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The Chairman
Victor Niederhoffer
12/06/04
10 Variations on Theme of Fibonacci and Markets, by Victor Niederhoffer
Often one is playing a game in play, life or markets where the best or worst score could come at any point . For example, during a week, Monday could be the highest price for the market, or Tuesday or Wed or Thur or Friday. Or you could be playing dice for 8 hours and you wonder exactly how likely are you to be at your maximum of wealth at the end of 4 hours? Or perhaps like Professor Pennington, the tennis playing physicist who studies individual stocks here, you're wondering what happens if you divide the Globex S and P price moves from 1 am to 930 am until 1 am into 51 10 minute intervals, exactly how likely is it to end at a minimum at 930 by chance alone? He observes that it happens fully 18% of all days In the last 7 years and wonders at this seemingly high number. Similar seeming non-randomness has been observed by the Eisenstadt student Mr. Downing who notes that Friday's max and minima in the market occur much more than they should taking account of the drift. Both the Professor and Mr. Downing appeal to simulation to find the answer. They wonder if a closed form solution is readily at hand. If you divide the Globex S and P price moves from 1 am to 930 am until 1 am into >51 10-minute intervals, exactly how likely is it to end at a minimum at 930 by chance alone?
Dr. Castaldo addresses the problem:
"If you divide the Globex S and P price moves from 1 am to 930 am until 1 am
into
51 10-minute intervals, exactly how likely is it to end at a minimum at 930 by
chance alone?"
Definition. A 'simple random walk' starts at S0 = 0 and increases or decreases
by one at each step. The values of an N-step walk can be written S0, S1, ...,
SN.
We want to know the probability that S0 is the maximum (which by symmetry and
time-reversal is the same as the probability that SN is the minimum).
The answer is given in Feller, Volume 1, 3d edition, on Page 89.
The probability is either Choose(N, N/2) / 2^N or Choose(N, (N+1)/2) / 2^N
depending whether N/2 or (N+1)/2 is an integer (that is whichever of these two
expressions makes sense). Choose(N, k) is the number of combinations of k objects chosen from N objects.
For N=51 we have Choose(51,26)/2^51and this evaluates to 0.110116