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In his book "The Misbehavior of Markets", Mandelbrot introduces the Hurst exponent with a coin toss example. He says that when you toss coins for a large number of times, you can get consecutive runs of heads or tails. For instance, in a $100$ tosses game, you can get $8$ Heads one after the other and you can get $3$ Tails one after the other. The range in this example is $8 + 3 = 11$.

Now, Mandelbrot says that in a $10,000$ tosses game, which has $100$ times more tosses than the $100$ tosses game, the range is expected to be $110$, because the range grows as the square root of the growth in the number of tosses. In other words, as the number of tosses grows from $100$ to $10,000$, the range is expected to grow by ten times, from $11$ to $110$.

I tried this in R and, no. I simulated the ranges in $100, 1,000, 10,000, 100,000$, and $1,000,000$ tosses, $10,000$ simulations each. The modes of those simulations are:

$$100 \to 11; 1,000 \to 18; 10,000 \to 24; 100,000 \to 31; 1,000,000 \to 38$$

which grow in no way as the square root of the number of tosses. For instance, from a $100$ tosses to a million tosses, a growth of $10^4$, the range should grow by a factor of $100$ according to the book, but it grows by a factor of $\sim 3.5$, so the growth factor is more like a log $\quad(\log 10,000 = 4)$ than a square root ($\sqrt{10,000} = 100$).

Happy to provide the R code.

Thanks much!

Mete

Ricky
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Mete V
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    The length of the longest run of Heads (say) grows like $\log_2 \left(\frac n2\right)$ (with a similar result for the longest run of either type). See, e.g., this. Now, the longest strings of heads and tails are not perfectly independent, but I'd expect them to be kind of close. I'm surprised by the $\sqrt n$ claim. – lulu Aug 31 '23 at 16:00
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    Are you sure he wasn't looking at the the max and min of the running total $#H-#T$ or something like that? Longest runs just don't grow all that fast. – lulu Aug 31 '23 at 16:07
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    You have misread, @lulu is correct:"simple formula for it: The range between Harry’s biggest winnings at one moment of the game and his worst loss at another time varies by the square root of the number of tosses. For instance, say the game lasts 100 tosses, and Harry’s biggest gain was eight and his worst loss was three. The range from best to worst score was eleven" – user619894 Aug 31 '23 at 16:12
  • @user619894, actually that is what I simulated as well: "The range from best to worst score", which in Mandelbrot's example is 11 and according to Mandelbrot it would grow to 110 if the number of tosses grows 100-fold. But the simulations give a totally different result, with a growth that is much slower. – Mete V Sep 01 '23 at 05:21
  • @lulu, Mandelbrot adds the max running total H with the max running total T, and he calls the sum "the range from best to worst score". My intuition and the simulations I ran contradict the idea of a square root growth of this range with increasing number of tosses. – Mete V Sep 01 '23 at 05:24
  • Mandelbrot was discussing the running score of heads-tails, not the length of an all heads sequence – user619894 Sep 01 '23 at 05:59
  • @user619894, I don't think anyone is discussing the length of an all heads sequence. – Mete V Sep 01 '23 at 06:22
  • @lulu, the formula you mentioned ( 2 * log2(n/2) ) matches the results of the simulations, and shows that there is indeed a mistake in the book. – Mete V Sep 01 '23 at 06:26
  • As I (and others) say, I expect you are misreading. I doubt the author intended to look at consecutive runs....nobody thinks they grow like \sqrt n$. – lulu Sep 01 '23 at 10:00
  • I am sure I am misreading as Mandelbrot wouldn't make such a mistake. I wonder what the correct reading is though. – Mete V Sep 01 '23 at 13:49
  • I got it. It's not the sum of the max head runs and max tail runs. It's the sum of the max head accumulations and max tail accumulations. That DOES grow by the square root. – Mete V Sep 02 '23 at 13:31
  • By the way the paper by Schilling (1990) is really enjoyable, thanks – Mete V Sep 02 '23 at 13:31

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