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THE DERIVATIVES 'ZINE^TM     November 2001

Derivative Games^TM from 1996

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Crossings (I) (11/22/96)

Joe does equity pairs trading for a hedge fund. Each year he takes a major ski trip. This year he went to Blackcomb-Whistler with a one-mile vertical.

He started for the top at 11:00 a.m. Sometimes he waited in line, but usually moved upward over time, and never slipped downhill. He reached the top at noon. There, he had a mostly liquid lunch and a snooze in the sun. At one p.m. he started back down, gingerly, and weaving a bit more than most skiers of his ability. He reached the bottom at 2:00 p.m.

• If we make all the convenient assumptions, and consider only the minutes, seconds, and smaller units of time (not the hour, and larger time units), how many times was Joe at the same altitude at the same time on his way down as on his way up?

Crossings (I) – Solution (3/10/97)

Congratulations, again, to Christopher Merrill, who solved this one, too.

"It's clear that Joe is at a certain altitude twice (going up, then coming down) at exactly the same time (modulo hours). Because we start timing from time xx:00 and because the altitude on the way up is a non-decreasing function of time: 1) a non-increasing altitude on the way down is sufficient to guarantee at least one cross point, and 2) a monotonically decreasing altitude on the way down is necessary to guarantee only one cross point. (No uphill coasting allowed, only Schussing!)"

Crossings (II) (11/22/96)

When he got back from his trip, Joe learned that one day his favorite stock had opened at 100 and closed at 200. His favorite short had opened at 200 and closed at 100. What can you say about the number of times for which the two stock prices were the same? Consider the following highly artificial cases:

1. The price processes were both log binomial, with a common, constant drift and a common, constant volatility over the day, and 4,000 binomial periods of equal duration each day. They started 1,001 steps apart.
2. Each price process was continuous, with a continuous first derivative, monotonic (increasing or decreasing), and with finite variation.
3. Each price process was log normal diffusion with constant drift and volatility over the two-day period.

Crossings (II) – Solution (4/5/97)

No one submitted a complete solution to this problem, although Christopher Merrill did get parts one and two correct. Part 3 turns on a counterintuitive technical point that you've either studied and remembered, or you haven't.

1. In part 1, you can model both price processes simultaneously on precisely the same binomial tree. They start 1,001 steps apart at time zero. One step later, we have four possibilities, as either price might have stepped up or down: both up, the first up and the second down, the first down and the second up, and both down. However, the key point is that in each of these four cases the prices are still an odd number of steps apart. Thus, they will always remain an odd number of steps apart and can never have the same price.
2. Part 2 is mathematically almost the same as Crossings (I). The difference is that here, neither of the functions is ever flat. Hence, the prices will cross at a unique time.
3. Part 3 requires some advanced mathematics of stochastic processes. After taking logarithms of both prices, we get two normal diffusions. Their difference is a normal diffusion. Because of the violent motion of a normal diffusion – its infinite variation – if it crosses a level once, it crosses that level infinitely many times within any finite time interval. We know that the difference in logarithms is zero at least once, since the diffusion is continuous and the prices start and end as they do. Hence, the number of crossings is infinite. (I'll stick my neck out and say countably infinite.)

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