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Risk Management puzzles, solved step by step

Puzzles
100
Traced to a firm
17
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13
Hard
30
Topic
All topicsCapital and leverage6Compounding and drawdowns8Correlation and diversification8Counterparty exposure and collateral7Credit risk arithmetic10Duration and rates7Liquidity and balance sheet7Logic, estimation and brainteasers7Operational loss and fraud7Options and Greeks7Probability and base rates8Statistics and estimation10VaR and expected shortfall8
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Showing 81–90 of 100
  1. 081A bond has a modified duration of 7 and convexity of 60. Yields rise 100 basis points. Estimate the price change with and without convexity.Duration and ratesCoreTreasury and ALMBank market risk

    Try it first

    Does convexity make the loss bigger or smaller than the duration estimate?

    Show the worked solution

    Duration alone gives a -7.00% fall; adding convexity gives -6.70%. Duration is the straight-line estimate: minus 7 times 1%. Convexity adds half of 60 times 1% squared, which is 0.30%. So the bond falls about 6.7%, not 7%. The gap looks small at 100 basis points, but it grows with the square of the move.

    What does each number measure?

    Think of a car's speedometer and its acceleration. Speed tells you how far you will go in the next minute if nothing changes; acceleration tells you how the speed itself is changing. Duration is the speedometer of a bond: the percentage price change for a small yield move. Convexity is the acceleration: how duration itself shifts as yields move. A modified duration of 7 means about 7% of price per 1% of yield, and a convexityThe second-order sensitivity of a bond price to yield; it measures how much the price curve bends away from the duration line. of 60 says the price curve bends away from that straight line.

    The relationship
    ΔPP≈−D Δy+12C(Δy)2=−7(0.01)+12(60)(0.01)2=−7.00%+0.30%\frac{\Delta P}{P} \approx -D\,\Delta y + \tfrac{1}{2} C (\Delta y)^2 = -7(0.01) + \tfrac{1}{2}(60)(0.01)^2 = -7.00\% + 0.30\%
    Dmodified duration, 7
    Cconvexity, 60
    Delta ychange in yield, 0.01
    What it says in wordsThe price change is the straight-line duration estimate plus a convexity term that grows with the square of the yield move.
    Duration draws a straight line; the bond's price curve bows above it-4%-2%0+2%+4%80100120Change in yieldPrice, from 100bond price curve:bows above the lineduration line:straight, too pessimistic+100 bpAt +100 bp, zoomed (axis starts at -5%)Duration only-7.00%Convexity adds+0.30%Estimate-6.70%-5%-7 x 1% + 0.5 x 60 x (1%) squared= -7.00% + 0.30% = -6.70%
    The bond's price curve bows above the straight duration line whichever way yields move; at plus 100 basis points, duration predicts -7.00% and convexity adds back 0.30%, an estimated fall of -6.70%.

    When does the convexity term stop being a rounding error?

    Square the move and see. At 100 basis points convexity is worth 0.30%; at 300 basis points it is nine times that, 2.70%, against a duration effect of 21%. Because the convexity term grows with the square of the move, it is small for daily risk and large in a stress scenario. A treasury stress test that uses duration alone will overstate losses on a large rate rise for a plain bond and understate them for a callable bond or a mortgage book, where convexity is negative.

    Close with the sign. Positive convexity is something a bond holder pays for through a slightly lower yield, and negative convexity, from options sold to borrowers, is what makes prepayable loans harder to hedge.

    Where candidates lose it

    The usual slip is forgetting the half in the convexity term and adding 0.60%, which gives minus 6.4%. Write the formula before the numbers so the half is on the page.

    The second is subtracting convexity because yields rose. The convexity term is a square, so it is positive for a rise and a fall alike.

    What the interviewer asks next

    • What is the price change if yields fall 100 basis points instead?
    • Why does a callable bond have negative convexity at low yields?
    • How would you hedge the convexity of a mortgage book?
  2. 082A Rs 200 crore bond position has a one-day 99% VaR of Rs 3 crore. The bid-ask spread on the bond is 40 basis points. What is the liquidity-adjusted VaR?Liquidity and balance sheetCoreBank market riskTreasury and ALM

    Try it first

    How much does the spread add to the Rs 3 crore VaR?

    Show the worked solution

    About Rs 3.4 crore: the Rs 3 crore VaR plus Rs 0.4 crore to get out. The position is marked at mid, but selling means taking the bid, half the spread below mid. Half of 40 basis points is 0.20%, and 0.20% of Rs 200 crore is Rs 0.4 crore. The exit cost adds 13% to the risk number, and more if spreads widen in a stress.

    Why is ordinary VaR missing a cost?

    A second-hand car dealer will quote you two prices for the same car: what he pays and what he sells for. Your car is worth the middle on paper, but if you need cash today you get the lower one. VaR is computed on mid prices, so it measures how far the value might move, not what it costs to actually leave the position. For a liquid government bond the difference is small; for a corporate bond the spread can be a meaningful share of the risk.

    Getting out costs half the spread, even on a day prices do not moveask 100.20mid 100.00bid 99.80you sell hereOne bond quote, price per 100Half spread = 0.40% / 2 = 0.20%0.20% x Rs 200 cr = Rs 0.4 cr3.03.0Market VaR3.03.4Normal spread3.0+1.24.2Stressed spread+0.4Rs crore, one day, 99%
    Selling at the bid costs half the 40 basis point spread, 0.20% of Rs 200 crore or Rs 0.4 crore, which lifts a Rs 3.0 crore market VaR to a liquidity-adjusted Rs 3.4 crore, and to Rs 4.2 crore if the spread widens to 120 basis points.
    The relationship
    LVaR=VaR+12×s×P=3.0+0.5×0.0040×200=3.4\text{LVaR} = \text{VaR} + \tfrac{1}{2} \times s \times P = 3.0 + 0.5 \times 0.0040 \times 200 = 3.4
    VaRmarket VaR on mid prices, Rs 3 crore
    sbid-ask spread as a fraction of price, 0.40%
    Pposition value, Rs 200 crore
    What it says in wordsAdd the cost of crossing half the spread to the market VaR, because that cost is paid even if prices do not move.

    What makes the adjustment larger than it looks?

    Two things, and both arrive in a stress. Spreads widen exactly when you need to sell, and a large position moves the price against you as you sell it. If the spread triples to 120 basis points, the exit cost is Rs 1.2 crore and the adjusted figure is Rs 4.2 crore, 40% above plain VaR. A risk team would also ask how many days it takes to exit Rs 200 crore without moving the market; if the answer is five days rather than one, the market VaR itself should be scaled to that horizon.

    Say the limit of the simple version: it treats the spread as a fixed number. A fuller treatment uses the spread's own volatility, adding a multiple of its standard deviation, so the adjustment reflects how bad the spread gets in a bad week, not how it looks on an average one.

    Where candidates lose it

    The common mistake is adding the full 40 basis points, Rs 0.8 crore. Positions are marked at mid, so selling costs only the distance from mid to bid, half the spread.

    The second is saying VaR already includes liquidity because it uses market prices. It uses mid prices and a one-day horizon, and assumes you could exit at mid; the question is testing whether you see that gap.

    What the interviewer asks next

    • The bond takes five days to sell without moving the price. How would you change the VaR?
    • Why might a desk argue against a liquidity add-on for government bonds?
    • How would you estimate a bid-ask spread for a bond that rarely trades?
  3. 083You have eight identical-looking balls, one of them slightly heavier, and a two-pan balance. What is the fewest number of weighings that guarantees you find the heavy ball?Logic, estimation and brainteasersCoreBank market riskRisk GCC

    Try it first

    What is the minimum number of weighings that always works?

    Show the worked solution

    Two weighings. Put three balls on each pan and leave two aside. If one pan drops, the heavy ball is among its three: weigh one against another, and a balance points to the third. If the pans balance, weigh the two set aside against each other. A balance has three outcomes, so two weighings separate up to nine balls.

    Why is halving the wrong instinct?

    A guessing game where someone answers only yes or no halves the possibilities with each question. A balance answers in three ways: left heavier, right heavier or level. Each weighing has three outcomes, so the best split is into three groups, and the balanced outcome is information, not a wasted turn. Four against four uses only two of the three outcomes, because the pans can never balance, and so it wastes a third of what the scale can tell you.

    Three outcomes per weighing, so split into three groupsWeighing 1: 1 2 3 against 4 5 67 and 8 wait on the tableleft dropsWeigh 1 against 2heavy in 1, 2, 3ball 1left dropsball 3balanceball 2right dropsbalanceWeigh 7 against 8heavy is 7 or 8ball 7left dropsball 8right dropsright dropsWeigh 4 against 5heavy in 4, 5, 6ball 4left dropsball 6balanceball 5right drops8 balls, 9 possible endings after two weighings: 3 x 3 = 9, which is at least 8
    Weighing 1, 2 and 3 against 4, 5 and 6 sends every case down one of three branches, and a second weighing within each branch names the heavy ball, so eight balls are always solved in two weighings.

    How do you prove two is the minimum and find the limit?

    One weighing has three outcomes, and eight balls need eight different answers, so one weighing cannot be enough. Two weighings have 3 times 3, nine outcomes. With w weighings you can separate at most 3 to the power w balls, so nine balls also need only two, and a tenth ball needs a third weighing. Saying the bound, not only the procedure, is what turns a trick into reasoning.

    The relationship
    3w≥n⇒w=⌈log⁡38⌉=23^{w} \ge n \quad\Rightarrow\quad w = \lceil \log_3 8 \rceil = 2
    wnumber of weighings
    nnumber of balls, 8
    3outcomes of one weighing
    What it says in wordsYou need enough weighings that three to the power of that number covers every ball.

    Why would a risk interviewer ask this?

    It tests whether you count what a test can tell you before you run it. The same habit applies to a control check: a review that can only return pass or fail tells you less than one that also flags items for a closer look. Name the three outcomes first and the answer follows.

    Where candidates lose it

    Most candidates split four against four, then two against two, then one against one, and answer three. It works, but it is not the minimum, because a balance never happens in the first two steps.

    The other loss is finding two weighings without saying why one is impossible. Give the counting bound: one weighing, three outcomes, fewer than eight balls.

    What the interviewer asks next

    • What is the most balls you can handle in three weighings?
    • Now you do not know whether the odd ball is heavier or lighter. How many weighings for twelve balls?
    • What if the scale shows the weight difference rather than just which side drops?
  4. 084An erroneous payment must pass three independent checkers, each of whom catches 80% of errors. What share of errors gets through, and why is the real figure likely to be higher?Operational loss and fraudCoreOperational risk

    Try it first

    Out of 100 errors, how many get past all three checkers if they really are independent?

    Show the worked solution

    About 0.8% get through if the checkers are truly independent. Each misses 20% of what reaches them, so 100 errors become 20, then 4, then 0.8. The real figure is likely higher because the checkers are not independent: later checkers relax because someone already looked. If checkers two and three catch only 40%, 7.2% get through, nine times as many.

    Why do independent checks multiply?

    Think of three sieves stacked on top of each other. If each lets one grain in five through, the first passes 20 of 100 grains, the second 4 of those 20, and the third 0.8 of the 4. When checks are independent, the chance of an error getting past all of them is the product of each one's miss rate. That is why a three-check process looks almost watertight on paper: 0.2 cubed is 0.008.

    Independent checks multiply; checkers who lean on each other do notIndependent: each catches 80%1002040.8Checkers 2 and 3 lean on checker 1: they catch 40%100Errors made20After checker 112After checker 27.2After checker 30.2 x 0.2 x 0.2= 0.8%get through0.2 x 0.6 x 0.6= 7.2%get through
    Independent checkers catching 80% each let 100 errors shrink to 20, 4 and then 0.8, but if the second and third checkers lean on the first and catch only 40%, 100 errors shrink to 20, 12 and 7.2, nine times as many.

    Why is the real leakage higher than 0.8%?

    Because people who know others are checking check less carefully. The second checker assumes the first caught the obvious problems, and the third signs off because two colleagues already did. Checks that rely on each other are not independent, and once they are correlated the multiplication rule overstates how much they catch. There is also a common cause: an error that looks like a normal payment, a correct-looking amount to a familiar name, fools all three checkers for the same reason.

    The relationship
    P(leak)=∏i(1−ci)=0.2×0.6×0.6=0.072P(\text{leak}) = \prod_i (1 - c_i) = 0.2 \times 0.6 \times 0.6 = 0.072
    c_ithe catch rate of checker i
    0.2the first checker's miss rate
    0.6the miss rate of a checker who leans on the one before
    What it says in wordsThe share of errors that gets through is the product of the miss rates, so a drop in the later checkers' care multiplies straight into the leak.

    What would an operational risk team do about it?

    Measure each check on its own, not the chain. Seed known errors into the flow and count what each checker catches; rotate who checks first; give each checker a different thing to look for rather than the same whole payment. The goal is to make the checks genuinely independent, because three checks that behave like one are more dangerous than one check everyone knows is the only one.

    Where candidates lose it

    The first slip is adding: three checks at 80% sound like 240% coverage, or candidates subtract 3 times 20% and get a negative. Multiply the miss rates, not the catch rates.

    The bigger miss is stopping at 0.8%. The interviewer asked why the real figure is higher; name reliance between checkers and a common cause, and give a number for how much it matters.

    What the interviewer asks next

    • How would you test whether your checkers really are independent?
    • Would you rather have three checkers at 80% or one automated check at 99%?
    • An error gets through all three checks. How do you run the review?
  5. 085A rule of thumb says an at-the-money option is worth about 0.4 times volatility times the square root of time times the price. What is a three-month at-the-money call on a Rs 1,000 stock with 20% volatility worth?Options and GreeksWarm upBank market riskQuant risk

    Try it first

    Roughly what is the premium?

    Show the worked solution

    About Rs 40. The typical one-year move is 20% of Rs 1,000, Rs 200. Over three months it is Rs 200 times the square root of 0.25, Rs 100. The call is worth about 0.4 of that typical move, Rs 40. A full Black-Scholes calculation at zero rates gives Rs 39.88, so the rule is within a rupee.

    Why is the premium a slice of the typical move?

    Imagine insuring a shop's daily takings against a bad day. The premium depends on how much the takings usually swing, not on how large the takings are. An at-the-money option pays out on the upside half of the stock's moves, so its value is proportional to the size of a typical move over the option's life, not to the stock price itself. The price enters only through converting volatility into rupees.

    An at-the-money premium is a slice of the typical move, not of the priceShare priceRs 1,000x volatility 20%Rs 200a typical move over one yearx sqrt(0.25) = 0.5Rs 100a typical move over three monthsx 0.4Rs 40the at-the-money call premiumWhy 0.4? The average payoff of the upside half of a normal move is 1 / sqrt(2 x pi) = 0.399 of one standard deviation.Black-Scholes at zero rates: Rs 39.88. The rule is within 0.3%.
    Rs 1,000 times 20% volatility is a Rs 200 one-year move, times the square root of 0.25 is a Rs 100 three-month move, and 0.4 of that is a Rs 40 premium, within 0.12 of the Black-Scholes value of Rs 39.88.
    The relationship
    CATM≈0.4 σT S=0.4×0.20×0.25×1000=40C_{ATM} \approx 0.4\, \sigma \sqrt{T}\, S = 0.4 \times 0.20 \times \sqrt{0.25} \times 1000 = 40
    sigmaannual volatility, 20%
    Ttime to expiry in years, 0.25
    Sshare price, Rs 1,000
    0.4about 1 over the square root of 2 pi
    What it says in wordsMultiply the price by volatility and the square root of time to get a typical move, then take 0.4 of it.

    Where does 0.4 come from, and when does the rule fail?

    If the stock's move is roughly normal with standard deviation of one typical move, the average payoff from the upside half is that deviation divided by the square root of 2 pi, 0.399. The 0.4 is not a fudge factor; it is the average size of the positive half of a normal move. The rule is built for at-the-money options with short expiries and low rates. It fails for options far in or out of the money, where the payoff is mostly intrinsic value or mostly zero, and over long horizons, where interest rates and the skew of returns start to matter.

    A risk manager uses this to sanity-check a trader's mark in ten seconds. If a three-month at-the-money call on a Rs 1,000 stock is marked at Rs 70, the mark implies volatility of about 35%, which is a question worth asking.

    Where candidates lose it

    The common error is scaling time linearly: a quarter of a year, so a quarter of Rs 200, then 0.4 of Rs 50 gives Rs 20. Volatility scales with the square root of time, so three months is half a year's move.

    The other slip is applying the rule to a deep out-of-the-money option. Say it is an at-the-money shortcut and name where it breaks.

    What the interviewer asks next

    • What is the matching at-the-money put worth at zero rates?
    • How much does the premium rise if volatility doubles, and if time to expiry doubles?
    • A trader marks the call at Rs 70. What implied volatility is that?
  6. 086Ten fund managers each have a 50% chance of beating the index in any year, with no skill at all. What is the probability that at least one of them beats it five years running?Probability and base ratesCoreAsset manager risk

    Try it first

    Roughly how likely is it that at least one of ten coin-flipping managers has a five-year streak?

    Show the worked solution

    About 27.2%. One manager beats the index five years running with probability one half to the fifth, 1 in 32. The chance that none of the ten does it is 31/32 to the power 10, about 0.728. So at least one streak appears 27.2% of the time with no skill anywhere. With 100 managers it is 95.8%.

    Why is a streak among many managers weak evidence?

    Ask a hall of 300 people to toss a coin five times, and about nine will throw five heads. Nobody would call them skilled tossers. The chance that one named person has a streak is small, but the chance that someone in a crowd has one is large, and a fund manager with a five-year record is usually picked from a crowd. Which question you are answering decides everything: the probability for manager G, or the probability for whoever turned out to have the streak.

    With enough managers, a five-year streak is what luck looks likeY1Y2Y3Y4Y5ABCDEFGfive in a rowHIJbeat indexdid notChance of a five-year streak by luck aloneOne manager1 in 32 = 3.1%At least one of 1027.2%At least one of 10095.8%1 - (31/32) to the power 10= 1 - 0.728 = 27.2%
    Among ten coin-flipping managers, one happens to beat the index all five years; the chance for any single manager is only 1 in 32, but at least one of ten does it 27.2% of the time and at least one of 100 does it 95.8% of the time.
    The relationship
    P(at least one)=1−(1−132)10=1−0.728=0.272P(\text{at least one}) = 1 - \left(1 - \tfrac{1}{32}\right)^{10} = 1 - 0.728 = 0.272
    1/32one manager beating the index five years in a row by chance
    10the number of managers
    What it says in wordsWork out the chance that nobody has a streak, then take it away from one.

    How do you get there fast without a calculator?

    Use the expected count first: ten managers times 1 in 32 is about 0.31 streaks. When the expected count is small, the chance of at least one is a little below it, so 31% is an upper bound and the exact answer is about 27%. The gap comes from the chance of two or more streaks, which the simple sum counts twice. Saying the bound and then the exact figure shows the interviewer you can check your own work.

    What does an asset manager's risk team do with this?

    It sets the bar for evidence. A five-year record of beating the index is common among hundreds of funds even if none has skill, so the team looks at how the returns were earned: the size of the edge against its volatility, whether it comes from one bet or many, and whether the process explains the outcome. Survivorship matters too: the managers whose streaks broke often closed their funds, so the crowd you see is already filtered toward winners.

    Where candidates lose it

    The fast wrong answer is 3.1%, the chance for one named manager. The question asks about any of ten, and the interviewer is testing whether you notice the difference.

    The other miss is adding ten times 3.1% and calling it 31%. That overcounts the cases with two streaks; give it as a bound and then the exact figure.

    What the interviewer asks next

    • How many managers would you need before a ten-year streak by luck is more likely than not?
    • A fund's marketing shows five straight years of beating the index. What questions do you ask?
    • How does survivorship bias change these numbers?
  7. 087A desk's daily returns have a volatility of 1%. Over 250 days its average daily return is 0.05%. How precisely is that mean estimated, and can you say the desk has skill?Statistics and estimationCoreAsset manager riskQuant risk

    Try it first

    Is 0.05% a day, measured over one year, clearly different from zero?

    Show the worked solution

    Not precisely enough to claim skill. The standard error of the mean is the daily volatility over the square root of the number of days: 1% over the square root of 250, about 0.063%. The estimate of 0.05% is only 0.79 standard errors from zero, and a two standard error band runs from -0.076% to 0.176%. You would need about 1,600 days, 6.4 years, to clear zero.

    Why is a year of daily data not enough?

    Weigh yourself on a bathroom scale that jumps by two kilos each time you step on it. If you lost 100 grams last month, a week of readings will not show it; the jumps drown the signal. The precision of an average improves only with the square root of the number of observations, so a small edge buried in large daily noise takes years to show. Here the daily noise is 1% and the edge is 0.05%, twenty times smaller.

    The relationship
    SE=σn=1%250=0.063%t=0.050.063=0.79SE = \frac{\sigma}{\sqrt{n}} = \frac{1\%}{\sqrt{250}} = 0.063\% \qquad t = \frac{0.05}{0.063} = 0.79
    sigmadaily return volatility, 1%
    nnumber of daily observations, 250
    thow many standard errors the mean is from zero
    What it says in wordsThe mean's uncertainty is the daily volatility divided by the square root of the number of days, and the edge is less than one of those units from zero.
    One year of data cannot tell a 0.05% edge from zerozero: no skill250 daysone year-0.076%+0.176%1,600 daysabout 6.4 years0.000%+0.100%estimate 0.05%-0.10%0+0.10%+0.20%Average daily return, with a two standard error band
    After 250 days the two standard error band around the 0.05% daily mean runs from -0.076% to 0.176% and straddles zero; after about 1,600 days, 6.4 years, the band narrows to 0.000% to 0.100% and only just clears zero.

    How long would it take, and what does that mean for judging desks?

    Set the t-statistic to 2 and solve for n: n equals (2 times 1% over 0.05%) squared, which is 1,600 days. An annual Sharpe ratio of about 0.79 needs more than six years of data before it is statistically distinguishable from zero. That is longer than most desks keep the same strategy, so a risk team cannot rely on the P&L record alone; it looks at whether the edge has a reason, whether it survives out of sample, and how much the result depends on a few days.

    State the assumptions. The calculation treats daily returns as independent with constant volatility. Fat tails and volatility clustering make the true uncertainty larger, so six years is a floor, not a promise.

    Where candidates lose it

    The common error is annualising the mean to 12.5% and declaring skill, as if a big annual number were proof. The annual volatility grows too, to about 16%, and the ratio of the two is what matters.

    The other slip is dividing by 250 instead of its square root, which makes the mean look fifty times more precise than it is.

    What the interviewer asks next

    • The desk's volatility is 0.5% instead. How many days now?
    • Why is the mean so much harder to estimate than the volatility?
    • How would you judge a new desk that has only six months of history?
  8. 088A treasury holds Rs 500 crore of government bonds with a modified duration of 6. Daily changes in yield have a standard deviation of 6 basis points. What is the one-day 99% VaR?VaR and expected shortfallCoreUBSAnonymous employee in · 2020

    Try it first

    What is the position's DV01, the loss for a one basis point rise in yield?

    Show the worked solution

    About Rs 4.19 crore. The position loses Rs 500 crore times 6 times 0.0001, Rs 30 lakh, for each basis point rise in yield. A 99% one-day rise is 2.33 times 6 basis points, about 14 basis points. Rs 30 lakh times 13.96 is Rs 4.19 crore. It assumes normal yield changes and a linear price response.

    Why start from DV01 instead of a price volatility?

    A taxi fare is a rate per kilometre times the distance. You would not guess the fare directly; you would multiply. For a bond, DV01 is the rupee rate per basis point and the yield move is the distance, so VaR is DV01 times the yield move at the chosen confidence. Yield volatility is what the market data gives you, and duration converts it to rupees. Guessing a price volatility for the bond skips the step the interviewer wants to see.

    Bond VaR = rupees per basis point x the 99% yield movePositionRs 500 cr, duration 6DV01Rs 30 lakh per bp99% yield move2.33 x 6 bp = 14.0 bpOne-day 99% VaRRs 4.19 crore500 x 6 x 0.0001 = 0.30 crore = Rs 30 lakh a bp0.30 x 13.96 = 4.19+14.0 bpworst 1% of days-18-12-60+6+12+18Daily change in yield, bp (standard deviation 6)yields up = bond price down,so the right tail is the loss
    A Rs 500 crore position with duration 6 loses Rs 30 lakh per basis point; a 99% daily yield rise is 2.33 times 6 basis points, 14.0 basis points, so the one-day 99% VaR is Rs 4.19 crore, the loss on the worst 1% of days.
    The relationship
    VaR99=P⋅D⋅0.0001⋅z99⋅σbp=500×6×0.0001×2.33×6=4.19\text{VaR}_{99} = P \cdot D \cdot 0.0001 \cdot z_{99} \cdot \sigma_{bp} = 500 \times 6 \times 0.0001 \times 2.33 \times 6 = 4.19
    Pposition value, Rs 500 crore
    Dmodified duration, 6
    z2.33, the one-sided 99% point
    sigma_bpdaily standard deviation of yield, 6 bp
    What it says in wordsMultiply rupees lost per basis point by the yield rise that is exceeded only one day in a hundred.

    What does this number leave out?

    Three things, and a treasury risk manager names them unprompted. The estimate assumes yield changes are normal, that the price responds in a straight line, and that every bond in the book moves with the same yield. Fat tails make a 14 basis point day more common than the normal says. Convexity makes the true loss slightly smaller than the linear figure. And a book spread along the curve has curve risk: if short yields rise and long yields do not, one DV01 figure misses it. Over ten days, the square-root rule would scale this to about Rs 13.2 crore, if daily moves are independent.

    Also say which way hurts. A holder of bonds loses when yields rise, so the one-sided 99% point on the upside of yields is the one that matters, which is why the figure uses 2.33 and not the two-sided 2.58.

    Where candidates lose it

    The usual slip is a units error: forgetting that a basis point is 0.0001 and producing a VaR a hundred times too large or too small. Say DV01 out loud first, Rs 30 lakh a basis point, and the rest follows.

    The other is using 2.58 because 99% sounds like a two-sided number. VaR is a one-sided loss measure, so the multiplier is 2.33.

    What the interviewer asks next

    • What is the 99% expected shortfall under the same normal assumption?
    • The book holds 2-year and 10-year bonds with the same total DV01. What risk does one number hide?
    • How would convexity change the VaR for a 300 basis point stress?

    Asked at UBS, Risk Management, Anonymous employee in, 2020 (Wall Street Oasis): Calculate VAR

  9. 089Bank A reports a CET1 ratio of 14%, with risk-weighted assets equal to 25% of its total assets. Bank B reports 11%, with risk-weighted assets at 55% of total assets. Which bank has more equity per rupee of assets?Capital and leverageHardBank credit riskRating agency

    Try it first

    Which bank holds more equity for every rupee of assets?

    Show the worked solution

    Bank B, with about 6.05% of assets in equity against 3.5% for bank A. Equity to assets is the CET1 ratio times the share of assets that count as risk-weighted: 14% times 25% for A, 11% times 55% for B. A's higher ratio rests on low risk weights, so it is levered about 29 times against B's 16.5 times.

    Why can a higher capital ratio mean less capital?

    Two people each say they save 20% of their income. One counts only the salary left after rent; the other counts the whole salary. The first sounds equally thrifty but saves far fewer rupees. A CET1 ratio divides equity by risk-weighted assets, not total assets, so a bank that assigns low risk weights to its loans can show a high ratio on a thin equity base. The share of assets that ends up risk-weighted is called {term('RWA density', 'Risk-weighted assets divided by total assets; a low figure means the bank judges most of its assets to be low risk.')}, and it is the number that reconciles the two views.

    The relationship
    EA=ERWA×RWAAA:14%×25%=3.5%B:11%×55%=6.05%\frac{E}{A} = \frac{E}{RWA} \times \frac{RWA}{A} \qquad A: 14\% \times 25\% = 3.5\% \qquad B: 11\% \times 55\% = 6.05\%
    ECET1 equity
    RWArisk-weighted assets
    Atotal assets
    What it says in wordsEquity per rupee of assets is the CET1 ratio multiplied by the RWA density.
    Divide by assets instead of risk-weighted assets and the ranking flips14.00%Bank Ahigher11%Bank BCET1 ratio (equity / RWA)3.50%Bank A6.05%Bank BhigherEquity / total assetsRisk weightsA: RWA = 25% of assetsB: RWA = 55% of assets14% x 25% = 3.5%
    Bank A's CET1 ratio of 14% beats bank B's 11%, but because A's risk-weighted assets are only 25% of its assets against B's 55%, A holds 3.5% of assets in equity and B holds 6.05%, and the ranking flips.

    Which number should a credit analyst trust?

    Both, because they answer different questions. The CET1 ratio is right if the risk weights are right. Equity to assets makes no judgement about risk, so it is the check on whether the risk weights are doing too much of the work. Test A's number: if its true risk density were 40% rather than 25%, perhaps because its internal models are optimistic, its CET1 ratio would be 3.5% over 40%, only 8.75%. That is why regulators set a leverage ratio floor alongside risk-based ratios; the current minimum is set by the regulator, so confirm it before quoting it.

    What would make bank A's low density legitimate?

    A book of home loans with low loan-to-value ratios, or large holdings of government bonds, genuinely carries low risk weights, and a 25% density can be honest. The question is whether the weights are earned. An analyst asks what the assets are, whether the density has fallen while the book stayed the same, and how the bank's model-based weights compare with the standardised ones.

    Where candidates lose it

    Most candidates answer bank A, reading the higher ratio as more capital. The question is built to see whether you ask what the denominator is.

    The second miss is getting B and then calling A unsafe. A low density can be legitimate; say what you would check before concluding the weights are too low.

    What the interviewer asks next

    • What happens to bank A's CET1 ratio if its risk weights rise to the level of bank B's?
    • Why do regulators set a leverage ratio at all if risk-weighted ratios exist?
    • Which kinds of assets carry low risk weights, and which of them surprised people in past crises?
  10. 090A fund's NAV over six observations is 100, 120, 90, 130, 100 and 140. What is its maximum drawdown?Compounding and drawdownsWarm upAsset manager risk

    Try it first

    What is the maximum drawdown?

    Show the worked solution

    25%. Drawdown is measured from the highest value reached so far. The fund peaks at 120 and falls to 90, a 25% drop. It then peaks at 130 and falls to 100, a 23.1% drop. The larger of the two, 25%, is the maximum drawdown. Measured from the start the worst point looks like only 10%, which understates the pain.

    Why measure from the running peak?

    If your savings climbed to Rs 1.2 lakh and then fell to Rs 90,000, you would not console yourself that you started with Rs 1 lakh. You lost Rs 30,000 of money you had. Drawdown measures the fall from the highest value an investor has held so far, because that is the loss an investor who bought at the top actually suffers. The running peak resets upward each time the fund makes a new high, and each drawdown is measured against it.

    Drawdown is measured from the running peak, not from the start8010012014010012090130100140120 to 90-25.0%130 to 100-23.1%from the start: only -10%running peak (dashed)t0t1t2t3t4t5ObservationNAV
    The fund's NAV falls 25.0% from its peak of 120 to 90 and 23.1% from its later peak of 130 to 100, so the maximum drawdown is 25%, although the worst point measured from the start of 100 is only 10% down.
    The relationship
    DDt=Vtmax⁡s≤tVs−1MDD=min⁡tDDt=90120−1=−25%DD_t = \frac{V_t}{\max_{s \le t} V_s} - 1 \qquad MDD = \min_t DD_t = \frac{90}{120} - 1 = -25\%
    V_tNAV at time t
    max V_sthe running peak up to time t
    MDDmaximum drawdown, the deepest fall from a running peak
    What it says in wordsEach point's drawdown is how far it sits below the best value seen so far; the maximum drawdown is the deepest of them.

    What does the number not tell you?

    Two things worth saying. Maximum drawdown is a single worst episode, so it depends heavily on the sample: a longer history can only make it larger, never smaller. It also ignores time. The fall from 120 to 90 took one period and the recovery took one more; a fund that takes three years to climb out of a 25% hole is a different experience from one that recovers in a quarter. A risk team reports duration of the drawdown and time to recovery alongside the depth, and remembers that a 25% fall needs a 33.3% gain to get back.

    Where candidates lose it

    The fast wrong answer is 10%, measuring the lowest point, 90, against the start, 100. It ignores that investors held the fund at 120.

    The other slip is picking the most recent fall, 23.1%, because it is fresh, or measuring 130 to 90, which mixes a later peak with an earlier trough. The peak must come before the trough.

    What the interviewer asks next

    • What gain does the fund need to recover from its maximum drawdown?
    • Why is maximum drawdown hard to compare across funds with different track record lengths?
    • How would you combine drawdown with volatility in one risk-adjusted measure?
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