Hedge Funds puzzles, solved step by step

Bet 20% of your current money each time; bet 40% and the long-run growth is gone. For an even-money bet the Kelly fraction is the win probability minus the loss probability, 0.6 minus 0.4. At 20% the typical path grows about 2.0% a bet. At double that, the average log growth is about -0.2% a bet, so the same edge now earns nothing over time.

Why not bet everything when the odds are in your favour?

Think of a shopkeeper who puts all of today's takings into tomorrow's stock. Six days in ten the stock sells and she doubles her money; on the other four she is back to zero, and one zero ends the shop. Wealth compounds, so a single wipe-out is permanent, and the stake that maximises the average payoff of one round is not the stake that maximises what you end with after many. Betting 100% has the highest expected value per round and a certainty of eventual ruin.

How do you find the best fraction?

Maximise the growth rate, which is the average of the log of what each bet does to your money. Bet a fraction f and a win multiplies your wealth by 1 + f, a loss by 1 - f. The growth rate 0.6 ln(1 + f) + 0.4 ln(1 - f) peaks where f equals the edge, 0.6 - 0.4 = 20%, at about 2.01% a bet. Over 100 bets that turns Rs 1,000 into about Rs 7,490 on the typical path. This rule is the Kelly criterionA sizing rule that picks the bet size maximising the long-run growth rate of wealth, the expected log return per bet..

What happens at twice the Kelly bet?

Near the peak the growth curve is close to a parabola centred on 20%. Double the Kelly fraction and you sit as far down the far side as not betting at all sits on the near side: growth of about zero, -0.24% a bet. Over 100 bets the typical path ends near Rs 783, below where you started, although every single bet had a positive expected value. The extra swings cost more in compounding than the extra stake earns.

Say the limitation too. Kelly assumes you know the 60% exactly. On a desk the edge is an estimate, and overestimating it pushes you to the right of the peak, where the curve falls fastest. That is why many traders bet half Kelly: 10% here keeps about 75% of the growth with half the swings.

The one-die game is worth more: Rs 15.17 against Rs 12.25, a gap of 35/12, about Rs 2.92. With two independent dice the average product is the product of the averages, 3.5 x 3.5 = 12.25. With one die the average square is (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6. The gap between the two is exactly the variance of one die.

Why is averaging a square not the same as squaring an average?

Take two students who score 2 and 8 in a test. Their average is 5, and 5 squared is 25. Square each score first and then average, and you get (4 + 64)/2 = 34. Squaring rewards the high value more than it penalises the low one, so the average of the squares is always at least the square of the average, and the gap is the spread. Here the gap, 9, is exactly the variance of the two scores.

How do the two games come out?

In game A the dice are independent, so the average of the product is the product of the averages. Two independent dice give 3.5 x 3.5 = Rs 12.25, because a high roll on one die is as likely to meet a low roll on the other as a high one. In game B one number is multiplied by itself, so a 6 always meets a 6 and a 1 always meets a 1. The average of the six squares is 91/6, about Rs 15.17.

What is the desk lesson?

A payoff that is the square of a move gains from dispersion; one built from two independent pieces does not. Correlation is what turns a product into a square: if the second die always copied the first, game A would be game B. Flip it and make the second die show 7 minus the first, and the average product falls to 9.33. Say that range out loud and the interviewer knows you see the payoff as a bet on varianceThe average squared distance of an outcome from its mean; for one fair die it is 35/12, about 2.92. as well as on the average.

They are exactly equal: 40 ml of wine sits in the water glass and 40 ml of water sits in the wine glass. After the first pour the water glass holds 200 ml of water and 50 ml of wine. A 50 ml pour of that mix is one fifth wine, so 10 ml of wine and 40 ml of water go back. Each glass ends at 200 ml, which forces the two amounts to match.

What is the argument that needs no arithmetic?

Two classrooms hold 30 students each. Five walk from room A to room B, and then any five people at all walk back. Both rooms hold 30 again. Every seat in room A left empty by a room A student who stayed away must now be filled by a room B student, so the room B students in room A always equal the room A students in room B. The glasses work the same way, because both finish at 200 ml.

How do the numbers confirm it?

After the first pour the water glass holds 250 ml, of which 50 ml, one fifth, is wine. Stirred evenly, a 50 ml pour back carries 10 ml of wine and 40 ml of water. The wine glass ends with 150 plus 10, 160 ml of wine, and 40 ml of water. The water glass keeps 200 minus 40, 160 ml of water, and 40 ml of wine. Forty and forty.

What does the interviewer learn from how you answer?

Whether you look for something that stays fixed before you reach for arithmetic. The conservation argument survives every variation: uneven stirring, several pours back and forth, any spoon size, as long as both glasses finish at their starting volumes. Desks use the same move on inventory: if two accounts hold the same totals after a string of transfers, whatever left one must have been replaced from the other, whatever the route. Give the one-line argument first, then the 40 and 40 as the check.

Let X be -1, 0 or 1 with equal chances and let Y = X squared. Y is fixed completely once you know X, yet the correlation is exactly zero. The covariance is E[XY] minus E[X]E[Y]; E[X] is 0 and E[XY] is the average of -1, 0 and 1, which is also 0. Correlation measures only straight-line co-movement, and a symmetric U shape has none.

What does correlation actually measure?

Think of a household's electricity bill against the outside temperature. It is high in the coldest months, when the heater runs, high in the hottest, when the air conditioner runs, and low in between. Correlation asks only whether one variable tends to rise along a straight line as the other rises, so a U-shaped link, however tight, can score zero. Over a year balanced around a mild middle, temperature explains the bill almost completely and a straight-line measure misses all of it.

How do you prove the covariance is zero?

Write out the three cases. X averages 0, Y averages 2/3, and XY takes the values -1, 0 and 1, which also average 0, so the covariance E[XY] - E[X]E[Y] is exactly 0. Symmetry does the work: every point to the right of the axis has a mirror image on the left with the same Y, so the upward slope on one side cancels the downward slope on the other. A continuous version works the same way: X normal with mean zero and Y equal to X squared.

Where does this bite on a desk?

Anywhere a payoff depends on the size of a move rather than its direction. A long straddleA call and a put bought at the same strike and expiry, which gains from a large move in either direction. gains from a big move either way, so over moves balanced around zero its return shows little correlation with the stock's return while being driven entirely by it. A risk report built only on correlations would call that position unrelated to the stock. Zero correlation means no straight-line link; only independence means no link at all.

Six flips on average for two heads in a row, and four for a head then a tail. Track how far along the pattern you are. For HH, a tail at any point sends you back to the start, including right after a head. For HT, a head after a head keeps you one step away, so progress is never lost. Solving the two small chains gives 6 and 4.

Why do two equally likely patterns take different times?

Think of two ladders where a slip costs you differently. On one, slipping from the first rung drops you to the ground; on the other, you can only ever slip back to the first rung. Both patterns are equally likely in any given pair of flips, but after one head the wrong next flip costs you everything for HH and nothing for HT. A head then a tail breaks HH and restarts it; a head then a head is still a perfect start for HT.

How do you solve the chain?

Let E0 be the expected flips still needed from the start and E1 after one head. For HH, E0 = 1 + E1/2 + E0/2 and E1 = 1 + E0/2, because a tail from one head sends you back, and these solve to E1 = 4 and E0 = 6. For HT, the one-head state just waits for a tail, which takes 2 flips on average, and reaching the first head takes 2 more, giving 4. Say the states out loud before the algebra; the interviewer wants to hear them named.

What is the general pattern?

Patterns that can fail back to nothing take longer. The expected wait for n heads in a row is 2 to the power n + 1, minus 2: 2, 6 and 14 flips for one, two and three heads. This is a Markov chainA process whose next step depends only on the current state, not on how it got there, so it can be solved state by state. at heart, and the same state-by-state method handles anything that depends on a path: a streak, a barrier, a drawdown rule on a trading book.

No. The square root of the unbiased variance underestimates the standard deviation on average. The square root is concave, so by Jensen's inequality the average of the square roots is below the square root of the average. For normal data with two observations the estimate averages about 0.80 sigma; the bias shrinks as the sample grows, to about 6% at five observations and under 1% at thirty.

Why does taking a square root break unbiasedness?

Two square rooms have floor areas of 4 and 16 square metres, so their sides are 2 and 4 metres. Average the areas, 10, and take the root: 3.16 metres. Average the sides instead: 3 metres. Averaging and then taking a square root gives a bigger answer than taking square roots and then averaging, because the square root bends downwards. The sample variance is right on average, so the average of its square roots must fall short of the true standard deviation.

How big is the bias?

It depends on the sample size and on the distribution. For normal data the expected sample standard deviation is c4 times sigma, with c4 about 0.80 at n = 2, 0.94 at n = 5, 0.97 at n = 10 and 0.99 at n = 30. At n = 2 you can check it directly: the sample standard deviation is the gap between the two draws divided by the square root of 2, and the average gap between two normal draws is 2 sigma over the square root of pi, which leaves the square root of 2/pi, about 0.798.

Does it matter in practice?

Sometimes. With a year of daily returns the bias is a rounding error; with a handful of monthly returns for a new fund it is not. A manager with five monthly returns has a volatility estimate that averages about 6% too low under normality, which flatters a Sharpe ratioAverage excess return divided by the standard deviation of returns, a measure of return per unit of risk. before anyone has looked at fat tails. Dividing by c4 removes the bias for normal data, but the fix depends on the distribution, so name the assumption. And unbiased is not the same as most accurate.

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.

About 88%. Let r be the ratio of losing to winning odds, 0.45/0.55 = 9/11. The chance of reaching Rs 20 from Rs 10 is (1 - r to the 10th)/(1 - r to the 20th), which simplifies to 1/(1 + r to the 10th). Since r to the 10th is about 0.134, the answer is 1/1.134, about 88.1%. A fair game would give exactly 50%.

Why does a small edge per bet become a large edge on the game?

Think of a tug of war between two teams, one a shade stronger. A single pull is close to a coin toss, but the rope has to travel a long way before either side wins, and every pull leans the same way. Reaching Rs 20 or Rs 0 takes many Rs 1 bets, and the 0.55 edge applies to every one of them, so the chance of winning the whole game rises far above 0.55. Staking all Rs 10 on one bet would give only 55%; betting Rs 1 at a time gives about 88%.

How do you get the formula?

Let P(i) be the chance of reaching 20 from a stake of i. One bet later you are at i + 1 with chance p or i - 1 with chance q, so P(i) = p P(i + 1) + q P(i - 1), with P(0) = 0 and P(20) = 1, and the solution is P(i) = (1 - r to the i)/(1 - r to the 20), where r = q/p. For a fair game the formula collapses to a straight line, P(i) = i/20, because a fair game keeps your expected wealth at 10, so 20 times P must equal 10. Checking the fair case is the fastest way to trust the biased one.

How do you compute r to the tenth in your head, and what is the lesson?

Square repeatedly: 9/11 squared is 81/121, about 0.669; squared again about 0.448; again about 0.201; times 0.669 gives about 0.134. The desk lesson is about sizing: with an edge, make many small bets so the edge compounds; without one, the same arithmetic works against you, and at 0.45 the Rs 1 strategy reaches Rs 20 only 12% of the time against 45% for one bold bet. This is the classic gambler's ruinThe problem of a gambler betting fixed amounts until reaching a target or losing everything, solved as a random walk with two absorbing ends. result, and it is why a trader with a real but thin edge wants volume, not size.