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Derivative GamesTM Last revised: March 02, 2002Revised: 8/28/00. He's Got Butterflies. Warning:
The Following Problem Is Extremely Difficult, Unless ... Down-and-out
on Wall Street –
Solution. Down-and-out
on Wall Street –
Revised. Want to challenge the world with a Derivative Game? Send it along to the attention of "Derivative Games". If we use your material, we will acknowledge your contribution. All puzzles and solutions submitted become the property of the William Margrabe Group, Inc. Derivative Games from past years. 1999 1998 1997 1996 8/28/00 He's Got Butterflies (8/28/00)Dr. Risk is always on the lookout for the fast buck. A reader sent him a question, and Dr. Risk couldn't help finding the inevitable arbitrage opportunity in the numerical example?
Dear Dr. Risk – I have a question relating to options, to be precise long butterfly strategy. How do I determine the maximum loss for a long butterfly strategy, I have heard that it is the cost of the spread, but am unsure as to what that means. To be more precise, say I created a long butterfly position with long put A, short put B, short call B and long call C.
P r e m i u m – Vinny 8/28/00 Warning: The Following Problem Is Extremely Difficult, Unless ... (8/28/00)Dr. Risk was trying to help Crédit Mayonnaise sort out some of its messy modeling problems when he stopped by the office of the global head dollar derivatives controller, just off the trading floor. The controller was trying to get a special report on spread-lock positions to the global risk manager in fifteen minutes, but was in a bit of a muddle. He had imbibed too much Beaujolais at lunch, had left his position report under the table with his drinking companions, and had a shaky memory of the notional amounts in the report. Never one to give up, Dr. Risk asked, "How can we help?" The head controller said, "I don't see how you can. I spent all last week collecting information for end-of-quarter from all our controllers. We have desks in New York, Chicago, Los Angeles, and Bermuda. The notional amount for each desk was $1 million, $2 million, etc. No fractions of a million dollars. No odd pennies, dollars, or even hundreds of thousands of dollars. No $1,234,567.89." "I hope you can tell me more." "I know for certain that notional amounts for New York, Chicago, and LA were the same number, to the penny. The notional amount for Bermuda was 81 million." "Anything else?" "The notional amount for each product was $1 million, $2 million, etc. Again, no fractions of a million. For interest rate swaps the notional amount was either 200 million or twenty million dollars. The swaptions were either eighty million or eight million dollars. Caps and floors, five million ... no, maybe, fifty billion. I'm certain of this. Spread locks, more than five million, less than fifteen." Dr. Risk started making marks on his yellow pad. Five minutes later, he said, "Can't be done. Do you remember anything else?" The controller looked glum for a moment, then brightened. "The two biggest spread lock producers were exactly tied in production, and the only other producer sold spread locks with $1 million notional value. Does that help." "We'll see." Dr. Risk made some more marks on his pad. After a few minutes, he said "I know what your spread lock figure is, but I have no idea about swaps, swaptions, and caps and floors." What is the notional amount for spread locks, and how did Dr. Risk know it? (No brute force. Not required.) Housing Crisis (7/28/00)David Friedman, head of real estate finance at at Stayer Burns, has just moved into the Martin Frankel mansion (in Greenwich, CT) for which he paid $2.7 million cash from savings. Similarly, David Johnson, on the security staff at Stayer Burns, has just moved into a co-op (in New Rochelle, NY), for which he paid $45,000 from savings. Shortly after Friedman moved in, the value of his mansion rose 30%. Right after Johnson moved in, the value of his co-op dropped 30%.
For simplicity, assume about the Davids the sort of stuff that neoclassical economists would assume to reach a conclusions in finite time: Their working lives are infinite and they are in what economists like to call static equilibrium (Conditions at this time will remain as they are. They will not get better jobs, divorce or marry, retire and move to Florida, etc.), and taxes and transaction costs are zero. They did not anticipate the change in price and don't anticipate any further change in price. Each of the two has a steady job with stable earnings, as well as some steady investment income. My Three Sons (7/28/00)Chris Merrill presents us with a problem that he found in a book that we'll cite after we give readers a chance to solve the problem We've modified it slightly to give it a "derivatives" slant. "A mathematics professor meets an old friend on the street, and they begin to walk together and chat. [After talking obnoxiously and at length about how much more money he makes as a 'rocket scientist' on Wall Street', t]he friend tells the professor: 'You know today is a very special day for me: all three of my sons celebrate their birthday this very day! So, can you tell me how old each of them is?' "'Sure,' answers the mathematician, 'but you'll have to tell me something about them.' "'OK, I'll give you some hints,' replies the father of the three sons. 'The product of their ages is 36.' "'That's fine,' says the mathematician, 'but I'll need more information.' "'The sum of their ages is equal to the number of windows in that building,' says the father pointing to a structure next to them. "The mathematician thinks for some time and replies, 'Still, I need an additional hint to solve your puzzle.' "'My oldest son has blue eyes,' says the father. "'Oh that is sufficient!' exclaims the mathematician and proceeds to give the father the correct answer, the ages of his three sons. "How old is each of the sons?" How did the friend deduce the answer? Juan Cardenas offers the following, similar challenge: "Andy Clifford the swap trader tells you: 'The notionals (in Millions) of three of my swaps add up to your house number, but their product is 72.' "You still don't know their notionals. However, he adds: 'The largest of my swaps is in Yen.' "'Aha', you say. 'Now I know the notionals!' What are they? More Rage Against the Machine (7/28/00)Juan Cardenas proposes the following challenges: Continuing with the handy "Rage Against the Machine" swap evaluator, suppose you had to be able to construct exotic swaps of all values (by fine-tuning notionals e.g.) ranging from $1M to $50M in increments of $1M, and you have some benchmark instruments. What is the minimum number of benchmarks you need? For example, with a set of three benchmarks worth 1, 2, and 5M one can price 1, 2, 3(=2+1), 4(=5-1), 5, 6, 7, 8 and so on. Similar
situation as above, except now you don't have to construct the exotic swaps, but
are given the swaps of unknown value. However, you know their values are
integral multiples of 1M. Can you do with less benchmark instruments than
in the previous case in order to get all up to 50M? For example, with benchmarks
worth 1 and 4 one can get Down-and-out on Wall Street (6/28/00)Charlie Costello had drunk one too many bottles of Irish whiskey – on the job, while trading spot dollar-yen – and had lost a bundle of his bank's money. His boss would have fired him, except that Charlie's attorney convinced the bank's attorneys that his drinking was protected under the Americans with Disabilities Act. In order to get rid of Charlie, management got him drunk one more time, so he would sign a paper agreeing to be a derivatives quant. They planned to give him a problem that he would muff, so they could then fire him for incompetence. When the fateful day came, the head currency option trader walked over to where the quants sat, so he would have witnesses, and said, "Charlie, I need you to work out the price on a yen call, cuevo put barrier option on one billion yen. The strike is 0.9 cuevos per yen, and that's the barrier, too. Both rates of interest are zero, and they're trading at parity – one to one – in the spot market. Charlie asked, "So, right now, the spot rate is one cuevo per yen, and if it gets to 0.9 cuevos per yen, the option's history and worthless?" The boss said, "That's it, Charlie." Charlie's main rival for leadership of the quant group said, "Get me cuevo-yen vol, and I can get you a price from the calculator I whipped up last week." Charlie thought, "I sure could use a drink right about now, to straighten me out." However, with all the people around him, that wasn't practical. So he focused all the attention he could on the problem. Before anybody could find cuevo-yen vol, Charlie gave a price for the option. To everybody's amazement, he was right. What was the price, and how did Charlie do it? 8/28/00 Down-and-out on Wall Street – SOLUTION (8/28/00)Juan Cardenas solves the problem, as follows: "I'll say the option is worth 100 million cuevos, and suggest that Charlie's reasoning could have been: Since they didn't tell me an expiration date for the option, it must mean its value is independent of time. Therefore I can assume the time to expiration is 0, hence its value is (spot minus strike) * 1 billion = (1 - .9) * 10^9 = 100,000,000 cuevos. 8/28/00 Down-and-out on Wall Street – Revised (8/28/00)Change the original "Down-and-out on Wall Street" problem, so the option expires in six months. CUSIP Obsession (7/28/00)Vinnie diFazio, Fixed Income Division Controller at Stanley Dean Morgan Witless, was looking over his desk's position report and noticed the following CUSIP numbers, issuers, and security issues:
That evening, he couldn't get those numbers out of his head, but he didn't know why. All night, he lay awake thinking about them, playing with them. Some was suspicious about them, but he couldn't put his finger on the problem. What -- if anything -- is the matter with them? Cosi Fan Tutte (6/28/00)Sean O'Neil challenges us with the following: "Fifteen arbs and their fifteen girlfriends – one girl per boy – went down to a resort in Boca to celebrate another profitable quarter. Like most traders, they had a penchant for fooling around, usually with each other's girlfriends. In fact, each arb had slept with every girl, though that was something that they never disclosed to any girlfriend who didn't have an obvious 'need to know'. (Not to be 'male-centric', that also means that each girl had slept with every arb.) Each arb promptly shared his latest conquest will every other arb – except for that girl's boyfriend. Thus, each arb knew (a) that all the other arbs' girlfriends were unfaithful and (b) that all the other arbs 'knew' the same thing. The only girlfriend that each arb was uncertain about was his own, since everyone was very careful to keep the facts a secret from the cuckold in question. "This was due to the fact that the arbs also had an uncompromising, if somewhat hypocritical, code --THEY could fool around as much as they wanted, but if any of them ever found out that his girlfriend had been unfaithful, he would kill her before the night was over. "One of the arbs was different, however. Besides being the only arb who didn't bring his girlfriend along, for years he had been considering quitting this most materialistic profession and taking his vows. The first night at the resort he was so disgusted by the shenanigans he saw his comrades (and their mates) engaging in, he made an irrevocable decision. The next morning he packed his bags and, in a fit of pique, announced to the assembled arbs who were razzing him, 'At least one of your girlfriends is unfaithful.' Then he headed to the airport and boarded the first flight for Rome. "The remaining 14 arbs didn't doubt his word, inasmuch as he was the only honest man in the bunch. Being arbs, they were masters of logic. They, along with their girlfriends, planned to stay at the resort for another 2 weeks. What happened after the disillusioned arb left?" Cosi Fan Tutte – SOLUTION (7/28/00)Kishor Laud offers the following solution: "If
the arb has heard of 0 'conquests' other than your own, Juan Cardenas writes: "The Cosi Fan Tutte type of problem was analyzed by Ian Stewart in his Mathematical Recreations column of the August 1998 issue of Scientific American (so I can't take credit for the solution). "It
looks like there's a typo since first it states there are fifteen girlfriends,
but it says the arb who left didn't bring his girlfriend. To make matters work
better, it is implied that every so often whoever The
argument is by induction. Sean O'Neil's submitted with his problem a solution that was a somewhat more elaborate version of the above approach: "What happens is that on their last night at the resort, each of the arbs kills his girlfriend (and in general, if there are N arbs with unfaithful girlfriends, they will kill them after N nights). "To see why this is, proceed by induction. If there are only two arbs, each one knows the other one's girlfriend is unfaithful, but he is uncertain about his own. Arb 1 says to himself, 'Assume my girlfriend is faithful. Then the other guy, knowing that he has never slept with my girlfriend, knows it is his girlfriend who is unfaithful, and will therefore kill her tonight.' When he wakes up the next morning and sees Arb 2's girlfriend flirting with some guy by the pool, he knows that indeed his girlfriend has strayed, and kills his girlfriend that night. The other arb of course does the same that same (second) night. "If there are three arbs, Arb 1 again assumes his girlfriend is faithful (and knows the other two aren't), and then deduces what the other two would do if that were so. From Arb 1's position, he imagines Arbs 2 & 3 going through the same chain of logic as above, but since he is assuming his girlfriend is faithful, and knows that the other two aren't, then they should act as in the case above of just two arbs. When he wakes up after the second night and both of the others' girlfriends are still breathing, he knows that his girlfriend cannot be faithful, with the usual consequences. "Proceed from there." Dr. Risk's explanation:
Thus, the N=14 remaining arbs all kill their girlfriends on night 14. I hope everybody had a good time, until then. Rage Against the Machine (6/28/00)Sean O'Neil challenges us with the following: A junior analyst at Sure Thing Risk Consultants was eyeing his computer balefully when his hotshot director stuck his head in and asked him if he wanted to grab a quick drink. Seeing that the analyst seemed on the verge of a breakdown, the director made inquiries. It turned out the analyst had received a call that one of the firm's most important clients was on the verge of buying a portfolio of structured notes, twelve in all, put together by the firm's most senior, if somewhat boozy, analyst, "Close Enough" Charlie. However, senior management was concerned, based on some comments overheard from Charlie, that possibly one (but only one) of the notes was priced either too high or too low, though they had no idea which one or indeed if there *was* any mispricing. Since Charlie was nowhere to be found, they had ordered the junior analyst to check it out prior to the deal going through. As the analyst explained, however, the notes "Close Enough" had designed were of an especially complex nature, and so could only be priced accurately via extensive Monte Carlo simulation. All the analyst had available on the spur of the moment for something like this was Charlie's note pricing program, which he called the ASymmetric SWaps and Interest Payment Evaluator, custom-designed and coded by Charlie himself and incomprehensible to everyone else. All the analyst had figured out about the program was how to run the "Compare" routine, which took two baskets of any kind of interest-rate derivatives and output "<", "=", or ">", depending on whether Basket One was worth less then, the same as, or more than Basket Two. His dilemma, which seemed insoluble, was that he had 3 hours in which to give an answer to senior management, and the program took fifty minutes to run for each comparison. That meant that his strategy, which was the only one he could think of, to compare each of the notes (suitably weighted according to its' declared price) to a particular one, which would require the program being run 11 times, was infeasible. In fact he would only be able to run the program three times. The analyst slid open the window and declared that he was throwing the computer through the opening, to be followed shortly by himself. The hotshot director smiled, and calmly told the analyst what he should do, and said he would return in two-and-a-half hours. Was the analyst able to determine which, if any, of the notes was mispriced? If so, how? EXTRA CREDIT: (a) Suppose the analyst was faced with the same dilemma, except now there are thirteen notes, all declared to be worth $100M. Can he still solve the problem, and if so, how? (b) Suppose the analyst had to find the outlier among fourteen notes. Pancakes, Lies, and Plastic Cups (5/28/00)Eddie Van Mechelen, a compliance officer, is enjoying breakfast one Sunday morning at the International House of Pancakes with his wife and son. His wife is a psychotherapist who had trained in Vienna. His eight-year-old son had finished his orange juice and was working on a small stack of pancakes. His parents were reading the writing on his juice cup, which was part of a series, each cup giving facts and figures about a country. The cup read: “Germany What “lies” triggered their outburst? ILOVEYOU (5/28/00)Vicki, a homesick and lovelorn, Filipina secretary at a hedge fund, was the first person at her fund to receive an infected e-mail message with “ILOVEYOU” in the subject line. She quickly opened it, but didn’t find love. Instead, her computer had contracted the Love-Bug virus from a friend of a friend back in Manila. Her e-mail program quickly sent an infected message to everyone in her address book, namely, everyone else at her hedge fund, at least four persons. Half of the recipients of her messages fell for the trick and the virus infected their e-mail software. Each of these infected recipients then infected additional computers at number greater than two of firms outside the hedge fund. (Curiously, half of those computers were Macs and half ran some version of Windows.) By a miraculous coincidence, each of the other affected firms had the same number (more than two) of computers that were infected by someone that Vicki had infected. Then the world learned about the Love-Bug virus and nobody else was infected. The number of computers outside Vicki’s hedge fund that people in Vicki’s hedge fund infected was at least 250 and no more than 300. Clearly, that’s not enough information to let you compute how many companies had computers that Vicki infected, indirectly. However, if I told you the total number of computers infected, indirectly, then you could then compute the number of companies affected. Initial question: How many companies had computers that Vicki’s computer infected, directly? ILOVEYOU – Solution (7/28/00)Anthony Cambeiro read the initial question
carefully, which had the last word, "directly", and solved the problem
correctly, as follows: ILOVEYOU – Revised (7/28/00)How many companies did Vicki's computer infect indirectly? Power to the People (5/28/00)A quantitative hedge fund manager had a bad year, and is shutting down his firm. He has put $200,000 in Treasury securities in escrow to cover termination costs. He offered the 50 people on his support staff the following choice of termination packages:
Which bonus should a rational employee select? Power to the People – Solution (7/28/00)Anthony Cambeiro carefully read the original
question – which did not require the number of dollars or cents to exceed 1"
– and did the best he could in a flawed situation, as follows: Power to the People – Revised (7/28/00)A quantitative hedge fund manager had a bad year, and is shutting down his firm. He has put $200,000 in Treasury securities in escrow to pay for termination packages. He offered the 50 people on his support staff the following choice of termination packages:
Which bonus should a rational employee select? G-Whiz (4/28/00)
Too Smart by Half (4/28/00)When they're not outsmarting the opposition, the genius traders of Mensa Partners are usually either drinking heavily or sleeping it off. Their drink of choice is the fifth of Scotch. Consequently, the teetotalling managing partner spends much of his time interviewing for potential new geniuses from the Ivies, MIT, Cal Tech, and peer institutions to replace the old geniuses whose livers and brains fail them. Out for drinks – non alcoholic – with a candidate one evening, he posed this brain teaser: "If a genius-and-a-half can drink a fifth-and-a-half in an hour-and-a-half, how many geniuses would it take to drink two fifths in 45 minutes? " Can you do it without pencil and paper? Too Smart by Half – Solution (5/28/00)Sean D. O'Neil solved this puzzle, as follows: "Like all these rate problems, the key here is to reduce to rate per genius. If a genius-and-a-half can drink a fifth-and-a-half in an hour-and-a-half, that means a genius-and-a-half can drink a fifth in an hour, which means a genius can drink 2/3 of a fifth per hour. To drink two fifths in 45 minutes means drinking 2 2/3 fifths per hour, a rate requiring four geniuses. Q.E.D." Dr. Risk can’t improve on that solution. Big Game (4/28/00)Stephen Gould offers the following challenge to our readers: A derivatives trader and his auditor are told that they are going to be dropped at a random point on the globe, and will be given guns and survival gear. They are told also that no one else will be near them. When they land, they head off in different directions. After a while, the trader sees a large moving object in the distance, and he instinctively raises his gun and shoots it. He then walks one mile south, one mile east, and one mile north, and finds himself back where he first fired the shot. What is the probability that he shot the auditor rather than a polar bear? Big Game – Solution (5/28/00)Sean D. O'Neil solved "Big Game", as follows: "Neither is likely (shooting the auditor or a polar bear). Furthermore, what's probably going to happen next is that our intrepid explorer will face criminal charges in an American courtroom. For what, you say? For shooting at a member of the crew stationed at our base at....(drumroll, please)...the South Pole, Amundsen-Scott, which is the only source of large moving objects in that area. "I know the original questioner is thinking that our explorers landed somewhere near the North Pole, hence the polar bear reference, but in fact it is far more likely that they landed near the South Pole. Infinitely more likely. Why? Because there is only one point from whence one can walk South for one mile, East for one mile, then North for one mile, anywhere near the North Pole, and that is the pole itself. "However, there is an infinity of such points lying near the South Pole, all lying between 1 mile North of the South Pole and about 1 + 1/(2*pi) miles North of the Pole. Take the larger value. The explorer heads South from that point 1 mile until he is merely 1/(2*pi) miles from the pole, then circumnavigates the pole once by heading East for a mile, then heads back North 1 mile until he is at his starting point. Given the way the question is phrased, the likelihood of who he shot is proportional to the number of points from which one can execute this curious maneuver." Sean's right, except in for what he assumed about Stephen Gould's solution. Stephen anticipated and proposed essentially the same solution as Sean. Big Game Follow-up (5/28/00)Sean O'Neil proposed the following challenge: "One follow-up to the above question: I said 'about 1 + 1/(2*pi) miles' above because the exact value is a little different. Extra credit for whoever can compute the exact value. Assume the Earth is a perfect sphere with a circumference of 40,000 km." Let's hope there's an easy solution to this one, because the way Dr. Risk solved it was a real bear – forgive the pun. Big Game Follow-up – Solution (5/28/00)Juan Cardenas solves Sean's Big Game Follow-up, as follows: The exact value is 1 + R * arcsin( 1 / ( 2 * Pi *
R )), where R is the radius of the earth (assumed to be 40,000/1.609344 miles).
A simple derivation is as follows: Comp Time (4/28/00)At bonus time the new head trader, Midas Touche, was trying to come up with a compensation formula for his quants. Lacking any better index than time served, he asked the head quant, Buzzy Propellerhead, Ph.D. (Cal Tech), how long his two junior quants had been been the firm. The head quant answered, as follows: "Mary has been with us for two years. She has been with us twice as long as Ann had been with us when Mary had been with us as long as Ann has been with us." The Midas thought for a moment, did a quick calculation, then thanked the head quant and said, "Your group's bonus pool is $450,000. Give each of the junior quants $10,000 per month, time served. The rest is yours." How much bonus did Buzzy get? Comp time – Solution (4/28/00)Sean D. O'Neil solved this, as follows: "Not much. 30 grand. Ann has been there for a year-and-a-half. "As to your question of how I solved the Comp Time problem, it's called trial and error. Not really trial and error, but sort of a discrete version of Newton-Raphson in my head. Try a value, see if it works, figure out whether it needs to be increased or decreased. Not pretty, but it's the best way to solve an off-the-cuff problem with a limited solution space and a fast way to check the answer." Dr. Risk reduced this information to four simultaneous equations in four variables: MaryNow
= 2 Dr. Risk couldn't discover these in his head or hold them in his head, much less solve them there. So he put them down on paper, along with the following: AnnThen
= 1/2 MaryNow = 1/2 x 2 = 1 That means compensation for the junior quants was $420,000. That left $30,000 = $450,000 - $420,000 for Buzzy. Five minutes after Buzzy figured this out, he felt a sudden urge to reconnect with some of his old buddies at other firms. |
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