Hedge Funds puzzles, solved step by step

The book's volatility is about Rs 23.1 crore a year, and the Rs 60 crore position carries about 74% of it on 60% of the capital. Stand-alone risks are Rs 18 crore and Rs 8 crore. Book variance is 18 squared plus 8 squared plus 2 x 0.5 x 18 x 8, which is 532, so volatility is Rs 23.07 crore. Each position's contribution is its covariance with the book over the book's volatility: Rs 17.17 crore and Rs 5.90 crore, which add back to the total.

Why is risk not shared out like capital?

Two friends share a taxi. One rides twice as far, straight through the traffic jam; the other gets off after a short hop. Splitting the fare by the number of bags each carries would be absurd. Risk belongs to a position in proportion to how much it moves and how much it moves with everything else, not to how much money sits in it. Here the first stock is larger, more volatile and positively correlated with the second, so it carries far more than its 60% of the capital.

Why do the contributions add up exactly to the total?

Split the variance. The cross term, 2 x 0.5 x 18 x 8 = 144, is shared equally, 72 to each position. So position A owns 324 + 72 = 396 of the 532 of variance and position B owns 64 + 72 = 136, and dividing each by the book's volatility of 23.07 gives rupee contributions that sum exactly to Rs 23.07 crore. The diversification benefit is the gap between the stand-alone total of Rs 26 crore and the book's Rs 23.07 crore.

What does a risk manager do with the split?

Cut where the risk is, not where the money is. Each rupee in position A carries 28.6 paise of marginal risk against 14.7 paise in position B, so trimming Rs 10 crore from A lowers book volatility by roughly Rs 2.9 crore. Recomputing exactly gives Rs 2.84 crore, close to the estimate. The limitation: the split is a snapshot at one correlation, and when correlations move, both the total and the split move with them.

The MLE is 9.0, the largest observation; it is biased low, and the unbiased correction is 5/4 x 9.0 = 11.25. The likelihood is 1 over theta to the fourth for any theta of at least 9.0 and zero below it, so it peaks at the sample maximum. But the maximum of n draws averages n/(n + 1) of theta, here 4/5, so scaling by (n + 1)/n removes the bias.

Why is the MLE the largest observation?

A friend draws four raffle tickets numbered from 1 up to some unknown top number, and the highest you see is 90. The top number is at least 90; guessing higher only spreads your belief over tickets nobody drew. The likelihood, 1 over theta to the n, is zero for any theta below the largest observation and falls as theta rises above it, so it is maximised exactly at the sample maximum, 9.0. This is a case where you do not differentiate: the maximum sits on a boundary, not where a slope is zero.

Why is it biased, and by how much?

The largest draw can never be above theta and is almost always below it. Four points dropped at random on 0 to theta cut it into five gaps of equal expected length, so the largest point sits on average four fifths of the way up, and the MLE underestimates theta by a fifth on average. Multiplying by 5/4 fixes it: 9.0 becomes 11.25. The bias shrinks as n grows, since n/(n + 1) tends to 1, but with four points it is large.

How does it compare with the obvious alternative?

The method of moments doubles the sample mean, since a uniform on 0 to theta averages theta/2: the mean here is 6.175, giving 12.35. Both 11.25 and 12.35 are unbiased, but the corrected maximum has a much smaller variance, theta squared over n(n + 2) against theta squared over 3n, because the largest draw carries the most information about the top of the range. With four points that is theta squared over 24 against theta squared over 12: half the variance.

3/4, so fair odds are 3 to 1 on black. Two black draws rule out bags with 0 or 1 black. The ways to draw two blacks in order are 2, 6 and 12 for bags with 2, 3 and 4 black, so those bags now carry 10%, 30% and 60%. The next stone is black with chance 0, 1/2 and 1 in them, which averages to 3/4. A contract paying 100 if black is worth 75.

How do the two black draws change your view of the bag?

If a friend pulls two red sweets from a jar you have never seen, you start to suspect it is mostly red. Each possible bag is reweighted by how likely it was to produce what you saw: equal priors times the chance of two blacks. Drawing two blacks in order has 0 ways from bags with 0 or 1 black, 2 x 1 = 2 ways with 2 black, 3 x 2 = 6 with 3 black and 4 x 3 = 12 with 4 black. Out of 20 in total, that is 10%, 30% and 60%: the all-black bag is now the favourite.

How do you turn the posterior into a price?

Average the chance of black over the bags you still believe in. In the 2-black bag both remaining stones are white; in the 3-black bag one of two is black; in the 4-black bag both are, so the answer is 0.1 x 0 + 0.3 x 1/2 + 0.6 x 1 = 3/4. A contract paying 100 if the next stone is black is worth 75. You would buy it below 75 and sell it above; as a market maker you might quote 70 at 80. Offered even money on black, your expected profit per rupee staked is 0.75 minus 0.25, which is 50 paise.

Check it with Laplace rule of successionWith a uniform prior, after s successes in n trials, the chance the next trial succeeds is (s + 1) / (n + 2).: after 2 blacks in 2 draws the next is black with chance (2 + 1)/(2 + 2) = 3/4, and the rule holds exactly for this finite bag. Two routes to 3/4 is what separates a solid answer from a lucky one. The limitation: everything rests on the flat prior. If you had reason to think mixed bags were more common, 3/4 would fall.

1 minus 2/e, about 26.4%. The total wait is the sum of two independent exponential waits. Convolving the two densities gives t e^-t, a gamma shape that starts at zero because two steps cannot both be instant. Its area from 0 to 1 is 1 minus e^-1 (1 + 1), which is 1 minus 2/e. A second route: it is the chance that a Poisson process with rate 1 produces at least two arrivals in one minute.

Why is this not the chance of one wait, squared?

Squaring would be right if both orders were racing from the same start line. Here they queue: the second clock only starts when the first order lands. It is like two buses where you must take the first to reach the stop for the second. The event is that the sum of the two waits is under one minute, which is stricter than each wait being under one minute. The chance one wait is under a minute is 1 minus 1/e, about 63%; squaring gives 40.0%, which answers a different question.

How do you get the density of the sum?

Add up every way to split the total t between the two waits. The density of a sum of independent waits is the convolutionThe density of a sum of two independent variables, found by integrating one density against the other shifted over every possible split of the total. of their densities, and for two exponentials it is t e^-t. Every split of t into s and t minus s has density e^-s times e^-(t minus s), which is e^-t whatever s is, and there is a length t of possible splits. Integrate t e^-t from 0 to 1 by parts and you get 1 minus 2/e.

Check it with counting. Exponential waits are the gaps of a Poisson process, so both orders arrive within a minute exactly when the process makes at least two arrivals in that minute. With one arrival expected per minute, the chance of zero is e^-1 and of exactly one is e^-1, so at least two is 1 minus 2/e, the same number. Say both routes and the interviewer will usually skip ahead.

About 3.28 loops. With k strands in the bowl there are 2k free ends. Pick any end; the other end you pick is one of the remaining 2k minus 1, and exactly one of those belongs to the same strand, so this tie closes a loop with chance 1/(2k minus 1). Either way the number of strands falls by one. Adding 1/199 + 1/197 + ... + 1/3 + 1 gives about 3.284.

What does one tie do, whatever happens?

Start with the bookkeeping, because it makes the rest easy. Every tie reduces the number of loose strands by exactly one: either it closes a strand into a loop, or it joins two strands into one longer strand. So there are always exactly 100 ties, and with k strands left there are 2k free ends. The question becomes how many of those 100 ties happen to close a loop.

What is the chance a given tie closes a loop?

Think of a room of dancers holding hands in lines: grab one free hand, then pick a second free hand at random, and a circle forms only if the second hand is at the other end of the same line. With 2k free ends, the second end is one of 2k minus 1, and exactly one of them is the other end of the strand you picked, so the chance is 1/(2k minus 1). Give each tie an indicator that is 1 if it closes a loop; by linearity of expectation the expected number of loops is the sum of the chances, from 1/199 for the first tie up to 1 for the last.

For a sense check without a calculator: the sum of odd reciprocals up to 1/(2n minus 1) is about half of ln n plus ln 2 plus half of Euler's constant, which for n = 100 gives 3.28. The number of loops grows only like the logarithm of the number of noodles: a million noodles would give only about 7.9 loops.

Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.

Why start from the last roll?

Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.

How do the thresholds come out?

With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.

Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.

The lift says they saw at least two heads, so the fair value is now at least 5.75, not 5; requote around 5.75 bid, 6.5 offered. The other seven flips are worth 3.5 heads, so the buyer's value is heads seen plus 3.5. Paying 5.5 only makes sense with two heads (5.5) or three (6.5). Those are 3 to 1 likely, giving 5.75; a buyer who needs a strict edge saw three heads, worth 6.5.

What is the trader's view before they trade?

A friend offers to buy your raffle ticket after the first few numbers are drawn. The offer itself is the warning. The informed trader values the contract at heads already seen plus 3.5, the expected heads in the seven unseen flips, so their value is 3.5, 4.5, 5.5 or 6.5 with chances 1, 3, 3 and 1 in 8. Against your market of 4.5 bid and 5.5 offered, they buy only if their value is at least 5.5, and sell to you at 4.5 only if it is 4.5 or less. The flat 5 you quoted around is right only for someone who has seen nothing.

How do you turn the trade into a new fair value?

Condition on the fact that they traded. Only the two-head and three-head worlds produce a buy at 5.5, and they are 3/8 and 1/8 likely, so given a lift the value is (3 x 5.5 + 1 x 6.5) / 4 = 5.75. If you assume they would not bother trading at zero edge, only the three-head world is left and the value is 6.50. Either way you sold too cheaply: this is adverse selectionThe tendency of a market maker to trade most with the people who know more, so the trades that happen are the ones that lose money for the market maker., and it is the cost every market maker prices into the spread.

Now requote. Your bid should not be below what you now believe the floor is, and your offer should sit where even the best informed buyer has no edge. Something like 5.75 bid, 6.5 offered does both: a buyer who saw three heads is indifferent at 6.5, and you are no longer selling below value. Cut your size too, because you know someone is trading with more information than you, and say you would ask whether they could see the flips before quoting again.

The probability is 1/6, not 1/5. Ending on a 2 rules out a first roll of 2, but it does not leave the other five faces equally likely. A first roll of 4 or 6 has already cleared one rival, so 2 then finishes last half the time; after an odd first roll it finishes last only a third of the time. Weighting the faces by those chances gives 1/6 for a 1.

Why does ruling out one face not spread its weight evenly?

Suppose you hear that a friend arrived late to a meeting. Before, the bus, the train and the car were equally likely ways she travelled. Learning how things ended shifts weight towards the starts that make that ending more likely, in proportion to how strongly each one leads to it. If the bus is late twice as often as the train, the bus now carries twice the train's weight. The 1/5 answer treats the ending as if it only ruled a face out and said nothing else.

How likely is a 2 to finish last after each first roll?

Odd rolls never change the order in which the even numbers first appear, so after an odd first roll the three evens are still symmetric and 2 is last with chance 1/3; after a first roll of 4 or 6 only two evens remain and 2 is last with chance 1/2; after a first roll of 2 it can never be the last new even. Multiply each by the 1/6 chance of that first roll: the joint chances are 1/18 for each odd face, 0 for a 2 and 1/12 for each of 4 and 6. They add to 1/3.

How do you check the answer?

Make the six posterior chances add up. Three odd faces at 1/6 each and two even faces, 4 and 6, at 1/4 each give 1/2 plus 1/2, which is 1, with nothing left for a 2. The odd faces keep exactly their starting weight because an odd roll tells you nothing about the evens; all the weight removed from the 2 goes to 4 and 6. Saying that sentence shows the interviewer you understand Bayes ruleThe rule for updating a probability after new information: the prior times the likelihood of the information, divided by the overall chance of the information. rather than recite it.

Overall volatility is 2% a day and the kurtosis is 9.75, against 3 for a normal distribution. Variances mix in proportion: 0.8 x 1 + 0.2 x 16 = 4, so volatility is 2%. Fourth moments mix the same way, and each normal contributes 3 times its volatility to the fourth: 0.8 x 3 + 0.2 x 768 = 156. Dividing by the variance squared, 16, gives 9.75.

Why is a mixture of normals not normal?

Think of a commute that takes 30 minutes on most days and two hours on strike days. The average trip hides the shape: most days cluster tightly and a few days sit far out. Mixing a calm regime with a wild one gives more small moves than a normal with the same overall spread, fewer medium ones, and far more big ones. The peak is taller, the shoulders thinner and the tails fatter, which is what {term('kurtosis', 'The fourth moment of a distribution divided by the variance squared; 3 for a normal distribution, higher when tails are fatter.')} measures.

How do you get the two numbers?

Work with moments, because they mix in proportion to the weights. The variance is 0.8 x 1 + 0.2 x 16 = 4, so volatility is 2%; the fourth moment is 0.8 x 3 x 1 + 0.2 x 3 x 256 = 2.4 + 153.6 = 156, and kurtosis is 156 / 4 squared = 9.75. Look at where the 156 comes from: 153.6 of it is the wild days, which occur only one day in five. The fourth power makes rare large moves dominate.

What does this mean for a risk model?

A model that fits a normal to the 2% volatility is right about the average day and wrong about the days that matter. It says a move beyond 6% happens on about 0.27% of days, roughly once in 370 trading days; the mixture says 2.7%, roughly once in 37. That is the usual story of market returns: calm stretches and volatile stretches, each close to normal, adding up to fat tails. A model that lets volatility change over time captures much of it.

Both players should show heads 3/8 of the time, and the game is worth minus Rs 0.125 a round to you. Choose your mix so the opponent gains nothing by switching: 3p - 2(1 - p) = -2p + (1 - p) gives p = 3/8. The opponent's mix solves the same balance from your side, also 3/8. At those mixes you lose an eighth of a rupee a round, although the payoffs look even.

Why is a fair coin the wrong strategy?

Think of a penalty taker and a goalkeeper. If the taker always shoots left, the keeper dives left; the taker's only defence is to mix so the keeper cannot profit from guessing. A mix is right only if it leaves the other side indifferent, and a fair coin here does not: against it the opponent's tails pays you 0.5 x (-2) + 0.5 x 1 = -0.50 a round. So the opponent always shows tails, and you lose Rs 0.50 a round rather than breaking even.

How do you find the equilibrium mix?

Set up your payoff for each of your choices as the opponent's chance of heads, q, varies. Showing heads pays 3q - 2(1 - q) = 5q - 2; showing tails pays -2q + (1 - q) = 1 - 3q; they are equal at q = 3/8, where both pay -1/8. The opponent plays 3/8 heads so that nothing you do beats -1/8. By the same balance from the opponent's side, you play 3/8 heads so that nothing the opponent does pushes you below -1/8. That pair is the Nash equilibriumA pair of strategies in which neither player can do better by changing only their own choice..

Why is the game negative when the payoffs look even?

Because the balance is in the totals, not in the play. Your two winning cells need coordination the opponent will not give you, while their winning cells, the mismatches, pay the same 2 either way. The opponent can lean towards tails, starving your big Rs 3 cell, and the only price is feeding your small Rs 1 cell. Say what a desk would do with it: ask to be paid about 13 paise a round to play, or ask to swap sides.