Quant puzzles, solved step by step

About 28.9%, roughly 29%. Under 2 and 20 the manager earns Rs 20 crore of management fee plus 20% of the remaining Rs 80 crore gain, Rs 36 crore in all. At 1% the management fee is Rs 10 crore and the gain left is Rs 90 crore, so the performance fee must bring in Rs 26 crore: 26 / 90 = 28.9%. It only balances at a 10% return.

Why is the answer not simply 30%?

Think of a shopkeeper who takes a fixed rent from a stall plus a fifth of whatever the stall-holder clears after paying that rent. If the rent is halved, the stall-holder clears more, and the shopkeeper's fifth is a fifth of a larger amount. Cutting the management fee enlarges the gain on which the performance fee is charged, so part of the lost management revenue comes back automatically. Here the old fee of 20% on Rs 90 crore earns Rs 18 crore, up from Rs 16 crore, and the true shortfall is Rs 8 crore, not Rs 10 crore. Adding 10 points to get 30% repays the shortfall twice over in part.

How do you set it up in one line?

Write revenue as management fee plus performance rate times the gain after the management fee, and hold it equal. The new rate is the old revenue less the new management fee, divided by the new fee base: (36 - 10) / 90 = 28.9%. The investor's position is the mirror image: gross gain Rs 100 crore less Rs 36 crore of fees leaves Rs 64 crore, a 6.4% net return under either set of terms, which is a good check that you have kept the total fixed.

What does the swap change, even when revenue matches?

The match holds at one return only. Swapping management fee for performance fee moves risk onto the manager: less is guaranteed, more depends on the year. At a 5% gross return the old terms pay Rs 26 crore and the new ones Rs 21.6 crore; at 20% the old terms pay Rs 56 crore and the new ones Rs 64.9 crore. State the convention too: if the performance fee were charged on the gross gain instead, the old revenue is Rs 40 crore and the new rate comes to exactly 30%. Say which convention you assume before giving a number.

Macaulay duration is 1.943 years and modified duration is 1.833. At par the yield equals the coupon, 6%. The year 1 coupon is worth 6 / 1.06 = 5.660 and the final payment 106 / 1.06^2 = 94.340, adding to 100. Weight each time by its share: 0.0566 x 1 + 0.9434 x 2 = 1.943. Divide by 1.06 for modified duration: a 1 point rise in yield cuts the price by about 1.83%.

What is Macaulay duration actually measuring?

Think of a seesaw with weights placed along it at the dates money arrives, each weight equal to what that payment is worth today. Macaulay duration is the point where that seesaw balances: the average time to your money, weighted by present value. A zero-coupon bond has all its weight at maturity, so its duration equals its maturity. Any coupon puts a little weight earlier and pulls the balance point forward. Here the coupon is small and the bond is short, so the pull is small, and that is the whole shape of the answer before any arithmetic.

How do you get the numbers fast in your head?

At par, the yield is the coupon, so discount at 6%. The coupon's present value, 6 / 1.06, is about 5.66, and the rest of the price, 94.34, sits at year 2, so duration is 2 minus the coupon's weight: 2 - 0.0566 = 1.943. That shortcut works for any two-period bond: start at the maturity and subtract the early weight times the time it saves. For a par bond there is also a closed form, (1 + y)/y times (1 - 1/(1 + y)^n), which gives the same 1.9434.

Why are there two durations, and how good is the estimate?

Macaulay duration is a time; modified duration is a slope. Modified duration is the percentage price change for a one point change in yield, so this bond loses about 1.83% if yields rise from 6% to 7%, about Rs 0.018 per Rs 100 for each basis point. Repricing exactly gives 98.192 at 7%, a fall of 1.808%, and 101.859 at 5%, a rise of 1.859%. The rise is bigger than the fall, which is convexityThe curvature of the price against yield: prices rise more when yields fall than they drop when yields rise by the same amount.; for a two-year bond it is a small correction, for a thirty-year bond it is not.

1/6. Let D be the distance to the nearest edge. D exceeds d exactly when the point lies in the inner square of side 1 - 2d, so P(D > d) = (1 - 2d)^2 for d up to 1/2. The expected value of a non-negative variable is the integral of its tail, and the integral of (1 - 2d)^2 from 0 to 1/2 is 1/6.

Why work with the chance of being far rather than the distance itself?

Think of a sandpit where a child stands at a random spot and the question is how far they are from the nearest edge. Writing the distance as min(x, 1 - x, y, 1 - y) and integrating a minimum of four things means splitting the square into four triangles. Asking instead when the point is farther than d from every edge has a one-picture answer: the point must lie in a smaller square, shrunk by d on every side. That square has side 1 - 2d, so its area, (1 - 2d)^2, is the tail probability. One formula replaces four cases.

How does the tail give the expectation?

For any non-negative random variable, the expected value equals the integral of the chance that it exceeds each level, E[D] = integral of P(D > d). Here that is the integral of (1 - 2d)^2 from 0 to 1/2. Substitute u = 1 - 2d and it becomes half the integral of u^2 from 0 to 1, which is 1/6. A simulation with 200,000 random points gives 0.1669, against the exact 0.1667.

How do you sanity-check 1/6 against simpler cases?

Build up the number of edges. The distance to one fixed edge averages 1/2, to the nearer of two opposite edges averages 1/4, and to the nearest of all four it falls to 1/6, so each added constraint pulls the minimum closer. The density of D is the slope of the tail, 4(1 - 2d), largest at the edge, which says most random points are near the boundary. That is the same reason most of the volume of a high-dimensional cube sits near its surface, a fact that matters when sampling scenarios in many risk factors at once.

110 paths avoid the block, so the probability is 110/210 = 11/21, about 52.4%. All shortest paths use 6 rights and 4 ups in some order: C(10,4) = 210. Paths through (3,2) combine 10 ways in with 10 ways out, 100 in all. Subtract, and 110 survive.

How do you count all the shortest paths?

A shortest path is a string of 10 moves with exactly 6 rights and 4 ups, like a ten-letter word made of R and U. Choosing which 4 of the 10 slots are ups fixes the path completely, so there are C(10,4) = 210 shortest paths. That is the whole sample space, and every one of those strings is equally likely by the question's rule.

Why multiply for the paths through the blocked point?

Think of a trip from home to the office with a stop at a coffee shop. If there are 10 routes to the shop and 10 routes from the shop to the office, there are 10 x 10 = 100 full trips, because every first half pairs with every second half. Paths through (3,2) split the same way: C(5,2) = 10 in and C(5,2) = 10 out, so 100 of the 210 pass through the block. Subtract and 110 remain.

The grid method in the figure is the check, and it is also what you would code. Each point's count is the count from the left plus the count from below, because the last step into any point came from one of those two neighbours. Setting the block to zero removes every path through it automatically, and the same method handles several blocks, where the subtraction formula needs inclusion and exclusion.

Does the answer change if the walker flips a coin at each step?

Yes, and interviewers often ask this next. If the walker flips a fair coin for right or up at each step, any visit to (3,2) happens on move five, and a walker still able to get there has not yet touched the top or right edge, so all five of those moves were free coin flips. The chance of standing on (3,2) after five flips is C(5,2)/2^5 = 10/32 = 5/16, so the coin-flip walker avoids the block with probability 11/16, about 68.8%, well above 11/21. Choosing uniformly among complete paths is not the same as flipping coins: conditioning on the end point (6,4) pulls paths towards the diagonal that leads there, and (3,2) sits on it. Say which model the question means before you answer.

Do not spin: pulling straight away survives with probability 3/4, against 2/3 with a spin. The click tells you the hammer sat on one of the four empty chambers. Because the two bullets are adjacent, the four empties form a run, and only the last empty in that run is followed by a bullet. A spin throws that information away and resets you to 4 empties out of 6.

What does the click actually tell you?

Think of a row of six houses where two neighbours keep dogs. You knocked at a random house and no dog barked. If you now try the next house along, you are only in trouble if you had knocked on the one house sitting just before the dogs. The click narrows the hammer's position to the four empty chambers, and the question becomes how many of those four have a bullet immediately after them. With the bullets side by side, the empties run 3, 4, 5, 6 in firing order, and only chamber 6 hands over to a bullet.

Does the answer depend on the bullets being adjacent?

Completely, and that is the follow-up most interviewers ask. If the two bullets are not next to each other, the empties split into two runs, two empties now sit in front of a bullet, and pulling straight away survives only 1/2, worse than the 2/3 of a spin. The same count works for bullets one apart or directly opposite: in both cases two of the four empties are followed by a bullet. So the rule is not spin or do not spin; it is count the empties that border a bullet and compare with a fresh spin.

What is the general lesson for a trading interview?

A random reset destroys information, and information has value only if the structure of the problem lets you use it. Here the structure is clustering: the bullets sit together, so a safe chamber is likely followed by another safe one. Markets have the same feature in volatility: a calm day tends to be followed by a calm day, so conditioning on what just happened beats assuming each day is a fresh draw. Say that link in one line after the arithmetic.

1 - 2/e, about 26.4%. Convolving two exponential densities gives the total the density x e^-x, which starts at zero and peaks at one minute. Its area below one minute is 1 - 2/e. The same number drops out of the Poisson view: the total is under a minute exactly when at least two arrivals land in the first minute of a rate-one Poisson process.

Why does adding two waits change the shape so much?

Suppose you need two buses, one after the other, and each arrives on average a minute after you reach its stop. Catching the first bus quickly is common; catching both quickly is rare, because both have to cooperate. A single exponential wait is most likely near zero, but a sum of two is almost never near zero, so its density starts at zero and rises into a hump. That shift of mass away from zero is why the answer is much smaller than the 63.2% chance that one wait is under a minute.

Is there a way to get 1 - 2/e without integrating?

Yes, and it is the cleaner answer to give aloud. Exponential waits with mean one are the gaps between arrivals of a Poisson process with rate one per minute. The second arrival comes before one minute exactly when at least two arrivals land in the first minute, and the Poisson chance of zero or one arrival is e^-1 + e^-1 = 2/e. So the answer is 1 - 2/e, about 26.4%, with no calculus at all.

Sanity-check the size. Both waits being under a minute has probability (1 - 1/e)^2, about 40.0%, and the total being under a minute is a stricter event, so the answer must be smaller: 26.4% is. The limitation is the independence assumption; if the two waits were driven by the same traffic, they would move together and the total would be more spread out.

The two largest drawdowns are 130 to 80, 38.5%, and 120 to 90, 25.0%. Walk through the prices once, keeping the highest price seen so far. At each step the drawdown is one minus price over that running peak; the maximum drawdown is the largest value seen. For the n largest, close an episode each time a new high is set, record its trough, and sort the episodes.

What exactly is a drawdown measured from?

Think of a hiker measuring how far below the highest point reached so far she has dropped. A drop only counts from a summit already climbed, not from a peak further along the trail. A drawdown is the fall from the running maximum to a later price, so the order of the prices matters and the overall high and low cannot simply be paired. Here the 140 comes after the 80, so the tempting 42.9% never happened.

How do you get the n largest drawdowns rather than just the worst?

Split the path into episodes. An episode opens at a running peak and closes when the price makes a new high; its size is the fall from that peak to the lowest price inside it. Here the 120 episode closes when the price reaches 130, with a trough of 90, so it is 25%. The 130 episode closes at 140 with a trough of 80, 38.5%. The 140 episode never closes, so report it as open at 20.0%. Sorting the episodes gives the n largest in one pass plus a sort, O(N log N) at worst, and a heap of size n keeps it at O(N log n).

Say the edge cases, because the interviewer is testing code judgement as much as arithmetic. Two drawdowns from the same peak must not be counted twice: the fall to 80 and the later level of 110 belong to one episode. An episode still open at the end of the data is real risk and should be reported with a flag. And drawdown on a price series is not the same as drawdown on a strategy's cumulative P&L, where you would use the equity curve, not the price.

About 4 years at a Sharpe of 1, and about 16 years at 0.5. The t-statistic for a mean return is the Sharpe ratio times the square root of the number of years, so reaching t = 2 needs (2 / SR)^2 years. Halving the Sharpe ratio quadruples the evidence you need, and sampling daily instead of yearly does not shorten it.

Why does the t-statistic come out as Sharpe times root years?

A t-test on a mean divides the average return by its standard error, which is the volatility over the square root of the number of observations. With annual observations that ratio is (mean / volatility) x sqrt(years), and mean over volatility is exactly the annual Sharpe ratio. So a Sharpe of 1 gives t = sqrt(years): 2 after 4 years. A Sharpe of 0.5 gives t = 0.5 x sqrt(years): 2 only after 16 years. With the textbook 1.96 in place of 2 the numbers are 3.8 and 15.4 years; the round figures are what you say in the room.

Can you shortcut it with daily data?

This is the follow-up that separates candidates. Sampling daily gives about 252 times as many observations a year, but the daily Sharpe ratio is smaller by the square root of 252, because daily mean scales with time and daily volatility with its square root. The two effects cancel exactly, so the t-statistic depends on calendar time, not on how finely you slice it. Think of estimating a river's average level: measuring every minute instead of every day does not help if the river's slow swings are the uncertainty.

What makes the real requirement even longer?

Three things, each worth one sentence. Returns are not independent from year to year, and positive autocorrelation inflates the true standard error. If you tested twenty strategies and kept the best, a t of 2 is easy to get by luck, so the bar has to rise with the number of ideas tried. And the Sharpe ratio itself drifts as markets change, so a sixteen-year record may be measuring two different strategies. A desk that says a Sharpe 0.5 strategy is proven after three years is reading noise.

About 9.4 million a day, and I would open at 8 bid, 11 offered, in millions. Assume about 700,000 births a year, 2.5 years in nappies, six changes a day and 90% disposable: 9.45 million. Multiplying the low and high ends of each input gives 5.5 to 15.0 million, so a quote of 8 at 11 is tight enough to trade and still honest about the uncertainty.

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

Chain it through things you can reason about. Children in nappies are roughly births a year times the years each child spends in them. Assume about 700,000 births a year, a round number worth checking against the latest official statistics, and 2.5 years in nappies: about 1.75 million children. Each child uses about six a day on average, more as a newborn and fewer as a toddler, and assume 90% of families use disposables: 1.75 million x 6 x 0.9 = 9.45 million a day. Say each assumption out loud and give it a range as you go.

Where should the width of the market come from?

From the ranges, multiplied. A shopkeeper who is unsure of both price and quantity is more unsure of revenue than of either. Put a low and a high on every input and multiply the lows together and the highs together: here 5.5 million to 15.0 million, about a factor of 1.6 either side of the centre. Centre the quote near the middle on a multiplicative scale, the geometric mean of the ends, 9.1 million, which sits close to the central estimate.

Then choose the width you will actually trade. A market as wide as the whole range, 5.5 at 15, is useless: nobody trades against it and it tells the interviewer you have no view. Quote tighter, 8 at 11, and move it as they trade: if they keep buying at 11, raise both sides, because their trades carry information. Say the scope questions too: does the count include adult incontinence products, and a school-age child in night-time pants? Those can move the answer more than any of the four inputs.

No die is best: red beats green, green beats blue and blue beats red, each with probability 5/9. So who picks first matters a great deal. Let me choose, then take the die that beats mine and win 5/9 of the time. If you are forced to pick first, every choice loses 5/9 of the time against an opponent who knows the cycle.

How do you work out who beats whom?

Each matchup has only nine equally likely pairs of faces, so write the 3 by 3 grid and count. Red against green: red's 2 beats only the 1, while its 6 and 7 each beat the 1 and the 5, for 1 + 2 + 2 = 5 wins out of 9. Do the same for the other two pairs and every matchup comes out 5 to 4: red over green, green over blue, blue over red. It is rock, paper, scissors built out of dice, and in rock, paper, scissors nobody asks which hand shape is best.

Why does green lose to red when green has the higher average?

The averages are red 5, green 6 and blue 5. Winning is about how often, not by how much. Green's 12 wins every time it shows, but it shows only a third of the time, and green's other two faces, 1 and 5, lose to both of red's high faces. A higher mean and a higher chance of winning are different things, and the gap between them is the whole puzzle. Change the rules so the winner collects the difference between the two numbers, and green's expected margin against red is 6 - 5 = +1: now you want green against red, and blue against red is a dead heat at 0.

Where does a trader meet the same thing?

Head-to-head comparisons need not line up into a ranking. Strategy A can beat strategy B on more days than not, B can beat C, and C can beat A, whenever one of them earns its money in rare large wins, as green does. Before choosing between strategies, decide whether you care about how often you win or how much you make, because under the first a cycle like this one means there may be no best choice at all. The limitation of the puzzle is that it is one roll; over many rolls with the total score counted, the mean matters more and green's 12 starts to pay.