Random Fractals and the Stock Market

Unifractal Cartoons - the Scaling Condition

Unique value of dt₂

The scaling condition is

(dt1)H - (dt2)H + (1 - dt1 - dt2)H = 1

To simplify the notation, set dt1 = a and dt2 = x.

We want to show the function

f(x) = aH - xH + (1 - a - x)H

satisfies f(x) = 1 for exactly one value of x in the range 0 < x < 1 - a.

First, the graph of f(x) is decreasing because the derivative

f '(x) = -bxb-1 - (1 - a - x)b-1

is negative.

To show f(x) = 1 has a unique solution, we must show f(0) > 1 and f(1 - a) < 1, then apply the intermediate value theorem.

Observe f(0) = aH + (1 - a)H.

Certainly, we must have 0 ≤ a ≤ 1 and H ≥ 0.

First, taking a = 0.5, in the graph below it is easy to see we must restrict H to the range 0 ≤ H ≤ 1 in order to guarantee f(0) > 1.

Plot of y = 0.5H + (1 - 0.5)H for 0 ≤ H ≤ 2. The line y = 1 is drawn for reference.

Proving aH + (1 - a)H ≥ 1 in the range 0 ≤ a ≤ 1 and 0 ≤ H ≤ 1 requires a bit of work.

Instead, we establish the plausibility of this result by plotting

y = f(0) = aH + (1 - a)H.

Note f(0) ≥ 0 for a and H in this range.

Finally, note f(1 - a) = aH - (1 - a)H, so to show f(1 - a) ≤ 1 for a and H in this range, examine this plot of y = f(1 - a).

So given dt1 in the range 0 ≤ dt1 ≤ 1 and H in the range 0 ≤ H ≤ 1, there is exactly one dt2 satisfying the scaling condition.

Return to the Scaling Condition.