
On a short walk the path you take wanders more than your average direction moves you; on a long journey the direction wins. Expected return grows in proportion to time while volatility grows with its square root, so over short horizons the mean is noise and over long ones it is material. At one day the mean is 12% / 252, about 0.048%, against a daily volatility of 1.26%; the VaR figures differ by 2.3%. At one year the gap is 36%.
Over one year a 12% expected return shifts the whole distribution right, moving the 5% cut from minus 32.9% to minus 20.9%, whereas over one day the same mean moves VaR from Rs 2.07 crore only to Rs 2.02 crore.
The relationship
$\text{VaR}_{95\%} = (1.645\,\sigma - \mu) \times W = (32.9\% - 12\%) \times 100 = 20.9$
\sigma annual volatility, 20%
\mu expected annual return, 12%
W portfolio value, Rs 100 crore
What it says in wordsVaR is how far the 5% worst outcome sits below zero: the volatility term pulls it down, the mean pushes it back up.
Which figure should a risk report quote?
Say which you are quoting, because both are used. Relative VaRVaR measured from the expected outcome rather than from zero, so it captures only the uncertainty and ignores the expected gain. ignores the mean and measures pure uncertainty; absolute VaR includes it and measures the loss from today's value. For a one-year horizon, many risk teams quote the zero-mean figure deliberately: a 12% expected return is an assumption, and counting it as a cushion lets an optimistic forecast shrink the risk number. Daily trading VaR usually ignores the mean for the simpler reason that it makes no difference.
Limits to name: over a year, compounding and fat tails matter, so a normal model with a constant 20% volatility understates the chance of a large loss, and volatility itself changes over the year. The one-year figure is a rough guide to the size of a bad year, not a promise about it.
Where candidates lose it
The most common slip is adding the mean to the loss, giving 44.9, which moves the cut the wrong way. The second is ignoring the question's own hint and giving only 32.9.
The quieter miss is not answering the when part. The square-root rule for volatility against the linear growth of the mean is the one sentence the interviewer is waiting for.
What the interviewer asks next
Over what horizon does the mean cut VaR by half?
Why might a regulator prefer the zero-mean figure?
How would you compute one-year VaR if returns were lognormal?
050An EWMA volatility model with a decay factor of 0.94 had yesterday's daily volatility estimate at 1.2%. Today's return is minus 3%. What is the updated volatility estimate?Statistics and estimationCoreBank market riskQuant risk
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Roughly where does the new daily volatility land?
Show the worked solution
About 1.38%. EWMA updates the variance, not the volatility: new variance is 0.94 x 1.2 squared plus 0.06 x 3 squared, which is 1.3536 plus 0.54, or 1.8936. The square root is 1.376%. The shock has only a 6% weight but enters squared, so it supplies 29% of the new variance.
Why blend variances rather than volatilities?
A household tracking how much its grocery bill swings would be misled if it averaged the size of the swings in rupees but ignored that one big swing matters far more than several small ones. Variance is the average squared move, so the model updates squared returns, and a move 2.5 times normal size counts 6.25 times as much. An EWMAExponentially weighted moving average: each day the estimate keeps a fixed share of yesterday and adds the rest from today, so older days fade geometrically. of variances is what makes one large shock move the estimate quickly. Blending the volatilities directly gives 1.31%, understating the jump.
Today's minus 3% return carries only a 6% weight in the EWMA update, but because it enters squared it supplies 28.5% of the new variance of 1.8936, lifting daily volatility from 1.20% to 1.38%.
The relationship
$\sigma_t^2 = \lambda\,\sigma_{t-1}^2 + (1-\lambda)\,r_t^2 = 0.94(1.44) + 0.06(9) = 1.8936, \quad \sigma_t = 1.376\%$
\lambda the decay factor, 0.94
\sigma_{t-1} yesterday's volatility estimate, 1.2%
r_t today's return, minus 3%
What it says in wordsKeep 94% of yesterday's variance, add 6% of today's squared return, then take the square root.
What happens next, and what are the model's limits?
If tomorrow is flat, the estimate decays to the square root of 0.94 x 1.8936, about 1.33%. Each day of calm keeps 94% of the variance, so a shock's influence halves in about 11 trading days: ln 0.5 over ln 0.94. That is the design choice behind 0.94: fast enough to react to a new regime within days, slow enough not to swing on every move.
The limits: EWMA has no pull towards a long-run average, so after a calm spell it can drift very low and understate risk just before volatility returns, which GARCH-type models address with a mean-reversion term. It also treats up and down moves alike, while falling markets often raise volatility more than rising ones. And the choice of 0.94 is a convention for daily data, not a law, so the estimate should be backtested against realised moves.
Where candidates lose it
The usual slip is to blend the two volatilities directly, 0.94 x 1.2 plus 0.06 x 3, and answer 1.31%. It looks like the formula but applies it to the wrong quantity and understates the effect of the shock.
The second is forgetting the square root at the end and quoting 1.89% as a volatility. Say the units at each step: variance in, variance out, then volatility.
What the interviewer asks next
What would the estimate be with lambda of 0.97 instead?
How many quiet days until the estimate is back below 1.25%?
Why might a risk manager prefer GARCH to EWMA for a ten-day VaR?

\sigma	standard deviation of the annual loss, Rs crore
0.02	default probability of each loan

N	face value of each leg, Rs crore
DV01	price change per 100 of face for a one basis point yield move

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

0.75	the share of the population who are adults
0.30	the share of adults who withdraw once a week
5	weekdays the withdrawals are spread across

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

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

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Why size by DV01 rather than by face value?

What does the trade make, and what can still go wrong?

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How do you structure the estimate so the interviewer can follow it?

How do you sanity check Rs 54 crore?

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Why does a threshold raise the recorded average?

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Why does a small daily edge grow so much?

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Why does dividing by n come out too low?

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Why does the mean matter at one year but not at one day?

Which figure should a risk report quote?

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Why blend variances rather than volatilities?

What happens next, and what are the model's limits?

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