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Hedge Funds puzzles, solved step by step

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Showing 31–35 of 35 · filtered from 100Clear filters
  1. 086A stock ticks up or down by Rs 1 each minute with equal probability, starting at Rs 50. On average, how many minutes pass before it first touches Rs 45 or Rs 55?Random walks and Markov chainsCoreQuant and systematic fundsProp and quant trading firms

    Try it first

    Pick your answer before setting up any equation.

    Show the worked solution

    25 minutes. Let E(k) be the expected minutes to exit from price k. Each minute costs one and moves the price up or down with equal chance, so E(k) = 1 + half E(k + 1) + half E(k - 1), with E(45) = E(55) = 0. The solution is E(k) = (k - 45)(55 - k), the product of the distances to the two barriers. From Rs 50 that is 5 x 5 = 25.

    Why is the answer not 5 minutes?

    Think of someone pacing a corridor, taking one step forward or back on each coin toss. After 25 steps they are not 25 steps away; typically they are about 5 away, because the steps keep undoing each other. A fair random walk covers distance like the square root of time, so reaching a barrier 5 away takes on the order of 5 squared, 25 steps, not 5.

    A fair walk wanders: 5 rupees of room takes 5 x 5 = 25 minutes on averageaverage exit: 25 minout at 55 after 11 minout at 45 after 23 minout at 55 after 45 min01020304050Rs 45Rs 50Rs 55Minutes since the startFrom price k, expected minutes = (k - 45) x (55 - k)from 50: 25from 48: 21from 46: 9
    Three sample paths from Rs 50 leave the Rs 45 to Rs 55 channel after 11, 23 and 45 minutes; averaged over all paths the exit takes (50 - 45) x (55 - 50) = 25 minutes, and from Rs 48 or Rs 46 it takes 21 or 9.

    How do you get exactly 25?

    Set up the one-step equation. From any price k strictly between the barriers, you spend one minute and then stand at k + 1 or k - 1 with equal chance. E(k) = 1 + half E(k + 1) + half E(k - 1) says the second difference of E is always minus 2, so E is a downward parabola that is zero at both barriers. The only such parabola is (k - 45)(55 - k). Check a point: from Rs 46 it gives 1 x 9 = 9 minutes, and it passes the one-step test, since 1 plus half of E(47), which is 16, plus half of E(45), which is 0, is 9.

    The relationship
    E(k)=(k−a)(b−k)E(50)=(50−45)(55−50)=25E(k) = (k - a)(b - k) \qquad E(50) = (50 - 45)(55 - 50) = 25
    a, bthe lower and upper barriers, Rs 45 and Rs 55
    kthe starting price
    E(k)the expected number of one-minute steps before either barrier is touched
    What it says in wordsFor a fair walk, the expected time to leave a channel is the distance to the floor times the distance to the ceiling.

    What does the shape tell a trader?

    Starting in the middle is the slowest place to be, and an off-centre start is much faster: from Rs 48 the answer is 3 x 7 = 21 minutes, and from Rs 46 only 9. Doubling both distances quadruples the expected time: barriers at Rs 40 and Rs 60 give 100 minutes. That is the arithmetic behind why widening a stop and a profit target together makes a trade live much longer. It holds only for a fair walk: if the stock ticks up 60% of the time, the exit comes sooner, about 19.2 minutes, and mostly at the top.

    Where candidates lose it

    The quick wrong answer is 5 minutes, which treats the walk as if it moved steadily towards one barrier. A fair walk wanders, and the interviewer is checking whether you know that distance grows with the square root of time.

    The second loss is reaching 25 from the square-root intuition without being able to show it. Write the one-step equation and the parabola; that is what turns a good guess into an answer.

    What the interviewer asks next

    • What is the probability the stock touches Rs 55 before Rs 45?
    • From Rs 50, the barriers move to Rs 40 and Rs 55. What is the expected time now?
    • The stock ticks up with probability 0.6. Why does the expected time fall?
  2. 088A stock's true model is: stock return = 0.5 x market return + 1.0 x sector return + noise. Regressing the sector's return on the market gives a slope of 0.4. If you regress the stock on the market alone, what slope do you get?Statistics and estimationCoreQuant and systematic funds

    Try it first

    What does the market-only regression report?

    Show the worked solution

    About 0.9. The market reaches the stock by two paths: directly, with a coefficient of 0.5, and through the sector, which moves 0.4 for each unit of market and passes all of it on with a coefficient of 1.0. A regression on the market alone cannot separate the two and reports the total, 0.5 + 1.0 x 0.4 = 0.9. The extra 0.4 is omitted variable bias.

    Why does leaving the sector out change the market slope?

    Suppose you measure how much ice cream sales rise on hot days, but hot days also tend to be holidays, and holidays sell ice cream too. Leave holidays out and the heat gets the credit for both. A regression gives a left-out variable's effect to whichever included variable moves with it, in proportion to how strongly the two move together. Here the sector moves with the market, so the market's slope absorbs part of the sector's effect.

    Leave the sector out and the market gets credit for both pathsMarketSectorStockdirect: 0.50.41.0via the sector: 0.4 x 1.0 = 0.4Regress stock onmarket alone0.5 direct0.4 borrowed= 0.900.50.9The extra 0.4 is omitted variable bias: the sector's effect, credited to the market
    The market reaches the stock directly with a coefficient of 0.5 and through the sector with 0.4 x 1.0 = 0.4, so a regression of the stock on the market alone reports 0.9, of which 0.4 is the sector's effect credited to the market.

    How do you compute the bias?

    Write the sector as 0.4 x market plus a part unrelated to the market, then substitute. Stock = 0.5 x market + 1.0 x (0.4 x market + other) + noise = 0.9 x market + (1.0 x other + noise). The bracket is unrelated to the market, so a regression on the market alone recovers 0.9. The bias is the omitted coefficient times the slope of the omitted variable on the included one, 1.0 x 0.4. A simulation of 20,000 days with these coefficients gives a slope of 0.897, matching the algebra.

    The relationship
    β^short=βM+βS δ=0.5+1.0×0.4=0.9\hat\beta_{\text{short}} = \beta_M + \beta_S\,\delta = 0.5 + 1.0 \times 0.4 = 0.9
    beta_Mthe stock's true direct loading on the market, 0.5
    beta_Sthe stock's loading on the sector that was left out, 1.0
    deltathe slope of the sector's return on the market's, 0.4
    What it says in wordsThe short regression's slope is the true slope plus the left-out variable's effect times how much that variable moves with the one you kept.

    Is 0.9 wrong, or answering a different question?

    It depends on what you use it for. If you want to hedge the stock with the market alone, 0.9 is the right hedge ratio, because it captures everything the market drags along with it. If you want the stock's exposure holding the sector fixed, say to build a sector-neutral book, 0.9 overstates it and 0.5 is the number you need. The bias can also run the other way: if the sector moved against the market, or the stock loaded negatively on the sector, the short slope would sit below 0.5. Naming both uses is what the interviewer is listening for.

    Where candidates lose it

    The fast wrong answer is 0.5: candidates assume a regression recovers the true coefficient whatever else is left out. It does so only when the omitted variable is unrelated to the included one.

    The second loss is getting 0.9 and calling it simply wrong. It is the correct total effect of the market and the right number for a market-only hedge; it is wrong only as an estimate of the direct effect.

    What the interviewer asks next

    • What slope do you get if the sector's slope on the market is minus 0.4?
    • You add the sector to the regression. What happens to the standard error of the market coefficient if the two are highly correlated?
    • How does this bias show up when you estimate a stock's factor exposures with too few factors?
  3. 091Cards are turned over one at a time from a well-shuffled 52-card deck. On average, at what position does the first ace appear?Counting and combinatoricsCoreQuant and systematic fundsProp and quant trading firms

    Try it first

    What is the expected position of the first ace?

    Show the worked solution

    Position 10.6. The four aces split the 48 non-aces into five gaps: before the first ace, between each pair of aces, and after the last. By symmetry, a given non-ace is equally likely to fall in any of the five gaps, so each gap holds 48/5 = 9.6 cards on average. The first ace comes straight after the first gap, at 9.6 + 1 = 10.6.

    Why do the gaps have the same size on average?

    Picture four red beads threaded at random among 48 white ones on a string. Pick any one white bead: it is as likely to sit before all four reds as between the first and second, or after the last, because a shuffle favours no order. Each non-ace lands in each of the five gaps with probability 1/5, so the expected size of every gap is 48/5 = 9.6 cards. Any single shuffle has uneven gaps; it is only the average that is equal.

    Four aces cut 48 other cards into five gaps of 9.6 on averageOne shuffle: the gaps are unevenAAAA45141114gaps 4 + 5 + 14 + 11 + 14 = 48Average over all shuffles: every gap is 48 / 5 = 9.69.6ace 1at 10.69.6ace 2at 21.29.6ace 3at 31.89.6ace 4at 42.49.6First ace: 9.6 + 1 = card 10.6
    One shuffle leaves uneven gaps of 4, 5, 14, 11, 14 non-aces around its four aces, but averaged over all shuffles every one of the five gaps holds 9.6 cards, so the aces sit at 10.6, 21.2, 31.8, 42.4.

    How do you turn gap sizes into a position?

    The first ace sits immediately after the first gap. Expected position = expected size of the first gap + 1 = 9.6 + 1 = 10.6. The same logic puts the second ace at 21.2, the third at 31.8 and the last at 42.4, evenly spaced 10.6 apart. A quick check: the last ace at 42.4 plus a final gap of 9.6 reaches exactly 52.

    The relationship
    E[first ace]=n−kk+1+1=n+1k+1=535=10.6E[\text{first ace}] = \frac{n - k}{k + 1} + 1 = \frac{n + 1}{k + 1} = \frac{53}{5} = 10.6
    ncards in the deck, 52
    kspecial cards, here the 4 aces
    (n - k)/(k + 1)the expected number of ordinary cards in each of the k + 1 gaps
    What it says in wordsWith n cards and k special ones, the first special card is expected at position (n + 1) divided by (k + 1).

    Why is 13 the wrong instinct?

    Thirteen is the average wait when every draw has a fresh 1-in-13 chance, as if each card were put back and the deck reshuffled. Drawing without replacement makes the aces easier to reach, because every non-ace you turn over raises the share of aces in what is left. The general rule, (n + 1)/(k + 1), is worth keeping: the first spade is expected at 53/14, about card 3.79, and the same shape answers how many trades you expect to check before finding the first of a handful of booking errors in a batch.

    Where candidates lose it

    The fast wrong answer is 13, borrowed from a wait with replacement. The interviewer wants to see you notice that the deck is finite, so the first ace must appear by card 49 at the latest and is pulled earlier than a with-replacement wait would suggest.

    The second loss is starting the direct sum, position times the chance the first ace is there, which is correct but slow and easy to get wrong at a whiteboard. The gap argument gets there in one line.

    What the interviewer asks next

    • What is the expected position of the last ace?
    • What is the expected number of cards between the first and second ace?
    • How many cards do you expect to turn before the first spade?
  4. 093Four people must cross a narrow bridge at night with one torch. They take 1, 2, 5 and 10 minutes to cross, at most two can cross at a time, a pair moves at the slower person's pace, and the torch must be carried on every crossing. What is the fastest time for all four to get across?Logic and brainteasersCoreProp and quant trading firmsLong-short equity funds

    Try it first

    What is the fastest crossing?

    Show the worked solution

    17 minutes. 1 and 2 cross (2 minutes), 1 returns (1), 5 and 10 cross together (10), 2 returns (2), and 1 and 2 cross again (2). The total is 17. The obvious plan, with the fastest person escorting everyone, takes 19, because the 5 and the 10 each cost a separate crossing. Pairing the two slowest hides the 5 inside the 10.

    Why is the obvious plan not the fastest?

    The natural plan uses the quickest person as a shuttle: 1 walks each person over and comes back. That costs 2 + 1 + 5 + 1 + 10 = 19. Every slow person who crosses on a separate trip pays their own time in full, so the 5 and the 10 together cost 15 minutes of crossing. Think of sending two slow parcels in one courier van instead of two: the second rides along for free.

    Pair the two slowest so the 5 rides free inside the 10Shuttle: 1 escorts all21+2 over11 back51+5 over11 back101+10 over19 minPair the slow two21+2 over11 back105+10 over22 back21+2 over17 min05101519the 5 on its own costs 5 minutes
    The shuttle plan spends 5 and 10 minutes on separate crossings and takes 19 minutes, while sending 5 and 10 together costs 10 minutes for both and brings the total to 17, even after the 2-minute person makes one extra return trip.

    How does the 17-minute plan pay for pairing the slow two?

    Pairing 5 and 10 costs 10 instead of 15, a saving of 5. But someone fast must already be waiting on the far side to bring the torch back afterwards. That setup costs one extra return by the 2-minute person and an extra crossing of the pair 1 and 2, so the net saving is 2 minutes: 19 down to 17. The general rule: pair the two slowest when twice the second-fastest time is less than the fastest plus the second-slowest, here 4 against 6.

    The relationship
    2+1+10+2+2⏟pair the slow two=172+1+5+1+10⏟shuttle=19\underbrace{2 + 1 + 10 + 2 + 2}_{\text{pair the slow two}} = 17 \qquad \underbrace{2 + 1 + 5 + 1 + 10}_{\text{shuttle}} = 19
    10the one crossing that carries both the 5-minute and the 10-minute person
    2 + 2the extra cost of having the 2-minute person bring the torch back and cross again
    What it says in wordsPairing the two slowest saves 5 minutes of crossing at a cost of 3 extra minutes of torch returns and re-crossing.

    How do you show 17 is the minimum?

    Five crossings are unavoidable: three over and two back. If the 5 and the 10 cross separately, the forward trips cost at least 10 + 5 + 2 and the two returns at least 1 each, which is 19; if they cross together, the person who returns next must already be across, which forces the returns to be the 1 and the 2, and the best total is 17. Talking through that bound is what separates a remembered answer from a reasoned one, and it is what lets you handle the follow-up with different times.

    Where candidates lose it

    19 minutes is the answer most people reach, and it comes so quickly that they stop there. The interviewer is waiting to see whether you ask what the 5-minute person costs and whether that cost can be hidden.

    The second loss is finding 17 by trial and error without the reason. Name the saving, 5 minutes from pairing, and its price, the extra return and re-crossing; that is what makes the answer hold when the interviewer changes the times.

    What the interviewer asks next

    • The times are 1, 5, 6 and 10. Does pairing the two slowest still help?
    • Add a fifth person who takes 20 minutes. What is the fastest crossing now?
    • State the general rule for when to pair the two slowest walkers.
  5. 099You borrow in yen at 1% and invest in rupee deposits at 7% for a year. By how much can the rupee's value in yen fall over the year before the trade loses money?Valuation, accounting and macro riddlesCoreLong-short equity fundsGlobal macro funds

    Try it first

    How big a fall in the rupee can the trade absorb?

    Show the worked solution

    About 5.6%. Borrow 1 of yen value, convert it, and the rupee deposit grows to 1.07 in yen terms at today's rate. You owe 1.01. If the rupee's yen value falls by d, the deposit is worth 1.07 x (1 - d), which equals 1.01 when d = 1 - 1.01/1.07, about 5.6%. The cushion is slightly below the 6-point gap because the fall hits the interest as well as the principal.

    Where does the carry trade's income come from?

    Think of borrowing from a relative at 1% to put the money in a deposit paying 7%, except that the relative wants to be repaid in a different currency. The 6 points look like easy income until the currency moves. A carry trade earns the interest gap only if the currency you hold does not fall by more than about that gap over the holding period. The rate gap is the cushion; the exchange rate is the risk.

    The 6-point rate gap cushions a rupee fall of only about 5.6%breakeven: 5.6% fall1.07 x (1 - d) = 1.01a 6% fall already loses Rs 0.42no fall: +Rs 610% fall: -Rs 4.700%2%4%6%8%10%Fall in the rupee's value in yen over the year-40+4Profit per Rs 100 borrowed
    Per Rs 100 of yen borrowed, the trade makes Rs 6 if the rupee holds its value, breaks even at a 5.6% fall, already loses Rs 0.42 at a 6% fall, and loses Rs 4.70 at a 10% fall.

    Why is the cushion 5.6% and not 6%?

    Work in yen per unit borrowed. At the end you hold 1.07 of rupee deposit, measured at today's exchange rate, and you owe 1.01 of yen; the rupee can lose value until 1.07 x (1 - d) = 1.01, so d = 5.6%. The fall applies to the whole 1.07, including the 0.07 of interest, which is why the cushion is a little smaller than 6. Borrow Rs 100 crore worth of yen and a 5.6% fall leaves you exactly even; a 10% fall loses about Rs 4.7 crore.

    The relationship
    1.07 (1−d)=1.01  ⇒  d=1−1.011.07=5.6%1.07\,(1 - d) = 1.01 \;\Rightarrow\; d = 1 - \frac{1.01}{1.07} = 5.6\%
    1.07what each unit becomes in the rupee deposit at 7%
    1.01what you owe on each unit of yen borrowed at 1%
    dthe fall in the rupee's value in yen over the year
    What it says in wordsThe trade breaks even when the grown rupee deposit, after the currency fall, buys back exactly the yen you owe.

    What does the forward market say about this?

    Covered interest parityThe rule that the forward exchange rate must offset the interest rate gap between two currencies, or else borrowing in one, lending in the other and locking in the forward would be an arbitrage. sets the one-year forward so that the rupee buys about 5.6% fewer yen forward than today, exactly the fall that wipes out the carry. So an unhedged carry trade is a bet that the rupee will fall by less than the forward already implies. The limitation to state: carry trades tend to earn small amounts steadily and then lose sharply when many holders unwind at once, so the spread of outcomes has a fat left tail that the breakeven alone does not show.

    Where candidates lose it

    The quick answer is 6%, the gap between the two rates. It misses that the currency loss applies to the interest earned as well as to the principal, so the true cushion is slightly smaller, and at exactly a 6% fall the trade is already losing.

    The second loss is treating the 6-point gap as a return and forgetting the currency altogether. Name the exchange rate as the risk before doing any arithmetic.

    What the interviewer asks next

    • The rupee weakens 10% against the yen over the year. What is the loss on Rs 100 crore of yen borrowed?
    • What one-year forward rate does covered interest parity imply, and what does it say about the trade?
    • Why do carry trades tend to lose money suddenly rather than gradually?
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