Hedge Funds puzzles, solved step by step

Replicate it with a tight call spread: buy 5 calls at 10,990 and sell 5 at 11,010. The position pays 0 below 10,990 and 100 above 11,010, a steep ramp standing in for the step. So the digital costs about 5 x (C at 10,990 minus C at 11,010). With calls at 212.40 and 203.60 that is 5 x 8.80 = Rs 44. In the limit, the price is minus 100 times the slope of call prices against strike.

Why does a call spread look like a step?

A steep enough ramp can stand in for a stair. A call spread's payoff is a ramp: nothing below the lower strike, rising point for point between the strikes, flat above the upper strike. Narrow the strikes and scale up the size, and the ramp tightens into the step a digital pays. Here the strikes are 20 points apart, so each spread pays at most 20, and five spreads pay at most 100, the digital's payout.

What does the price of the spread tell you?

The spread costs the difference in call prices, so the digital costs five times that. As the strikes close in, the price becomes minus 100 times the slope of the call price against strike, and that slope is the discounted market-implied chance of finishing above the strike. With the illustrative quotes, 8.80 across 20 points is a slope of 0.44, a digital worth Rs 44 and an implied chance of about 44% before discounting.

Which spread does a desk that sold the digital actually buy?

A desk that has sold the digital wants a hedge that pays at least 100 wherever the digital does. The centred spread overpays just below the strike and underpays just above it, so a seller hedges with five 10,980 / 11,000 spreads, which pay the full 100 by the strike and cost a little more. That difference is what the desk charges for an index that settles right at the strike. One more point marks a strong answer: the slope of call prices includes the change in implied volatility across strikes, so with the usual equity skew, where lower strikes carry higher volatility, the digital is worth more than a flat-volatility model says.

21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.

Why is it tempting to get 23.5?

Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.

How do the two routes agree?

Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.

Why would a trading firm ask this?

Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.

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.

About 31% to finish above Rs 110, and about 62% to touch it during the year. A 10% move is half of one year's 20% standard deviation, and a normal variable ends more than half a standard deviation up 30.9% of the time. By the reflection principle, every path that touches 110 and ends below has a mirror twin that ends above, so touching is twice as likely as finishing above: 61.7%.

Why is finishing above 110 about a one in three chance?

Scale the move by the volatility. Over one year the price spreads out with a standard deviation of about Rs 20, so Rs 110 is half a standard deviation above the start, and a normal variable finishes more than half a standard deviation up 30.9% of the time. Treating the price as an arithmetic random walk is close enough for a 10% move; a lognormal model, in which prices cannot go negative, gives a slightly lower figure, about 28%. Say you are approximating, and say which way the error runs.

Why is touching twice as likely as finishing above?

Think of a walker on a foggy path who is equally likely to step forward or back. Once she reaches a marker post, her remaining steps are a fair coin again: from the post she is as likely to end past it as short of it. So for every path that touches 110 and ends below, reflecting the part after the touch gives an equally likely path that ends above; touching paths split evenly between the two. Every path that ends above must have touched on the way, so the chance of touching is twice the chance of ending above.

Where does the factor of two matter on a desk?

Anything that triggers on a touch rather than on the finish. A stop-loss set 10% away is hit about twice as often as the price ends beyond it, and an option that pays on a touch is worth roughly twice one that pays only if the price finishes past the same level. The limitation: the factor of two holds for a driftless, continuously watched walk. Drift, jumps and checking the price only at the daily close all move it, and a checked-daily barrier is touched a little less often than a continuous one.

About 10.75% a year, for ever. In a constant-growth model the forward P/E equals the payout ratio divided by (required return minus growth). With a P/E of 40 and a payout of 0.5, r minus g must be 0.5/40 = 1.25%, so g = 12% - 1.25% = 10.75%. Retaining half its earnings, the company would need a return on equity of 21.5% for ever to fund that growth.

How does a P/E hide a growth assumption?

A flat that rents for Rs 30,000 a month and sells for Rs 1.2 crore is priced at 400 months of rent; a buyer paying that is quietly assuming the rent will grow. A price multiple is a compressed forecast: fix the return investors want and the share of earnings paid out, and the multiple pins down the growth the price needs. The constant-growth model, price equals next year's dividend over (r minus g), divided through by earnings, gives P/E = payout/(r - g).

Is 10.75% for ever plausible?

Test it against the business. Growth funded by retained earnings is return on equity times the share retained, so 10.75% growth with half the earnings kept needs a return on equity of 21.5%, held for ever. Few businesses hold returns like that for decades, and no company can outgrow the economy it sells into indefinitely, so compare the figure with the nominal growth you expect for that economy and say it as your assumption. The price is not wrong by arithmetic; it is demanding by assumption.

How sensitive is the answer?

Very. Because r - g is only 1.25%, every point on the required return moves the implied growth by a full point: at 11% the price implies 9.75%, at 13% it implies 11.75%. Using trailing rather than forward earnings shifts it too: 40 = 0.5(1 + g)/(0.12 - g) gives 10.62%. A high multiple rests on a thin gap between two large numbers, so small changes in either swing the value.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D's own vote is half, so he keeps all 100. With three, C buys E with 1 coin. With four, B buys D with 1 coin. With five, A needs two votes beyond his own and buys the two pirates who get nothing in the four-pirate split, C and E, for one coin each.

Why start from the end?

Planning a train journey, you work back from when you must arrive, not forward from when you wake up. Each pirate votes by comparing the offer with what he would get if the proposal failed, so you can only price a vote once you know the next round's outcome, which means solving the smallest game first and working upwards. That method, backward induction, is the whole puzzle.

How does each round play out?

Two pirates, D and E: D proposes 100 for himself, and his own vote is half, so it passes. Three pirates: C needs one more vote and buys E, who gets nothing in the two-pirate game, for 1 coin: 99, 0, 1. Four pirates: B needs one more vote and buys D, who gets nothing in the three-pirate game: 99, 0, 1, 0. Five pirates: A needs two more votes and buys C and E, both empty-handed in the four-pirate game: 98, 0, 1, 0, 1.

What is the interviewer really testing?

Whether you reason about the alternative each party faces rather than about fairness. A vote costs exactly one coin more than what the voter gets if the deal fails, so the cheapest supporters are the ones with the worst fallback. The same logic runs through any negotiation: a creditor backs a restructuring plan when it beats their recovery in liquidation, and support is cheapest from those whose alternative is worst. State the assumptions: perfect rationality, and a pirate who is indifferent votes against, which is why one coin, not zero, is needed.

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.