Hedge Funds puzzles, solved step by step

1,225 correlations and 1,275 parameters in all. A covariance matrix is symmetric, so only the cells above the diagonal carry new information: one per pair of stocks, 50 x 49 / 2 = 1,225. The diagonal holds the 50 variances. The count grows with the square of the number of names, which is why large books estimate risk through a handful of factors instead.

Why do you count pairs rather than cells?

In a class of 50, how many handshakes happen if everyone shakes everyone else's hand once? Each person shakes 49 hands, but every handshake has been counted twice, once from each side, so it is 50 x 49 / 2 = 1,225. A correlation is a handshake: it belongs to a pair, and the pair A and B is the same pair as B and A. The diagonal is each stock paired with itself. Its correlation is 1 and needs no estimating, but the diagonal of the covariance matrix holds each stock's variance, which does.

Why does the count become a problem for a big book?

Because it grows with the square of the names. Ten times as many stocks needs about a hundred times as many correlations: 500 names need 124,750 of them plus 500 variances, 125,250 parameters. A year of daily returns on 500 names is 250 x 500 = 125,000 numbers, fewer than the parameters being estimated. With fewer days than stocks the sample matrix is singular: some combinations of positions appear to carry zero risk, and an optimiser will pile into exactly those.

What does a factor model buy you?

A factor model says each stock's return is driven by a few shared drivers, such as the market, its sector and its size, plus noise of its own. With 5 factors, 50 stocks need 250 loadings, 50 specific variances and 15 factor covariances: 315 numbers instead of 1,275. At 500 names it is 3,015 instead of 125,250. The limitation is worth saying: any risk the factors do not name is assumed independent across stocks, and in a crowded unwind that assumption is the first to break.

Light rope one at both ends and rope two at one end at the same moment; when rope one burns out, light rope two's other end, and it burns out at 45 minutes. Two flames always meet after burning 60 minutes of rope between them, so rope one takes 30 minutes however uneven it is. Rope two then has 30 minutes left, which two flames finish in 15.

Why does lighting both ends halve the time on an uneven rope?

Picture two people eating a long, uneven sandwich from opposite ends, each chewing through whatever is at their end. However the filling is spread, they meet once the whole sandwich has been eaten between them, and together they finish in half the time one would take. Two flames consume the rope's total burn time twice as fast, so a 60 minute rope lit at both ends is gone in 30 minutes, wherever the flames happen to meet. Length tells you nothing here; burn time is the only quantity you can trust.

Why is this really a question about information?

The rope hides where its time is stored, much as an order book hides how much size is waiting behind a price. The solution uses only what is known, the total burn time, and never what is not, how it is spread along the rope. Anyone who cuts a rope in half is assuming evenness that the first sentence ruled out. Say that out loud before you give the method: naming what you may not assume is half of a good answer.

Expect the follow-up. The same trick measures 15 minutes as an interval, the gap between rope one going out and rope two going out. Each rope lit from its second end at a known moment halves whatever burn time it has left, and chaining those halvings is how you reach times such as 52.5 minutes with a third rope. Walk through the chain in order, one lighting at a time.

About 64.2%. A useless signal clears a 5% bar by luck one time in twenty. The chance that all 20 stay insignificant is 0.95 to the twentieth, about 35.8%, so the chance that at least one looks like a discovery is 64.2%. On average the search turns up one false signal, as 20 x 5% suggests, but at least one appears in roughly two searches out of three.

Why does testing more ideas manufacture a winner?

Ask twenty friends to flip a coin five times each. Any one of them flips five heads only once in 32 tries, yet the chance that at least one of the twenty does is 47.0%, and that friend will look gifted. Each test is a lottery ticket for a false discovery, and buying twenty tickets makes a win likely even when nothing works. A signal chosen because it looked best among twenty has not passed a 5% test; it has passed a 64.2% one.

How do you correct for it?

Tighten the bar to match the number of tries. The Bonferroni correction tests each signal at 5% divided by 20, which is 0.25%, and that brings the chance of any false discovery back to 4.9%. The cost is power: a real but modest signal now struggles to get through. The other defence is data the search never touched: choose the best signal on one period, then test it once on another.

What does a quant fund take from this?

Research teams run thousands of tests, and the ones that get presented are the survivors. Count every test, including the ones you ran and forgot, because the significance of the survivor depends on how many were tried. That is why systematic funds keep research logs and hold data back, and why a backtest with a t-statistic of 2 means much less after a large search than after one planned test. Say the limitation: the 64% assumes independent tests; correlated signals give a lower figure, but rarely a comfortable one.

About 25 trading days, roughly five weeks. The market trades 0.2% of the shares a day and you take at most a fifth of that, so you can sell 0.04% of the company a day. A 1% stake divided by 0.04% a day is 25 days. In the desk's language the position is five days of volume, and exiting it without moving the price takes a month.

What is the one division that answers it?

Emptying a water tank through a tap that you are only allowed to open a fifth of the way: how long it takes is the tank size divided by the flow you actually use. Days to exit equals position size divided by your daily selling capacity, and capacity is market volume times your participation cap. Here that is 1% over (0.2% x 20%), or 1% over 0.04%: 25 days. Keeping everything in percent of shares means you never need the share count or the price.

Why does a portfolio manager care about this number?

Because the price can move a long way in 25 days. A position you cannot exit quickly carries more risk than its daily volatility suggests: if the stock falls 2% a day for a week, you have sold only a fifth of it. That is why many desks cap a position at a set number of days of volume, often a few days, and why liquidity sits beside volatility in position sizing. The measure has a name, days to liquidatePosition size divided by the volume you can realistically trade in a day; a common liquidity limit on hedge fund books., and interviewers like hearing it.

Say the limitations. Volume is not steady; it dries up exactly when you most want to sell, and in a sell-off everyone is trying to do the same thing. A 20% participation rate also moves the price against you, so the real exit costs more than the screen price. A stronger answer adds that you would stress the calculation with half the normal volume, which gives 50 days.

About 8% a year, two points below the 10% average. Compound growth is roughly the average return minus half the variance: 10% minus 0.5 x 0.20 squared, which is 10% minus 2%. Check with two years of +30% and -10%: the average is 10% and the volatility 20%, but 1.30 x 0.90 is 1.17, which compounds at 8.17% a year. The drag grows with the square of volatility.

Why does the average return overstate what you end up with?

A shopkeeper whose sales rise 50% one month and fall 50% the next has not broken even: 100 becomes 150 and then 75. A loss is applied to a bigger base after a gain, and a gain to a smaller base after a loss, so swings always drag compound growth below the simple average. The bigger the swings, the bigger the drag. The simple average of yearly returns is called the {term('arithmetic mean', 'The plain average of the yearly returns, adding them up and dividing by the number of years.')}; the rate your money actually grows at is the geometric mean, and it is always the lower of the two when returns vary.

How big is the drag, and where does half the variance come from?

For returns that are not too large, the geometric mean is close to the arithmetic mean minus half the variance. With 20% volatility the variance is 0.04, half of that is 0.02, so a 10% average compounds at about 8%. The two-year example makes it concrete: +30% and -10% average 10% with a standard deviation of 20%, and 1.30 x 0.90 = 1.17, a compound rate of 8.17% a year. The rule of thumb says 8.00%, close enough to trust in an interview.

Add why an allocator asks this. Two funds with the same average return and different volatility do not leave investors with the same money. Cutting volatility from 20% to 10% raises compound growth by 1.5 points with no change in the average, which is part of why lower-volatility strategies can be worth more than their averages suggest. The limitation: the half-variance rule is an approximation that weakens for very volatile or fat-tailed returns.

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.

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.

1.5% a month compounds to about 19.6% a year, and 40% over three years is about 11.9% a year. Compounding multiplies growth factors rather than adding rates: 1.015 to the twelfth is 1.196. Going the other way, take the cube root of 1.40, which is 1.119. Simple arithmetic gives 18% and 13.3%, understating the first answer and overstating the second.

Why is twelve times 1.5% not the annual return?

A savings account that credits interest every month pays interest on last month's interest. Each month's 1.5% is earned on a base that already includes every earlier month's gain, so the growth factors multiply: 1.015 times itself twelve times. That is 1.196, a 19.6% year. The extra 1.6 points over 18% are interest on interest, tiny in any one month and not tiny over a year.

How do you go back from a total to an annual rate?

Take the root, not the division. A 40% total over three years means the yearly growth factor cubed is 1.40, so the factor is the cube root of 1.40, about 1.119, an annual rate of 11.9%. Dividing 40 by 3 gives 13.3%, and 1.133 cubed is 1.456, a 45.6% total: division overstates the yearly rate because later years grow on earlier gains. Allocators compare managers on the compound annual growth rateThe single yearly rate that, compounded over the period, turns the starting value into the ending value., so dividing can misrank two funds.

How do you do it in your head?

Add the square-term correction. Compounding adds roughly n(n - 1)/2 times r squared to n times r: for 12 months at 1.5% that is 66 x 0.000225, about 1.5 points, taking 18% to about 19.5%. That is close enough to show you know the direction and the size. For the cube root, guess and check: 1.12 cubed is about 1.405, a shade over 1.40, so the answer sits just under 12%. Checking by cubing is faster and safer than estimating a root directly.

4% of the book. If the stop is hit you lose 25% of the position, and that loss must not exceed 1% of the book. So the position times 25% equals 1%, and the position is 1% divided by 0.25, which is 4%. On a Rs 1,000 crore book that is a Rs 40 crore position and a Rs 10 crore loss at the stop.

Why is the answer not simply 1%?

Think of lending a friend money for a trip when you know the worst case is that a quarter of it never comes back. If you can stand to lose Rs 1,000, you can lend Rs 4,000, because only a quarter of the loan is at risk. The loss cap limits what you can lose, and the stop decides what fraction of the position you can lose, so the size is the cap divided by the stop distance. A trader who puts on 1% because the cap is 1% has confused the bet with the damage.

What happens to the size as the stop moves?

Tighten the stop and the position can grow; widen it and the position must shrink. With a 10% stop the same 1% cap allows a 10% position, and with a 50% stop only 2%. Every pair on the curve loses exactly 1% of the book when the stop is hit. This is why a trader whose thesis needs room to breathe, say through an earnings print, carries a smaller position than one working to a tight technical level.

What does the stop not protect you against?

The arithmetic assumes you get out at the stop price. A stock that gaps through the stop overnight, on results or a regulatory order, fills below it, and the loss exceeds the cap. If the stock opens 40% down, the 4% position loses 1.6% of the book, not 1%. Risk managers therefore size to the stop and then check the gap risk separately, often capping single-name positions whatever the stop says. Give that limitation straight after the number.

About 34.1%. Count the opposite. A hand with no ace is five cards from the 48 non-aces: 1,712,304 hands out of 2,598,960, or 65.9%. So at least one ace comes up about 34.1% of the time. Adding up exactly one, two, three and four aces reaches the same 886,656 hands, but takes four calculations instead of one.

Why count the hands without an ace?

If someone asks whether at least one of your five friends will be late to dinner, you do not add the chances of exactly one late, exactly two late and so on. You ask the chance that everyone is on time and subtract it from one. At least one is a collection of four separate cases, while none is a single case, so the complement turns four calculations into one.

Deal the cards one at a time. The first card misses the aces with chance 48/52, the second with 47/51, and so on down to 44/48. Multiply the five fractions and you get 0.6588. The same number is C(48,5) over C(52,5), which is 1,712,304 over 2,598,960.

Why is 5 x 4/52 wrong, and what does it actually measure?

Five times 4/52 is 38.5%, and it sounds reasonable. It adds up the chance that each card is an ace, which counts a hand with two aces twice and a hand with four aces four times. What it really gives is the expected number of aces in the hand, 0.385, which is always at least the chance of seeing one. The two drift further apart as the hand grows: deal fourteen cards and the same method gives more than 100%.

How do you check the answer the long way?

Exactly one ace: 4 ways to pick the ace times C(48,4) = 194,580 for the rest, 778,320 hands. Two aces: 6 x 17,296 = 103,776. Three: 4 x 1,128 = 4,512. Four: 48. The four cases sum to 886,656 hands, 34.1% of 2,598,960, matching the complement. Offer this as the check if there is time, never as the first route.