Hedge Funds puzzles, solved step by step

The normal model expects a 4% day about once every 63 years, so two in one year says the model is wrong, not that you were unlucky. A 4% move is four standard deviations, and the normal puts about 0.0063% of days beyond that in either direction, 0.016 such days a year. Under the model, two in a year has a chance of roughly 1 in 7,933. Real returns have fatter tails and volatility that clusters.

How rare is four standard deviations under a normal curve?

Adult heights are roughly normal. Someone four standard deviations above the average is so rare that you could meet many thousands of people without seeing one. The normal tail thins faster than exponentially, so each extra standard deviation makes an event far rarer: beyond 2 is about 1 day in 22, beyond 3 about 1 in 370, beyond 4 about 1 in 15,787. With 1% daily volatility, a 4% day is a four standard deviation day.

What do two such days in a year actually tell you?

Work out how surprising the evidence is under the model. With 0.016 expected a year, two or more has a probability of about 1 in 7,933 under the normal model, so either this was an extraordinarily rare year or the model is wrong, and the second is far more likely. Market returns have fatter tails than the normal and volatility that comes in clusters, so a 1% volatility estimated over calm months understates risk once the market turns. The honest conclusion is to re-estimate volatility with recent data and stop quoting tail odds from the normal curve.

What does a risk manager do with this?

Two things. Replace the normal tail with something that respects the data, a fatter-tailed distribution or historical scenarios, and let the volatility estimate react faster, for example by weighting recent days more heavily. Then ask the more useful question for the book: not how likely a 4% day is, but what the book loses if it happens twice in a month. A risk limit calibrated on the normal model is exactly the number this evidence has just discredited.

One in four. Three lengths make a triangle only if no piece is longer than the other two together, which for a stick of length 1 means no piece is longer than one half. Plot the two break points as a dot in the unit square. The dots that keep every piece under half fill two triangles of area 1/8 each, so the chance is 1/4.

What exactly has to be true for a triangle?

Try to make a triangle from three sticks where one is longer than the other two laid end to end: the short ones cannot reach each other across the long one. Three lengths form a triangle only if the longest is shorter than the sum of the other two, and with a total of 1 that means every piece must be under one half. That turns a geometry question into a single condition you can test on any pair of break points.

How do you turn two random breaks into an area?

Call the break points x and y, each uniform between 0 and 1. Every possible outcome is a dot in the unit square, and every dot is equally likely, so probability is area. Take the half where x is below y: the pieces are x, y minus x and 1 minus y, and all three are under half only when x is below 1/2, y is above 1/2 and y minus x is below 1/2. Those three lines cut out a triangle with corners (0, 1/2), (1/2, 1/2) and (1/2, 1), area 1/8. The half where y comes first gives its mirror image, another 1/8.

The red regions are also worth naming, because each one is a different piece being too long. Three failure regions of area 1/4 each are the first piece, the middle piece and the last piece exceeding half. They cannot overlap, since two pieces cannot both be longer than half, so the answer is 1 minus 3/4, which is 1/4, a second route to the same number.

Mention the variant, because interviewers often switch to it: break the stick once, then break the longer piece at a random point. The answer changes to about 0.386, which is 2 ln 2 minus 1, because the second break is no longer uniform over the whole stick. Knowing the setup changes the answer shows you are reading the question.

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.

7/16, about 43.8%. Plot one trader's arrival on each axis of a 60 by 60 minute square. They meet when the arrivals are within 15 minutes, a band around the diagonal. They miss in the two corner triangles, each 45 by 45 over two, which together cover 45 squared over 60 squared, or 9/16. So they meet 1 minus 9/16, which is 7/16 of the time.

Why turn two arrival times into a square?

Think of two friends who say they will meet at a cafe sometime in the lunch hour. Every possible pair of arrival times is one point: the first friend's time across, the second's up. Because both times are uniform and independent, every point in the square is equally likely. A probability question about two uniform times becomes an area question, and the areas of triangles need no calculus.

How do you get the area without integrating?

Find the region where they miss, because it is two clean triangles. They miss when A arrives more than 15 minutes after B, or B more than 15 minutes after A. Each miss region is a right triangle with legs of 45 minutes, so each covers 45 x 45 / 2 = 1,012.5 square minutes out of 3,600. Together that is 2,025 out of 3,600, or 9/16, and the band where they meet is the remaining 7/16.

How does the answer move with the waiting time?

The formula shows the shape. Waiting 30 minutes gives 1 minus (30/60) squared, which is 75%, not double 7/16. The chance rises quickly at first and then flattens, because the band is squeezed at the ends of the hour, where someone arriving at 9:55 has only 5 minutes left to be joined. The same geometry answers questions about two orders arriving within a latency window, or two news releases landing in the same trading hour.