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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 BlackcombWhistler
with a onemile 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 nondecreasing function of time: 1) a
nonincreasing 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:
 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.
 Each price process was continuous, with a continuous
first derivative, monotonic (increasing or decreasing),
and with finite variation.
 Each price process was log normal diffusion with constant
drift and volatility over the twoday 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.
 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.
 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.
 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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