Derivatives Foundation puzzles, solved step by step
- Puzzles
- 100
- Traced to a firm
- 66
- Topics
- 12
- Hard
- 29
094A broad equity index has returned an average of 11% a year over the past 30 years, with annual volatility of 16%. Give a 95% confidence interval for its true expected annual return.Old Mission CapitalChicago · 2025
Try it first
How wide is the 95% interval for the true expected return, either side of 11%?
Show the worked solution
About 5.3% to 16.7%. The standard error of a 30-year average is the yearly volatility over the square root of 30: 16/5.48 = 2.92 points. A 95% interval is 1.96 standard errors either side, about 5.7 points, so 11% plus or minus 5.7%. Thirty years of data leave the true expected return anywhere from modest to spectacular, and narrowing it to plus or minus one point would take nearly a thousand years.
Why is the interval so wide after 30 years?
Weigh yourself on a scale that is off by up to two kilos each time. One reading tells you little; the average of four readings is better, but only twice as good, not four times, because errors cancel in proportion to the square root of the count. The uncertainty in an average shrinks with the square root of the number of observations, so 30 noisy years cut a 16-point yearly spread only to about 2.9 points. That is the standard error. Multiply by 1.96 for 95% and you get roughly 5.7 points either side of 11%.
The 95% interval for the true expected return narrows only with the square root of the years of data, so with an 11% average and 16% volatility it runs from about -3% to 25% after 5 years, from 5.3% to 16.7% after 30 years, and would need about 983 years to shrink to plus or minus one point. What does the interviewer do with the answer?
Usually they push on what it means. A 30-year history cannot tell a 6% market from a 16% market with any confidence, which is why long-run return assumptions are judgements, not measurements. To get the interval down to plus or minus 2 points needs (1.96 x 16/2)^2 = 246 years, and to plus or minus 1 point, 983 years. Volatility, by contrast, is estimated far better from the same data, because daily or monthly returns give thousands of observations of spread, while there is only one 30-year path for the mean. Sampling more often does not help the mean: the average depends only on the first and last levels.
The relationshipr bar the sample average yearly return, 11% sigma the volatility of one year's return, 16% n the number of years, 30 1.96 the multiplier for a 95% normal interval What it says in wordsThe average of 30 noisy years is itself noisy, with an error of the volatility over the square root of 30, so the interval is about 5.7 points either side.Say the assumptions, because the interviewer will. The interval treats the 30 yearly returns as independent draws from one unchanging distribution, which real markets are not; regime shifts make the true uncertainty wider. The 11% and 16% are the inputs given for this exercise, so confirm any real index's figures before using them. And an arithmetic average of yearly returns is not the compound growth rate; with 16% volatility the compound rate is roughly 1.3 points lower, which is a separate question worth flagging rather than answering.
Where candidates lose it
The common loss is using 16% as the error, giving 11% plus or minus 31 points, which confuses the spread of one year with the uncertainty of an average. The opposite loss is a tiny interval from forgetting that only 30 independent years exist.
The second loss is reaching for daily data to shrink the interval. More frequent sampling sharpens the volatility estimate, not the mean: the total return over 30 years depends only on where the index started and where it ended.
What the interviewer asks next
- How many years of data would you need to tell a 6% expected return from an 8% one?
- Why does more frequent data help estimate volatility but not the expected return?
- If returns were autocorrelated, would the interval be wider or narrower?
Asked at Old Mission Capital, Prop Trading, Chicago, 2025 (Wall Street Oasis):
Confidence interval on S&P 500 return past 30 years
095Three people stand in a line facing forward, wearing hats drawn from 3 red and 2 blue. The back person sees the two hats ahead, the middle person sees only the front hat, and the front person sees none. The back says "I don't know my colour", then the middle says "I don't know my colour". What colour is the front person's hat?Wolverine Trading, Chicago, ILUSA · 2019
Try it first
What does the back person's "I don't know" rule out?
Show the worked solution
Red. The back person would know his hat if he saw two blues, since only two exist; his silence rules that out. The middle person now knows the two front hats are not both blue. If he saw blue on the front person, he would know his own was red, and he would say so. His silence means the front hat is not blue. So the front person, who sees nothing, deduces red from the two silences alone.
What does a silence tell everyone else?
If a friend who can see the scoreboard says she cannot tell who is winning, you learn that the scores are close. Her not knowing is information, because you know what she would have said if they were not. Each "I don't know" rules out every world in which that person would have known, and everyone behind and in front can use it. The back person would know only in one world: two blue hats ahead, which forces his own to be red. So his silence removes that world, and the middle and front people both hear it.
Of the four hat pairs the back person might see on the middle and front people, his silence crosses out blue and blue, the middle person's silence then crosses out a red middle with a blue front, and both rows that survive have a red hat on the front person, which is why the front person knows the answer without seeing anything. How does the middle person's silence finish it?
The middle person now knows that he and the front person are not both blue. He looks at the front hat. If it is blue, he cannot be blue too, so he is red, and he would say so. He does not. The middle person's silence can only mean he sees a red hat in front, because a blue one would have told him his own colour. The front person runs the same reasoning, does not need to see anything, and says red. The full check enumerates the seven possible hat triples, removes the ones where the back person would know, then the ones where the middle person would, and every survivor has red in front.
The relationshipM, F the middle and front hats B, R blue and red the arrow what each silence lets everyone conclude What it says in wordsThe first silence removes the case of two blues ahead, and the second removes a blue in front, which leaves only red.Then say what it rests on, because that is the interview point. The puzzle needs common knowledge: everyone knows the hat counts, everyone reasons perfectly, and everyone knows the others do too. If the middle person might simply be slow, his silence carries no information and the front person learns nothing. On a trading floor the same logic runs in the other direction: a counterparty who could have traded and chose not to has told you something, and reading those non-events is part of the job.
Where candidates lose it
The common loss is saying the front person cannot know anything because they see nothing. That ignores that the two silences are data. The question is built to see whether you treat a non-answer as information.
The second loss is running the logic from the front. Start with the person who has the most information, the back, ask what would have let him know, and strike that case. Then move forward one person at a time.
What the interviewer asks next
- Suppose the back person says "I know". What can the other two conclude?
- With 2 red and 3 blue hats, does the same chain of silences tell the front person anything?
- If only the back person speaks and says "I don't know", what can the middle person conclude about his own hat when he sees red in front?
Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis):
a brain teaser about the hat problem where 3 people go into a room with I think 3 red and 2 blue hats
098One trader in ten has a real edge and wins 60% of days; the rest win 50%. A new hire wins 8 of her first 10 days. What is the probability she has an edge?Belvedere TradingChicago · 2022
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Roughly how likely is it that she has an edge?
Show the worked solution
About 23%. With an edge, 8 wins in 10 has probability 45 x 0.6^8 x 0.4^2 = 12.1%; without, 45/1,024 = 4.4%. Out of 1,000 traders, 100 have an edge and about 12.1 of them go 8 of 10; 900 do not and about 39.6 of them go 8 of 10 by luck. So 12.1 of 51.6, about 23.4%, have an edge. A strong start raises the chance from 10% but leaves luck the likelier story.
Why is the answer so much lower than her win rate?
If a test for a rare condition comes back positive, most positives can still be false, simply because there are so many more people without the condition to produce them. Trading records work the same way. Ten days is a weak test and skill is rare, so most traders with a strong start are lucky members of the large unskilled group, not members of the small skilled one. The win rate of 80% is what she did, not what she is; the question is which kind of trader is likelier to produce that record, weighted by how many of each kind there are.
Of 1,000 new traders, the 100 with an edge produce about 12.1 who go 8 of 10 and the 900 without produce about 39.6, so only about 23% of traders with that start have an edge, even though the record is 2.75 times likelier with one. How do you run the numbers in the room?
Use natural frequencies. Of 1,000 hires, 100 have an edge. Each goes 8 of 10 with probability C(10,8) x 0.6^8 x 0.4^2 = 45 x 0.01680 x 0.16 = 0.1209, so about 12.1 of them. Of the 900 without, each goes 8 of 10 with probability 45/1,024 = 0.0439, so about 39.6. Among the 52 or so traders with 8 wins, about 12 have an edge, a posterior of 23.4%, because a likelihood ratio of 2.75 cannot overcome prior odds of 1 to 9. In odds form: 1/9 x 2.75 = 0.306, and 0.306/(1 + 0.306) = 23.4%.
The relationship0.1 the share of traders with an edge 0.1209 the chance of exactly 8 wins in 10 at 60% a day 0.0439 the chance of exactly 8 wins in 10 at 50% a day What it says in wordsWeight each explanation by how common it is and how well it fits the record, then take the edge's share.Two refinements the interviewer may ask for. If the record were 8 or more wins rather than exactly 8, the answer rises a little, to about 25%. And the practical point: a desk that promotes on a ten-day record mostly promotes luck. It takes many more days to separate a 60% trader from a 50% one, because the gap in win rate is small next to the day-to-day noise. The limitation of the model is the two-type world; real skill comes in shades, which spreads the posterior out but does not change the lesson about rare skill and short records.
Where candidates lose it
The common loss is answering 80% or 60%, confusing the trader's record with the probability that she is skilled. The interviewer is testing whether you start from the base rate.
The second loss is computing the likelihoods correctly and forgetting the prior, which gives 12.1/(12.1 + 4.4) = 73%. That is the answer for a world where half of all traders have an edge, which is not the world described.
What the interviewer asks next
- How many days would she need at an 80% win rate before you were 90% sure she had an edge?
- If one trader in three had an edge, what would 8 of 10 imply?
- Her next 10 days are 5 wins and 5 losses. Update the probability.
Asked at Belvedere Trading, Capital Markets, Chicago, 2022 (Wall Street Oasis):
The technical portion of the interview consisted of probability questions including one questions relating to Bayes' theorem
