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.

One in three if you learn at least one is a girl, and one in two if you learn the elder is a girl. List the four equally likely families by birth order: boy-boy, boy-girl, girl-boy, girl-girl. At least one girl rules out only boy-boy and leaves three, one of which is girl-girl. Naming the elder rules out two orders and leaves girl-boy and girl-girl.

Why is the answer not simply one half?

Picture a friend tossing two coins behind a screen and telling you that at least one came up heads. She has told you something about the pair, not about a particular coin, so you cannot treat the other coin as a fresh toss. The pair had four equally likely outcomes and her remark removes only one of them, tails-tails. Two children work the same way, provided each birth is equally likely to be a boy or a girl and the two births are independent.

What changes when you learn that the elder is a girl?

Now the information is about a named child. Naming the child removes a whole row of the grid, both orders in which the elder is a boy, so two cells survive and the answer becomes one half. The younger child's sex is untouched by what you learned, which is why it now behaves like a fresh toss. The same arithmetic, 1/4 divided by 1/2, gives 1/2.

Why would a fund interviewer ask this?

Because every piece of market news arrives through a filter, and the filter changes what the news means. How you came to learn a fact is part of the fact. A screen that says at least one of two stocks beat estimates tells you less about each stock than a screen that names the one that did. Draw the grid, give both answers, and say the assumption out loud: independent births, each equally likely to be a boy or a girl.

The smaller draw averages 1/3 and the larger averages 2/3. Two random points cut the stretch from 0 to 1 into three pieces, and nothing distinguishes one piece from another, so each has expected length 1/3. The minimum ends the first piece and the maximum ends the second. The two averages add to 1, as they must, because the minimum plus the maximum is just the sum of the two draws.

What is the one-line argument?

Imagine two friends who each arrive at a random minute within the same hour. The wait until the first arrival, the gap between the two arrivals and the time from the second arrival to the end of the hour are three stretches of time. No stretch is special, so each averages a third of the hour, and the first friend arrives, on average, 20 minutes in. The same symmetry puts the minimum of two uniform draws at 1/3 and the maximum at 2/3.

How do you check it with a calculation?

Work from the chance that the minimum sits above some level x. Both draws must be above x, which happens with probability (1 - x) squared. For a quantity that cannot be negative, the average equals the area under its chance of exceeding each level, and the area under (1 - x) squared from 0 to 1 is 1/3. The maximum then follows without a second integral: minimum plus maximum equals X plus Y, which averages 1, so the maximum averages 2/3.

Where does the pattern lead next?

With n draws, the points cut the interval into n + 1 pieces, so the smallest averages 1/(n + 1) and the largest n/(n + 1). The ordered values of uniform draws are evenly spaced on average, and that single fact answers a whole family of follow-ups. With ten bids drawn at random across a range, the highest averages 10/11 of the way up. The quick route here is also the one to say first: the order statisticsThe draws sorted from smallest to largest: the minimum is the first order statistic, the maximum the last. of uniform draws split the interval evenly.

The chance is 8 in 32, exactly 1/4. Sort the qualifying sequences by where their first run of three heads starts. Starting at flip 1: HHH then anything, 4 sequences. Starting at flip 2: flip 1 must be tails, then HHH, then anything, 2 sequences. Starting at flip 3: flip 2 must be tails and flip 1 is free, 2 sequences. Four plus two plus two is eight.

Why does the obvious count go wrong?

Suppose you count guests at a party by asking each room how many people it holds, while some guests stand in doorways. They get counted twice. A run of three can start in three places, but one sequence can contain a run starting at more than one of them, so three places times four fillings, 12, counts HHHHH three times and two other sequences twice. The fix is to give each sequence exactly one home: the position where its first run starts.

How do you make every sequence count once?

Force the flip just before the run to be tails. If the first run starts at flip 2, flip 1 must be tails; if it starts at flip 3, flip 2 must be tails, and flip 1 can be anything. That gives 4, then 2, then 2. Then check a second way by counting the complement: sequences with no run of three follow a Tribonacci-style rule, each term the sum of the three before, giving 1, 2, 4, 7, 13, 24 for 0 to 5 flips. And 32 minus 24 is 8.

Why would a quant fund ask this?

Streaks are what people notice in trading records and backtests. A run of three heads turns up in a quarter of all five-flip sequences of a fair coin, so a short winning streak is weak evidence of skill. The counting habit matters as much as the number: giving each outcome one home is the same discipline that stops you double counting overlapping signals when you tally how often a strategy fired.

40%. Measure everything in units of the average loss. Each trade wins 1.5 units with probability p and loses 1 unit otherwise, so the expected result per trade is 1.5p minus (1 - p). Setting that to zero gives p = 1/2.5 = 40%. At a 45% hit rate the book makes 0.125 units a trade, so a trader who is wrong more often than right can still run a good business.

Why is a hit rate below 50% not a problem on its own?

A street vendor selling umbrellas can stand idle most days and still do well, because the rainy days pay for the dry ones. What decides whether a trading book makes money is the hit rate times the size of the wins against the miss rate times the size of the losses, not the hit rate alone. A win-loss ratio of 1.5 moves the breakeven from 50% down to 40%.

What does the general rule look like?

For a win-loss ratio R the breakeven hit rate is 1 over (1 + R). At R = 1 it is 50%, at R = 1.5 it is 40%, at R = 2 it is 33% and at R = 3 it is 25%. This is why portfolio managers are reviewed on both numbers together. A falling hit rate is fine if the winners are running further, and a rising hit rate is a warning sign if it comes from cutting winners early and letting losers run.

What would you add before calling 45% a good business?

Costs and sample size. Commission, slippage and financing come off every trade, winners and losers alike, so they raise the breakeven. If costs run at 0.05 units a trade, the book needs 1.5p - (1 - p) = 0.05, a hit rate of 42%. And a ratio of 1.5 measured over twenty trades is noisy; say you would want a longer record before trusting either number.

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.

The spread must be Rs 0.60: a bid Rs 0.30 below the mid and an offer Rs 0.30 above it. Call the half-spread s. Each uninformed order earns her s on average. Each informed order trades on the side where the value really is Rs 1 away, so she loses 1 minus s. Breaking even needs 0.7s = 0.3(1 - s), which gives s = 0.30.

Why do some of the people who trade with you cost you money?

A shop that buys second-hand phones pays the same price to every seller, but the sellers who know their phone is faulty are the keenest to sell. A quote is an offer made to everyone, and the people who know more take it only when it hurts you. The spread is the fee charged to the whole crowd to pay for the few who know more, which is called adverse selectionThe tendency for the counterparties who choose to trade with you to be the ones who know the trade is good for them..

How do you set up the break-even?

Charge a half-spread s on each side. An uninformed order is as likely to be a buy as a sell and carries no information, so on average it earns you s; an informed order buys at your offer only when the value is Rs 1 above the mid, so it costs you 1 minus s. Weight each by how often it arrives and set the total to zero: 0.7s - 0.3(1 - s) = 0, so s = 0.30 and the spread is Rs 0.60.

What moves the spread?

Two things: how much of the flow is informed and how much the informed know. In general the half-spread is s = aV, where a is the informed share and V is how far the true value sits from the mid, so doubling either doubles the spread. Here 0.3 x 1 = 0.3. That is why quotes widen ahead of a results announcement, when more of the flow may know something, and in small stocks, where one piece of news moves the value a long way. The model leaves out inventory risk and competing quoters, both of which move a real spread; say so.

Yes: buy the call, sell the put, short the stock and lend Rs 970.87, which collects about Rs 14.13 today with nothing owed at expiry. Put-call parity says the call less the put should equal the stock less the present value of the strike, 1,000 minus 1,000/1.03, about Rs 29.13. The market prices that gap at Rs 15, so the call is cheap against the put by about Rs 14.

What does a call minus a put actually pay?

Imagine agreeing today to buy a used car in six months at a fixed price. If car prices rise you gain the difference; if they fall you lose it. A long call with a short put at the same strike does exactly that: at expiry it pays the stock price minus the strike, whichever way the stock moved. So it must cost the same as buying the stock today with money borrowed against the strike, 1,000 minus 970.87, which is Rs 29.13. That identity is put-call parityFor European options on the same stock, strike and expiry, call minus put equals the stock price minus the present value of the strike, less the value of any dividends..

How do you build the trade?

Buy what is cheap and sell what is dear. The option version of the forward, long call and short put, costs Rs 15; the cash version, long stock bought with borrowed money, is worth Rs 29.13, so buy the option version and sell the cash version. Buy the call for 60, sell the put for 45, short the stock for 1,000 and lend 970.87 at 3%. The cash left over today is Rs 14.13, worth about Rs 14.55 at expiry.

What could stop the arbitrage working?

Say what you assumed. The lock holds only if the options are European, the stock pays no dividend before expiry, and you can short the stock without a borrowing fee that eats the Rs 14. A dividend lowers the forward and so the fair gap; an American put can be exercised against you early; a hard-to-borrow stock charges a fee. A gap this size in a liquid name usually means one of those is present, so your first question back is which one.

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.