Hedge Funds puzzles, solved step by step

10 steps. Group the eight corners by how many edges separate them from the target: one corner at distance 3, three at 2, three at 1, and the target itself. From 3 the ant must move to 2; from 2 it slips back to 3 one time in three; from 1 it reaches the target one time in three. Solving the three equations gives 10 from the start, 9 from distance 2 and 7 from distance 1.

How do you turn eight corners into four states?

Ask a lost tourist how far they are from the station, not which street they are on. Every corner at the same distance from the target behaves the same way, so the distance is all you need to track. Collapsing the cube by distance turns eight corners into four states, 3, 2, 1 and 0, and the walk becomes a short chain. From distance 3 all three neighbours are at distance 2. From 2, one neighbour is back at 3 and two are at 1. From 1, two neighbours are at 2 and one is the target.

How do you solve the equations quickly?

Substitute from the two ends into the middle. Put E3 = 1 + E2 and E1 = 1 + 2E2/3 into the middle equation and it collapses to E2 = 2 + 7E2/9, so E2 = 9, E3 = 10 and E1 = 7. Then sanity-check the odd-looking one: from distance 1 the ant sits right next to the target yet still needs seven steps on average, because two of its three moves lead away. A number that surprises you is worth one sentence of explanation, not a recalculation.

What is the skill a fund is actually testing?

The same first-passage logic tells you how long a mean-reverting spread takes to reach a target or a stop, and how many moves a process needs to hit a barrier. Whenever many states behave alike, lump them, write one equation per lumped state, and solve: that is the skill, and the cube is only the costume. Candidates who write eight equations, one per corner, get the right answer too, but too slowly for the room.

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.

Two thirds of months are bull months, whatever the starting state. In the long run the number of months switching from bull to bear must equal the number switching back. 10% of bull months switch out and 20% of bear months switch in, so 0.1 x bull = 0.2 x bear, which makes bull twice as common as bear: 2/3 against 1/3. A second route: bull spells last 10 months on average and bear spells 5.

What has to balance in the long run?

Think of a shop with people walking in and out all day. Once the crowd inside stops growing or shrinking, the number walking in each minute must equal the number walking out. In the long run, the flow from bull to bear must equal the flow from bear to bull, because otherwise one state would keep filling up. The flow out of bull is 10% of bull months; the flow out of bear is 20% of bear months. Setting 0.1 x bull equal to 0.2 x bear, with bull plus bear equal to 1, gives bull = 2/3.

How do you check two thirds another way?

Use the length of each spell. A state you leave with probability p each month lasts 1/p months on average, so bull spells last 10 months and bear spells 5. Spells alternate, so over a long stretch the market spends 10 months bull for every 5 bear: 10 out of 15 is two thirds. Two methods that agree is what the interviewer is listening for.

Answer the part of the question people skip: why the starting state does not matter. The gap between today's odds and two thirds shrinks by a factor of 0.7 every month, so after 12 months only 1.4% of it is left. A model this sticky still forgets its starting point within about a year, which is why regime forecasts beyond a few months mostly return the long-run average.

25 minutes. Let E(k) be the expected minutes to exit from price k. Each minute costs one and moves the price up or down with equal chance, so E(k) = 1 + half E(k + 1) + half E(k - 1), with E(45) = E(55) = 0. The solution is E(k) = (k - 45)(55 - k), the product of the distances to the two barriers. From Rs 50 that is 5 x 5 = 25.

Why is the answer not 5 minutes?

Think of someone pacing a corridor, taking one step forward or back on each coin toss. After 25 steps they are not 25 steps away; typically they are about 5 away, because the steps keep undoing each other. A fair random walk covers distance like the square root of time, so reaching a barrier 5 away takes on the order of 5 squared, 25 steps, not 5.

How do you get exactly 25?

Set up the one-step equation. From any price k strictly between the barriers, you spend one minute and then stand at k + 1 or k - 1 with equal chance. E(k) = 1 + half E(k + 1) + half E(k - 1) says the second difference of E is always minus 2, so E is a downward parabola that is zero at both barriers. The only such parabola is (k - 45)(55 - k). Check a point: from Rs 46 it gives 1 x 9 = 9 minutes, and it passes the one-step test, since 1 plus half of E(47), which is 16, plus half of E(45), which is 0, is 9.

What does the shape tell a trader?

Starting in the middle is the slowest place to be, and an off-centre start is much faster: from Rs 48 the answer is 3 x 7 = 21 minutes, and from Rs 46 only 9. Doubling both distances quadruples the expected time: barriers at Rs 40 and Rs 60 give 100 minutes. That is the arithmetic behind why widening a stop and a profit target together makes a trade live much longer. It holds only for a fair walk: if the stock ticks up 60% of the time, the exit comes sooner, about 19.2 minutes, and mostly at the top.

About 0.289, and it settles at 2/7, about 0.286, for large targets. Let p(n) be the chance the total ever equals n. To hit n, the total must first land on one of n - 1 to n - 6 and then roll exactly the gap, each with chance 1/6, so p(n) is the average of the six values before it, with p(0) = 1. Working up gives p(10) = 0.2893. Totals advance 3.5 a roll on average, so they land on 1 number in 3.5.

How do you set up the recursion?

Think about the last roll before the total reaches n. The total can land exactly on n only by first landing on one of n - 1 down to n - 6 and then rolling exactly the gap, each with chance 1/6, so p(n) = (1/6)[p(n - 1) + ... + p(n - 6)]. Start with p(0) = 1, because you begin at zero, and p of any negative number = 0. Then p(1) = 1/6, p(2) = 7/36, p(3) = 0.227, and so on up to p(10) = 0.2893.

Why does the answer settle at 2/7?

Picture stepping stones across a river, where each stride covers 1 to 6 stones with equal chance. Over a long walk you touch about one stone in every 3.5, because that is your average stride. The running total advances 3.5 per roll on average, so in the long run it lands on a fraction 1/3.5 = 2/7 of all numbers, and each far-off target is hit with probability close to 2/7. The early values wobble: p(6) is the highest, 0.360, because 6 is the last total a single roll from zero can reach directly, and the wobbles die out by about 20.

Why would a quant interviewer ask for a table or code here?

Because the recursion is dynamic programmingSolving a problem by building up answers to smaller versions of it and reusing them, instead of recomputing from scratch.: each value reuses the six before it, so a table of ten numbers is faster and safer than listing every sequence of rolls that sums to 10. Say the recursion, compute a few terms out loud, and give the limit with its reason; that is the complete answer. Do not try to enumerate paths: there are 492 ordered ways to reach 10 with rolls of 1 to 6, each with its own probability.