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x(n+1) = 1 + 2.8*(x(n)-1)/(x(n)*x(n)-2*x(n)+2+y(n)*y(n))
y(n+1) = 2.8*y(n)/(x(n)*x(n)-2*x(n)+2+y(n)*y(n))
with probability q = ½.
We map x and y on a graph of two dimensions. If the coin flip comes up
heads, we iterate the system by the first two equations. This iteration
represents a simple 90-degree rotation about the origin (0,0). If the coin flip
comes up tails, we iterate the system by the second two equations. This
second type of iteration contracts or expands the current point with respect to
(1,0).
To see this Simple Stochastic Fractal system works in real time, be sure Java
is enabled on your web browser, and click here. [2]

Simple stochastic dynamical systems create simple fractals, like those we see
in nature and in financial markets. But in order to get from Bachelier to
Mandelbrot, which requires a change in the way we measure probability, it
will be useful for us to first think about something simpler, such as the way
we measure length.
One we™ve learned to measure length, we™ll find that probability is jam on
toast.
Sierpinski and Cantor Revisited
In Part 2, when we looked at Sierpinski carpet, we noted that a Sierpinski
carpet has a Hausdorff dimension D = log 8/log 3 = 1.8927¦ So if we have a
Sierpinski carpet with length 10 on each side, we get
N = rD = 10D = 101.8927 = 78.12
smaller copies of the original. (For a nice round number, we can take 9 feet
on a side, and get N = 91.8927 = 64 smaller copies.) Since each of these
smaller copies has a length of one foot on each side, we can call these "square
feet". But really they are "square Sierpinskis", because Sierpinski carpet is
not like ordinary carpet.


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So let™s ask the question: How much space (area) does Sierpinski carpet take
up relative to ordinary carpet? We have 78.12 smaller copies of the original.
So if we know how much area (in terms of ordinary carpet) each of these
smaller copies takes up, we can multiply that number by 78.12 and get the
answer.
Hmmm. To calculate an answer this question, let™s take the same approach
we did with Cantor dust. In the case of Cantor dust, we took a line of length
one and began cutting holes in it. We divided it into three parts and cut out
the middle third, like this:


0__________________________________________________1



0________________1/3 2/3_______________1


That left 2/3 of the original length. Then we cut out the middle thirds of each
of the two remaining lines, which left 2/3 of what was there before; that is, it
left (2/3)(2/3), or (2/3)2. And so on. After the n-th step of cutting out middle
thirds, the length of the remaining line is (2/3)n.
If we take the limit as n ’ ∞ (as n goes to infinity), we have (2/3)n ’ 0 (that
is, we keep multiplying the remaining length by 2/3, and by so doing, we
eventually reduce the remaining length to zero). [3] So Cantor dust has a
length of zero. What is left is an infinite number of disconnected points, each
with zero dimension. So we said Cantor dust had a topological dimension of
zero. Even though we started out with a line segment of length one (with a
dimension of one), before we began cutting holes in it.
Well. Now let™s do the same thing with Sierpinski carpet. We have an
ordinary square and divide the sides into three parts (divide by a scale factor
of 3), making 9 smaller squares. Then we throw out the middle square,
leaving 8 smaller squares, as in the figure below:




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So we have left 8/9 of the original area. Next, we divide
up each of the smaller squares and throw out the centers.
Each of them now has 8/9 of its original area, so the area
of the big square has been reduced to (8/9)(8/9) of its
original size, or to (8/9)2. At the n-th step of this process,
we have left (8/9)n of the original area. Taking the limit
as n ’ ∞ (as n goes to infinity), we have (8/9)n ’ 0 . So
the Sierpinski carpet has an area of zero.
What? This seems properly outrageous. The 78.12
smaller copies of the original Sierpinski carpet that
measured 10 x 10 (or 64 smaller copies of an original
Sierpinski carpet that measured 9 x 9), actually take up
zero area. By this argument, at least. By this way of
measuring things.
We can see what is happening, if we look at the
Sierpinski carpet construction again. Note in the graphic
above that the outside perimeter of the original big
square never acquires any holes as we create the
Sierpinski carpet. So the outside perimeter forms a loop:
a closed line in the shape of a square. A loop of one
dimension.
Next note that the border of the first center square we
remove also remains intact. This leaves a second smaller
(square) loop: a second closed line of one dimension,
inside the original loop. Next, the centers of the 8
smaller squares also form even smaller (square) loops. If
we continue this process forever, then in the limit we are
left with an infinite number of disconnected loops, each
of which is a line of one dimension. This is the
Sierpinski carpet.
Now, with respect to Cantor dust, we said we had an
infinite number of disconnected points, each with zero
dimension, and then chose to say that Cantor dust itself
had a topological dimension of zero. To be consistent,
then, we must say with respect to the Sierpinski carpet,
which is made up of an infinite number of disconnected
loops, each of one dimension, that it has a topological
dimension of one.
Hmm. Your eyebrows raise. Previously, in Part 2, I said
Sierpinski carpet had an ordinary (or topological)
dimension of 2 . That was because we started with a 10
by 10 square room we wanted to cover with carpet. So,
intuitively, the dimension we were working in was 2.

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The confusion lies in the phrase "topological or
ordinary" dimension. These are not the same. Or, better,
we need more precision. In the case of Sierpinski carpet,
we started in a context of two-dimensional floor space.
Let™s call this a Euclidean dimension of 2. It corresponds
to our intuitive notion that by covering a floor space with
carpet, we are doing things in a plane of 2 dimensions.
But, once we measure all the holes in the carpet, we
discovered that what we are left with is carpet that has
been entirely consumed by holes. It has zero area. What
is left over is an infinite number of disconnected closed
loops, each of which has a dimension of one. So, in this
respect, let™s say that Sierpinski carpet has a topological
dimension of one.
Thus we now have three different dimensions for
Sierpinski carpet: a Euclidean dimension (E) of 2, a
topological dimension (T) of 1, and a Hausdorff
dimension (D) of 1.8927¦
Similarly, to create Cantor dust, we start with a line of
one dimension. Our working space is one dimension. So
let™s say Cantor dust has a Euclidean dimension (E) of 1,
a topological dimension (T) of 0, and a Hausdorff
dimension (D) of log 2/log 3 = .6309¦
So here are three different ways [4] of looking at the
same thing: the Euclidean dimension (E), the topological
dimension (T), and the Hausdorff dimension (D). Which
way is best?
Blob Measures Are No Good
Somewhere (I can™t find the reference) I read about a
primitive tribe that had a counting system that went: 1, 2,
3, many. There were no names for numbers beyond 3.
Anything numbered beyond three was referred to as
"many".
"We™re being invaded by foreigners!" "How many of
them are there?" "Many!"
It™s not a very good number system, since it can™t
distinguish between an invading force of five and an
invading force of fifty.
(Of course, if the enemy was in sight, one could get
around the lack of numbers. Each individual from the
local tribe could pair himself with a invader, until there


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were no unpaired invaders left, and the result would be
an opposing force that matched in number the invading
force. George Cantor, the troublemaker who invented set
theory, would call this a one-to-one correspondence.)
"Many." A blob. Two other blob measures are: zero and
infinity. For example, Sierpinski carpet has zero area and
so does Cantor dust. But they are not the same thing.
We get a little more information if we know that Cantor
dust has a topological dimension of zero, while a
Sierpinski carpet has a topological dimension of one. But
topology often conceals more than it reveals. The
topological dimension of zero doesn™t tell us how Cantor
dust differs from a single point. The topological
dimension of one doesn™t tell us how a Sierpinski carpet
differs from a circle.
If we have a circle, for example, it is fairly easy to
measure its length. In fact, we can just measure the
radius r and use the formula that the length L (or
"circumference" C) is
L=C=2πr
where π = 3.141592653¦ is known accurately to
millions of decimal places. But suppose we attempt to
measure the length of a Sierpinski carpet? After all, we
just said a Sierpinski carpet has topological dimension of
one, like a line, so how long is it? What is the length of
this here Sierpinski carpet compared to the length of that
there circle?
To measure the Sierpinski carpet we began measuring
smaller and smaller squares, so we keep having to make
our measuring rod smaller and smaller. But as the
squares get smaller, there are more and more of them. If
we actually try to do the measurement, we discover the
length goes to infinity. (I™ve measured my Sierpinski
carpet; haven™t you measured yours yet?)
Infinity. A blob. "How long is it?" "Many!"
Coastlines and Koch Curves
If you look in the official surveys of the length of
borders between countries, such as that between Spain
and Portugal, or between Belgium and The Netherlands,
you will find they can differ by as much as 20 percent.
[5]

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Why is this? Because they used measuring rods that were
of different lengths. Consider: one way to measure the
length of something is to take a measuring rod of length
m, lay it alongside what you are measuring, mark the
end point of the measuring rod, and repeat the process
until you have the number N of measuring rod lengths.
Then for the total length L of the object, you have
L=mN
(where "m N" means "m times N").
For example, suppose we are measuring things in feet,
and we have a yardstick (m = 3). We lay the yardstick
down the side of a football field, and come up with N =
100 yardstick lengths. So the total length is
L = 3 (100) = 300 feet.
And, if instead of using a yardstick, we used a smaller
measuring rod”say a ruler that is one foot long, we
would still get the same answer. Using the ruler, m = 1
and N = 300, so L = 1 (300) = 300 feet. This may work
for the side of a football field, but does it work for the
coastline of Britain? Does it work for the border length
between Spain and Portugal?
Portugal is a smaller country than Spain, so naturally it
used a measuring rod of shorter length. And it came up
with an estimate of the length of the mutual border that
was longer than Spain™s estimate.
We can see why if we imagine measuring, say, the
coastline of Britain. If we take out a map, lay a string
around the west coast of Britain, and then multiply it by
the map scale, we™ll get an estimate of the "length" of the
western coastline. But if we come down from our
satellite view and actually visit the coast in person, then
we will see that there are a lot of ins and outs and
crooked jags in the area where the ocean meets the land.
The smaller the measuring rod we use, the longer will
our measure become, because we capture more of the
length of the irregularities. The difference between a
coastline and the side of a football field is the coastline is
fractal and the side of the football field isn™t.
To see the principles involved, let™s play with something
called a Koch curve. First we will construct it. Then we
will measure its length. You can think of a Koch curve

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as being as being a section of coastline.
We take a line segment. For future reference, let™s say its
length L is L = 1. Now we divide it into three parts (each
of length 1/3), and remove the middle third. But we
replace the middle third with two line segments (each of
length 1/3), which can be thought of as the other two
sides of an equilateral triangle. This is stage two (b) of
the construction in the graphic below:




At this point we have 4 smaller segments, each of length
1/3, so the total length is 4(1/3) = 4/3. Next we repeat
this process for each of the 4 smaller line segments. This
is stage three (c) in the graphic above. This gives us 16
even smaller line segments, each of length 1/9. So the
total length is now 16/9 or (4/3)2.
At the n-th stage the length is (4/3)n, so as n goes to
infinity, so does the length L of the curve. The final
result "at infinity" is called a Koch curve. At each of its
points it has a sharp angle. Just like, say, Brownian
motion seen at smaller and smaller intervals of time. (If
we were doing calculus, we would note there is no
tangent at any point, so the Koch curve has no
derivative. The same applies to the path of Brownian
motion.)
However, the Koch curve is continuous, because we can
imagine taking a pencil and tracing its (infinite) length
from one end to the other. So, from the topological point
of view, the Koch curve has a dimension of one, just like
the original line. Or, as a topologist would put it, we can
deform (stretch) the original line segment into a Koch
curve without tearing or breaking the original line at any

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point, so the result is still a "line", and has a topological
dimension T = 1.
To calculate a Hausdorff dimension, we note that at each
stage of the construction, we replace each line segment
with N = 4 segments, after dividing the original line
segment by a scale factor r = 3. So its Hausdorff
dimension D = log 4/log 3 = 1.2618¦
Finally, when we constructed the Koch curve, we did so
by viewing it in a Euclidean plane of two dimensions.
(We imagined replacing each middle line segment with
the other two sides of an equilateral triangle”which is a
figure of 2 dimensions.) So our working space is the
Euclidean dimension E = 2.
But here is the key point: as our measuring rod got
smaller and smaller (through repeated divisions by 3),
the measured length of the line got larger and larger. Just
like a coastline. (And just like the path of Brownian
motion.) The total length (4/3)n went to infinity as n
went to infinity. At the n-th stage of construction we had
N = 4n line segments, each of length m = (1/3)n, so the
total length L was:
L = m N = (1/3)n 4n = (4/3)n.
Well, there™s something wrong with measuring length
this way. Because it gives us a blob measure. Infinity.
"Many."
Which is longer, the coast of Britain or the coast of
France? Can™t say. They are both infinity. Or maybe they
have the same length: namely, infinity. They are both
"many" long. Well, how long is the coastline of Maui?
Exactly the same. Infinity. Maui is many long too. (Do
you feel like a primitive tribe trying to count yet?)
Using a Hausdorff Measure
The problem lies in our measuring rod m. We need to do
something to fix the problem that as m gets smaller, the
length L gets longer. Let™s try something. Instead of
L=mN,
let™s adjust m by raising it to some power d. That is,
replace m by md:
L = md N .


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This changes our way of measuring length L, because
only when d = 1 do we get the same measure of length
as previously.
If we do this, replace m by md, we discover that for
values of d that are too small, L still goes to infinity. For
values of d that are too large, L goes to zero. Blob
measures. There is only one value of d that is just right:
namely, the Hausdorff dimension d = D. So our measure
of length becomes:
L = mD N
How does this work for the Koch curve? We saw that for
a Koch curve the number of line segments at stage n was
N = 4n, while the length of a line segment m = (1/3)n. So
we get as our new measure of the length L of a Koch
curve (where D = log 4/log 3):
L = mD N = ((1/3)n)D (4n) = ((1/3)n)log 4/log 3 (4n) = 4-n
(4n) = 1.
Success. We™ve gotten rid of the blob. The length L of
the Koch curve under this measure turns out to be the
length of the original line segment. Namely, L = 1.
The Hausdorff dimension D is a natural measure
associated with our measuring rod m. If we are
measuring a football field, then letting D = 1 works just
fine to measure out 100 yards. But if we are dealing with
Koch curves or coastlines, then some other value of D
avoids the futile exercise having the measured length
fully dependent on the length of the measuring rod.
To make sure we understand how this works, let™s
calculate the length of a Sierpinski carpet constructed
from a square with a starting length of 1 on each side.
For the Sierpinski carpet, N gets multiplied by 8 at each
stage, while the measure rod gets divided by 3. So the
length at stage n is:
L = mD N = ((1/3)n)D (8n) = ((1/3)n)log 8/log 3 (8n) = 8-n
(8n) = 1.
Hey! We™ve just destroyed the blob again! We have a
finite length. It™s not zero and it™s not infinity. Under this
measure, as we go from the original square to the
ultimate Sierpinski carpet, the length stays the same. The
Hausdorff length (area) of a Sierpinski carpet is 1,


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assuming that we started with a square that was 1 on
each side. (We can informally choose to say that the
"area" covered by the Sierpinski carpet is "one square
Sierpinski", because we need a Euclidean square, the
length of each side of which is 1, in order to do the
construction.) [6]
[Note that if we use a d > D, such as d = 2, then the
length L of the Sierpinski carpet goes to zero, as n goes
to infinity. And if we use a d < D, such as d = 1, then the
length goes to infinity, as n goes to infinity. So, doing
calculations using the Euclidean dimension E =2 leads to
an "area" of zero, while calculations using the
topological dimension T=1 leads to a "length" of infinity.
Blob measures.]
If instead we have a Sierpinski carpet that is 9 on each
side, then to calculate the "area", we note that the
number of Sierpinski copies of the initial square which
has a side of length 1 is (dividing each side into r = 9
parts) N = rD = 9D = 64. Thus, using the number of
Sierpinski squares with a side of length 1, then, as the
basis for our measuresment, the Sierpinski carpet with 9
on each side has an "area" of N = 9D = 91.8927¦ = 64. A
Sierpinski carpet with 10 on each side has an "area" of N
= 101.8927¦ = 78.12. And so on.
The Hausdorff dimension, D = 1.8927¦, is closer to 2
than to 1, so having an "area" of 78.12 (which is in the
region of 102 = 100) for a side length of 10 is more
esthetically pleasing than saying the "area" is zero.
This way of looking at things lets us avoid having to say
of two Sierpinski carpets (one of side 9 and the other of
side 1): "Oh, they™re exactly the same. They both have
zero area. They both have infinite length!" Blah, blah,
blob, blob.
Indeed do "many" things come to pass.
To see a Sierpinski Carpet Fractal created in real time,
using probability, be sure Java is enabled on your web
browser, and click here.

Jam Session
One of the important points of the discussion above is
that the power (referring specifically to the Hausdorff D)
to which we raise things is crucial to the resulting


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measurement. If we "square" things (raise them to the
power 2) at times when 2 is not appropriate, we get blob
measures equivalent to, say, "this regression coefficient
is ˜many™ ".
Unfortunately, people who measure things using the
wrong dimension often think they are saying something
other than "many." They think their measurements mean
something. They are self-deluded. Many empirical and
other results in finance are an exercise in self-delusion,
because the wrong dimension has been used in the
calculations.
When Louis Bachelier gave the first mathematical
description of Brownian motion in 1900, he said the
probability of the price distribution changes with the
square root of time. We modified this to say that the
probability of the log of the price destribution changes
with the square root of time”and from now on,
without further discussion, we will pretend that that™s
what Bachelier said also.
The issue we want to consider is whether the appropriate
dimension for time is D = ½. In order to calculate
probability should we use T1/2, or TD, where D may take
values different from ½?
This was what Mandelbrot was talking about when he
said the empirical distribution of price changes was "too
peaked" to come from a normal distribution. Because the
dimension D = ½ is only appropriate in the context of a
normal distribution, which arises from simple Brownian
motion.
We will explore this issue in Part 4.
Notes
[1] David Bohm™s hidden-variable interpretation of the
quantum pilot wave (which obeys the rules of quantum
probability) is discussed in John Gribbin, Schrodinger™s
Kittens and the Search for Reality, Little, Brown and
Company, New York, 1995.
[2] If your computer monitor has much greater precision
than assumed here, you can see much more of the fractal
detail by using a larger area than 400 pixels by 400
pixels. Just replace "200" in the Java program by
one-half of your larger pixel width, and recompile the


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applet.
[3] Note that in Part 2, we measured the length of the
line segments that we cut out. Here, however, we are
measuring the length of the line segment that is left
behind. Both arguments, of course, lead to the same
conclusion. We cut out a total of length one from the
original line of length one, leaving behind a segment of
length zero.
[4] This three-fold classification corresponds to that in
Benoit B. Mandelbrot, The Fractal Geometry of Nature,
W.H. Freeman and Company, New York, 1983.
[5] L. F. Richardson, "The problem of contiguity: an
appendix of statistics of deadly quarrels," General
Systems Yearbook, 6, 139-187, 1961.
[6] Whether one refers to the resulting carpet as "1
square Sierpinski" or just "1 Sierpinski" or just "a carpet
with a side length of 1" is basically a matter of taste and
semantic convenience.

J. Orlin Grabbe is the author of International Financial
Markets, and is an internationally recognized derivatives
expert. He has recently branched out into cryptology,
banking security, and digital cash. His home page is
located at http://www.aci.net/kalliste/homepage.html .
-30-
from The Laissez Faire City Times, Vol 3, No 26, June
28, 1999




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A Simple Stochastic Fractal


A Simple Stochastic Fractal
If you can read this, then your browser is not set up for Java. . .

This applet generates a simple stochastic fractal. It maps points (x,y) on a plane. You can think of the
center of the applet screen as being (0,0). It plots 150,000 points in an area that covers 400 pixels x 400
pixels on your computer screen (i.e. a total of 160,000 pixels).
First a random number ("ran") between 0 and 1 is selected. If ran is greater than or equal to 1/2, then we
rotate the current (x,y) point 90 degress to the left, and plot a new point. If ran is less than 1/2, then,
depending on the distance of (x,y) from (1,0), we expand or contract the values of x and y with respect to
(1,0).

Here is the java source code.




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Sierpinski Fractal


Sierpinski Carpet Fractal
If you can read this, then your browser is not set up for Java. . .

This applet generates a Sierpinski carpet. If you look at the Java source code you will see that it generates
a random number, and then with probability 1/8 does the following:
q divides the current (x,y) value by 3,

q and may also add either 1/3 or 2/3 to x, or to y, or to both.

The key omission, in order to create the Sierpinski carpet holes, is that zero probability is assigned to the
possibility
Xnew = X/3 + .3333333333333333;
Ynew = Y/3 + .3333333333333333;
This omission creates a hole at each smaller level of the carpet.

Here is the java source code.




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Chaos and Fractals in Financial Markets, Part 4, by J. Orlin Grabbe


Chaos and Fractals in Financial Markets
Part 4
by J. Orlin Grabbe

Gamblers, Zero Sets, and Fractal Mountains
Henry and Thomas are flipping a fair coin and betting $1 on the outcome. If the coin
comes up heads, Henry wins a dollar from Thomas. If the coin comes up tails,
Thomas wins a dollar from Henry. Henry™s net winnings in dollars, then, are the total
number of heads minus the total number of tails.
But we saw all this before, in Part 3. If we let x(n) denote Henry™s net winnings, then
x(n) is determined by the dynamical system:
x(n) = x(n) + 1, with probability p = ½
x(n) = x(n) “ 1, with probability q = ½.
The graph of 10,000 coin tosses in Part 3 simply shows the fluctuations in Henry™s
wealth (starting from 0) over the course of the 10,000 coin tosses.
Let™s do this in real time, although we will restrict ourselves to 3200 coin tosses.
Let™s plot Henry™s winnings for a new game that lasts for 3200 flips of the coin. You
can quickly see the results of many games with a few clicks of your mouse. Make
sure Java is enabled on your web browser, and click here.

There are three things to note about this demonstration:
1. Even though the odds are even for each coin clip, winnings or losses can at
times add up significantly. Even though a head or a tail is equally probable for
each coin flip, there can be a series of "runs" that result in a large loss to either
Henry or Thomas. This fact is important in understanding the "gambler™s ruin"
problem discussed later.
2. The set of points where x(n) comes back to x(n) = 0 (that is, the points where
wins and losses are equalized), is called the zero set of the system. Using n as
our measure of time, the time intervals between each point of the zero set are
independent, but form clusters, much like Cantor dust. To see the zero set
plotted for the coin tossing game, make sure Java is enabled on your web
browser and click here. The zero set represents those times at which Henry has
broken even. (Make sure to run the series of coin flips multiple times, to
observe various patterns of the zero set.)
3. The fluctuations in Henry™s winnings form an outline that is suggestive of
mountains and valleys. In fact, this is a type of "Brownian landscapes" that we
see around us all the time. To create different "alien" landscapes, for, say, set


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decorations in a science fiction movie, we can change the probabilities. The
effects in three dimensions, with a little color shading, can be stunning.
Since we will later be discussing motions that are not Brownian, and distributions
that are not normal (not Gaussian), it is important to first point out an aspect of all
this that is somewhat independent of the probability distribution. It™s called the
Gambler™s Ruin Problem. You don™t need nonnormal distributions to encounter
gambler's ruin. Normal ones will do just fine.
Futures Trading and the Gambler™s Ruin Problem
This section explains how casinos make most of their money, as well as why the
traders at Goldman Sachs make more money speculating than you do. It™s not
necessarily because they are smarter than you. It™s because they have more money.
(However, we will show how the well-heeled can easily lose this advantage.)
Many people assume that the futures price of a stock index, bond, foreign currency,
or commodity like gold represents a fair bet. That is, they assume that the probability
of an upward movement in the futures price is equal to the probability of a
downward movement, and hence the mathematical expectation of a gain or loss is
zero. They use the analogy of flipping a fair coin. If you bet $1 on the outcome of the
flip, the probability of your winning $1 is one-half, while the probability of losing $1
is also one-half. Your expected gain or loss is zero. For the same reason, they
conclude, futures gains and futures losses will tend to offset each other in the long
run.
There is a hidden fallacy in such reasoning. Taking open positions in futures
contracts is not analogous to a single flip of a coin. Rather, the correct analogy is that
of a repeated series of coin flips with a stochastic termination point. Why? Because
of limited capital. Suppose you are flipping a coin with a friend and betting $1 on the
outcome of each flip. At some point either you or your friend will have a run of bad
luck and will lose several dollars in succession. If one player has run out of money,
the game will come to an end. The same is true in the futures market. If you have a
string of losses on a futures position, you will have to post more margin. If at some
point you cannot post the required margin, you will have to close out the contract.
You are forced out of the game, and thus you cannot win back what you lost. In a
similar way, in 1974, Franklin National and Bankhaus I. D. Herstatt had a string of
losses on their interbank foreign exchange trading positions. They did not break even
in the long run because there was no long run. They went broke in the intermediate
run. This phenomenon is referred to in probability theory as the gambler's ruin
problem [1].
What is a "fair" bet when viewed as a single flip of the coin, is, when viewed as a
series of flips with a stochastic ending point, really a different game entirely whose
odds are quite different. The probabilities of the game then depend on the relative
amounts of capital held by the different players.
Suppose we consider a betting process in which you will win $1 with probability p


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and lose $1 with probability q (where q = 1 - p). You start off with an amount of $W.
If your money drops to zero, the game stops. Your betting partner”the person on the
other side of your bet who wins when you lose and loses when you win”has an
amount of money $R. What is the probability you will eventually lose all of your
wealth W, given p and R? From probability theory [1] the answer is:

(q/p)W + R - (q/p)W
for p ≠ q
Ruin probability = ””””””””””””””””””,
(q/p)W + R - 1


Ruin probability = 1 - [W/(W + R)], for p = q = .5.

An Example
You have $10 and your friend has $100. You flip a fair coin. If heads comes up, he
pays you $1. If tails comes up, you pay him $1. The game ends when either player
runs out of money. What is the probability your friend will end up with all of your
money? From the second equation above, we have p = q = .5, W = $10, and R =
$100. Thus the probability of your losing everything is:
1 - (10/(10 + 100)) =.909.
You will lose all of your money with 91 percent probability in this supposedly "fair"
game.

Now you know how casinos make money. Their bank account is bigger than yours.
Eventually you will have a losing streak, and then you will have to stop playing
(since the casinos will not loan you infinite capital).
The gambler™s ruin odds are the important ones. True, the odds are stacked against
the player in each casino game: heavily against the player for kino, moderately
against the player for slots, marginally against the player for blackjack and craps.
(Rules such as "you can only double down on 10s and 11s" in blackjack are intended
to turn the odds against the player, as are the use of multiple card decks, etc.) But the
chief source of casino winnings is that people have to stop playing once they™ve had
a sufficiently large losing streak, which is inevitable. (Lots of "free" drinks served to
the players help out in this process. From the casino™s point of view, the investment
in free drinks plays off splendidly.)
Note here that "wealth" (W or R in the equation) is defined as the number of betting
units: $1 in the example. The more betting units you have, the less probability there
is you will be hit with the gambler™s ruin problem. So you if you sit at the blackjack
table at Harrah™s with a $1000 minimum bet, you will need to have 100 times the
total betting capital of someone who sits at the $10 minimum tables, in order to have
the same odds vis-à-vis the dealer.


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A person who has $1000 in capital and bets $10 at a time has a total of W = 1000/10
= 100 betting units. That™s a fairly good ratio.
While a person who has $10,000 in capital and bets $1000 at a time has W =
10000/1000 = 10 betting units. That™s lousy odds, no matter the game. It™s loser
odds.
Gauss vs. Cauchy
We measure probability with our one-pound jar of jam. We can distribute the jam in
any way we wish. If we put it all at the point x = 5, then we say "x = 5 with
certainty" or "x = 5 with probability 1."
Sometimes the way the jam is distributed is determined by a simple function. The
normal or Gaussian distribution distributes the jam (probability) across the real line
(from minus infinity to plus infinity) using the density function:
f(x) = [1/(2π )0.5] exp(-x2/2) , - ∞ < x < ∞
Here f(x) creates the nice bell-shaped curve we have seen before (x is on the
horizontal line, and f(x) is the blue curve above it):




The jam (probability) is smeared between the horizontal line and the curve, so the
height of the curve at each point (given by f(x)) indicates that point™s probability
relative to some other point. The curve f(x) is called the probability density.
So we can calculate the probability density for each value of x using the function
f(x). Here are some values:

x f(x)

-3 .0044

-2 .0540

-1 .2420

-.75 .3011

-.50 .3521



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-.25 .3867

0 .3989

.25 .3867

.50 .3521

.75 .3011

1 .2420

2 .0540

3 .0044

At the center value of x = 0, the probability density is highest, and has a value of f(x)
= .3989. Around 0, the probability density is spread out symmetrically in each
direction.
The entire one-pound jar of jam is smeared underneath the curve between “ ∞ and +
∞ . So the total probability, the total area under the curve, is 1. In calculus the area
under the curve is written as an integral, and since the total probability is one, the
integral from - ∞ to + ∞ of the jam-spreading function f(x) is 1.
The probability that x lies between a and b, a < x < b, is just the area under the curve
(the amount of jam) measured from a to b, as indicated by the red portion in the
graphic below, where a = -1 and b = +1:




Instead of writing this integral in the usual mathematical fashion, which requires
using a graphic in the html world of your web browser, I will simply denote the
integral from a to b of f(x) as:
I(a,b) f(x) dx.
I(a,b) f(x) dx, then, is the area under the f(x) curve from a to b. In the graphic above,

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we see pictured I(-1,1). And since the total probability (total area under the curve)
across all values of x ( from - ∞ to + ∞ ) is 1, we have
I(- ∞ ,∞ ) f(x) dx = 1.
A little more notation will be useful. We want a shorthand way of expressing the
probability that x < b. But the probability that x < b is the same as the probability
that -∞ < x < b. So this value is given by the area under the curve from -∞ to b. We
will write this as F(b):
F(b) = I(-∞ ,b) f(x) dx = area under curve from minus infinity to b.
Here is a picture of F(b) when b = 0:




For any value x, F(x) is the cumulative probability function. It represents the total
probability up to (and including) point x. It represents the probability of all values
smaller than (or equal to) x.
(Note that since the area under the curve at a single point is zero, whether we include
the point x itself in the cumulative probability function F(x), or whether we only
include all points less than x, does not change the value of F(x). However, our
understanding will be that the point x itself is included in the calculation of F(x).)
F(x) takes values between 0 and 1, corresponding to our one-pound jar of jam. Hence
F(-∞ ) = 0, while
F(+∞ ) = 1.
The probability between a and b, a <x < b, then, can be written simply as
F(b) “ F(a).
The probability x > b can be written as:
1 “ F(b).
Now. Here is a different function for spreading probability, called the Cauchy
density:



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g(x) = 1/[π (1 + x2)], - ∞ < x < ∞
Here is a picture of the resulting Cauchy curve:




It it nice and symmetric like the normal distribution, but is relatively more
concentrated around the center, and taller in the tails than the normal distribution.
We can see this more clearly by looking at the values for g(x):


x g(x)

-3 .0318

-2 .0637

-1 .1592

-.75 .2037

-.50 .2546

-.25 .2996

0 .3183

.25 .2996

.50 .2546

.75 .2037

1 .1592

2 .0637

3 .0318

At every value of x, the Cauchy density is lower than the normal density, until we

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get out into the extreme tails, such as 2 or 3 (+ or -).
Note that at “3, for example, the probability density of the Cauchy distribution is
g(-3) = .0318, while for the normal distribution, the value is f(-3) = .0044. There is
more than 7 times as much probability for this extreme value with the Cauchy
distribution than there is with the normal distribution! (The calculation is
.0318/.0044 = 7.2.) Relative to the normal, the Cauchy distribution is fat-tailed.
To see a more detailed plot of the normal density minus the Cauchy density, make
sure Java is enabled on your web browser and click here.
As we will see later, there are other distributions that have more probability in the
tails than the normal, and also more probability at the peak (in this case, around 0).
But since the total probability must add up to 1, there is, of course, less probability in
the intermediate ranges. Such distributions are called leptokurtic. Leptokurtic
distributions have more probability both in the tails and in the center than does the
normal distribution, and are to be found in all asset markets”in foreign exchange,
shares of stock, interest rates, and commodity prices. (People who pretend that the
empirical distributions of changes in log prices in these markets are normal, rather
than leptokurtic, are sadly deceiving themselves.)
Location and Scale
So far, as we have looked at the normal and the Cauchy densities, we have seen they
are centered around zero. However, since the density is defined for all values of x, -
∞ < x < ∞ , the center can be elsewhere. To move the center from zero to a location
m, we write the normal probability density as:
f(x) = [1/(2π )0.5] exp(-(x-m)2/2) , - ∞ < x < ∞ .
Here is a picture of the normal distribution after the location has been moved from m
= 0 (the blue curve) to m = 2 (the red curve):




For the Cauchy density, the corresponding alteration to include a location parameter
m is:
g(x) = 1/[π (1 + (x-m)2)], - ∞ < x < ∞
In each case, the distribution is now centered at m, instead of at 0. Note that I say
"location paramter m" and not "mean m". The reason is simple. For the Cauchy
distribution, a mean doesn™t exist. But a location parameter, which shows where the


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probability distribution is centered, does.
For the normal distribution, the location parameter m is the same as the mean of the
distribution. Thus a lot of people who are only familiar with the normal distribution
confuse the two. They are not the same.
Similarly, for the Cauchy distribution the standard deviation (or the variance, which
is the square of the standard deviation) doesn™t exist. But there is a scale parameter
c that shows how far you have to move in each direction from the location
parameter m, in order for the area under the curve to correspond to a given
probability. For the normal distribution, the scale parameter c corresponds to the
standard deviation. But a scale parameter c is defined for the Cauchy and for other,
leptokurtic distributions for which the variance and standard deviation don™t exist
("are infinite").
Here is the normal density written with the addition of a scale parameter c:
f(x) = [1/(c (2π )0.5)] exp(-((x-m)/c)2/2) , - ∞ < x < ∞ .
We divide (x-m) by c, and also multiply the entire density function by the reciprocal
of c.
Here is a picture of the normal distribution for difference values of c:




The blue curve represents c = 1, while the peaked red curves has c < 1, and the
flattened red curve has c > 1.
For the Cauchy density, the addition of a scale parameter gives us:
g(x) = 1/[cπ (1 + ((x-m)/c)2)], - ∞ < x < ∞
Just as we did with the normal distribution, we divide (x-m) by c, and also multiply
the entire density by the reciprocal of c.
Operations with location and scale are well-defined, whether or not the mean or the
variance exist.
Most of the probability distributions we are interested in in finance lie somewhere
between the normal and the Cauchy. These two distributions form the "boundaries",
so to speak, of our main area of interest. Just as the Sierpinski carpet has a Hausdorff
dimension that is a fraction which is greater than its topological dimension of 1, but


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less than its Euclidean dimension of 2, so do the probability distributions in which
we are chiefly interested have a dimension that is greater than the Cauchy dimension
of 1, but less than the normal dimension of 2. (What is meant here by the "Cauchy
dimension of 1" and the "normal dimension of 2" will be clarified as we go along.)
Notes
[1] See Chapter 14 in William Feller, An Introduction to Probability Theory and Its
Applications, Vol. I, 3rd ed., John Wiley & Sons, New York, 1968.

J. Orlin Grabbe is the author of International Financial Markets, and is an
internationally recognized derivatives expert. He has recently branched out into
cryptology, banking security, and digital cash. His home page is located at
http://www.aci.net/kalliste/homepage.html .
-30-
from The Laissez Faire City Times, Vol 3, No 27, July 5, 1999




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Coin Toss Plot


Coin Toss Plot
If you can read this, then your browser is not set up for Java. . .

The applet shows 3200 successive coin tosses, presented in 8 rows of 400 tosses each. The height of the
image is the number of heads minus the number of tails (the "winnings" of Henry who wins a dollar if
heads comes up, but losses a dollar if tails comes up.) Because of space constraints, the rows may
overlap if the cumulative wins or losses depart far from zero. (The eight rows are spaced 50 pixels apart,
so a win of 25 on one row will touch a loss of 25 on the previous row.) Often wins or losses become so
great that solid blocks of blue are formed.
Hit "Reload" or "Refresh Screen" on your browser to do a new series of 3200 coin tosses. Each series of
3200 coin tosses starts from a value of zero.

Here is the java source code.




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Coin Toss Zero Set Toss Plot


Coin Toss Zero Set Plot
If you can read this, then your browser is not set up for Java. . .

The applet shows the zero set of 3200 successive coin tosses, presented in 8 rows of 400 tosses each.
The zero set is the set of values of n where the number of heads and the number of tails are equal. It
corresponds to the times when Henry, a gambler, reaches a break-even point. Henry wins a dollar if
heads comes up, but losses a dollar if tails comes up, so the zero set corresponds to those times when his
wins and losses are equalized. Note that the zero set has a Cantor-dust-like structure.
Hit "Reload" or "Refresh Screen" on your browser to do a new series of 3200 coin tosses. Each series of
3200 coin tosses starts from a value of zero.

Here is the java source code.




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Normal Minus Cauchy Probability Density Plot


Normal Minus Cauchy Probability
Density
If you can read this, then your browser is not set up for Java. . .

This applet plots 400 points as x varies from x = -10 to x = +10. The vertical axis shows the value of f(x)
minus g(x), where f(x) is the normal density, and g(x) is the Cauchy density. The blue area corresponds
to the region f(x)>g(x), while the red area corresponds to the region g(x)>f(x). The Cauchy distribution
has much greater probability in the extreme tails.

Here is the java source code.




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Chaos and Fractals in Financial Markets, Part 5, by J. Orlin Grabbe


Chaos and Fractals in Financial Markets
Part 5
by J. Orlin Grabbe
Louis Bachelier Visits the New York Stock Exchange
Louis Bachelier, resurrected for the moment, recently
visited the New York Stock Exchange at the end of May
1999. He was somewhat puzzled by all the hideous
concrete barriers around the building at the corner of
Broad and Wall Streets. For a moment he thought he was
in Washington, D.C., on Pennsylvania Avenue.
Bachelier was accompanied by an angelic guide named
Pete. "The concrete blocks are there because of Osama
bin Ladin," Pete explained. "He™s a terrorist." Pete didn™t
bother to mention the blocks had been there for years. He
knew Bachelier wouldn™t know the difference.
"Terrorist?"
"You know, a ruffian, a scoundrel."
"Oh," Bachelier mused. "Bin Ladin. The son of Ladin."
"Yes, and before that, there was Abu Nidal."
"Abu Nidal. The father of Nidal. Hey! Ladin is just Nidal
spelled backwards. So we™ve gone from the father of
Nidal to the son of backwards-Nidal?"
"Yes," Pete said cryptically. "The spooks are never too
creative when they are manufacturing the boogeyman of
the moment. If you want to understand all this, read
about ˜Goldstein™ and the daily scheduled ˜Two Minutes
Hate™ in George Orwell™s book 1984."
"1984? Let™s see, that was fifteen years ago," Bachelier
said. "A historical work?"
"Actually, it™s futuristic. But he who controls the present
controls the past, and he who controls the past controls
the future."
Bachelier was mystified by the entire conversation, but
once they got inside and he saw the trading floor, he felt

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right at home. Buying, selling, changing prices. The
chalk boards were now electric, he saw, and that made
the air much fresher.
"Look," Bachelier said, "the Dow Jones average is still
around!"
"Yes," nodded Pete, "but there are a lot of others also.
Like the S&P500 and the New York Stock Exchange
Composite Index."
"I want some numbers!" Bachelier exclaimed
enthusiastically. Before they left, they managed to con
someone into giving them the closing prices for the
NYSE index for the past 11 days.
"You can write a book," Pete said. "Call it Eleven Days
in May. Apocalyptic titles are all the rage these
days”except in the stock market."
Bachelier didn™t pay him any mind. He had taken out a
pencil and paper and was attempting to calculate
logarithms through a series expansion. Pete watched in
silence for a while, before he took pity and pulled out a
pocket calculator.
"Let me show you a really neat invention," the angel
said.
Bachelier™s Scale for Stock Prices
Here is Bachelier™s data for eleven days in May. We
have the calendar date in the first column of the table;
the NYSE Composite Average, S(t), in the second
column; the log of S(t) in the third column; the change in
log prices, x(t) = log S(t) “ log S(t-1) in the fourth
column; and x(t)2 in the last column. The sum of the
variables in the last column is given at the bottom of the
table.

x(t)2
Date S(t) log S(t) x(t)

May 14 638.45 6.459043

May 17 636.92 6.456644 -.002399 .000005755

May 18 634.19 6.452348 -.004296 .000018456


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May 19 639.54 6.460749 .008401 .000070577

May 20 639.42 6.460561 -.000188 .000000035

May 21 636.87 6.456565 -.003996 .000015968

May 24 626.05 6.439430 -.017135 .000293608

May 25 617.34 6.425420 -.014010 .000196280

May 26 624.84 6.437495 .012075 .000145806

May 27 614.02 6.420027 -.017468 .000305131

May 28 622.26 6.433358 .013331 .000177716

sum of all x(t)2 = .001229332

What is the meaning of all this?
The variables x(t), which are the one-trading-day
changes in log prices, are the variables in which
Bachelier is interested for his theory of Brownian motion
as applied to the stock market:
x(t) = log S(t) “ log S(t-1).
Bachelier thinks these should have a normal distribution.
Recall from Part 4 that a normal distribution has a
location parameter m and a scale parameter c. So what
Bachelier is trying to do is to figure out what m and c
are, assuming that each day™s m and c are the same as
any other day™s.
The location parameter m is easy. It is zero, or pretty
close to zero.
In fact, it is not quite zero. Essentially there is a drift in
the movement of the stock index S(t), given by the
difference between the interest rate (such as the
broker-dealer loan rate) and the dividend yield on stocks
in the average.[1] But this is tiny over our eleven trading
days (which gives us ten values for x(t)). So Bachelier
just assumes m is zero.

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So what Bachelier is doing with the data is trying to
estimate c.
Recall from Part 2 that if today™s price is P, Bachelier
modeled the probability interval around the log of the
price change by
(log P “ a T0.5, log P + a T0.5), for some constant a.
But now, we are writing our stock index price as S, not
P; and the constant a is just our scale parameter c. So,
changing notation, Bachelier is interested in the
probability interval
(log S “ c T0.5, log S + c T0.5), for a given scale
parameter c.
One way of estimating the scale c (c is also called the
"standard deviation" in the context of the normal
distribution) is to add up all the squared values of x(t),
and take the average (by dividing by the number of
observations). This gives us an estimate of the variance,
or c2. Then we simply take the square root to get the
scale c itself. (This is called a maximum likelihood
estimator for the standard deviation.)
Adding up the terms in the right-hand column in the
table gives us a value of .001229332. And there are 10
observations. So we have
variance = c2 = .001229332/10 = .0001229332.
Taking the square root of this, we have
standard deviation = c = (.0001229332)0.5 = .0110875.
So Bachelier™s changing probability interval for log S
becomes:
(log S “ .0110875 T0.5, log S + .0110875 T0.5).
To get the probability interval for the price S itself, we
just take exponentials (raise to the power exp = e =
2.718281¦), and get
( S exp(“ .0110875 T0.5), S exp(.0110875 T0.5) ).
Since the current price on May 28, from the table, is
622.26, this interval becomes:


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(622.26 exp(“ .0110875 T0.5), 622.26 exp(.0110875
T0.5) ).
"This expression for the probability interval tells us the
probability distribution over the next T days," Bachelier
explained to Pete. "Now I understand what you meant.
He who controls the present controls the past, because he
can obtain past data. While he who masters this past data
controls the future, because he can calculate future
probabilities!"
"Umm. That wasn™t what I meant," the angel replied.
"But never mind."
Over the next 10 trading days, we have T0.5 = 100.5 =
3.162277. So substituting that into the probability
interval for price, we get
(622.26 (.965545), 622.26 (1.035683)) = (600.82,
644.46).
This probability interval gives a price range for plus or
minus one scale parameter (in logs) c. For the normal
distribution, that corresponds to 68 percent probability.
With 68 percent probability, the price will lie between
600.82 and 644.46 at the end of 10 more trading days,
according to this calculation.
To get a 95 percent probability interval, we use plus or
minus 2c,
(622.26 exp(“ (2) .0110875 T0.5), 622.26 exp( (2)
.0110875 T0.5) ),
which gives us a price interval over 10 trading days of
(580.12, 667.46).
Volatility
In the financial markets, the scale parameter c is often
called "volatility". Since a normal distribution is usually
assumed, "volatility" refers to the standard deviation.
Here we have measured the scale c, or volatility, on a
basis of one trading day. The value of c we calculated, c
= .0110875, was calculated over 10 trading days, so it
would be called in the markets "a 10-day historical
volatility." If calculated over 30 past trading days, it
would be "a 30-day historical volatility."

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However, market custom would dictate two criteria by
which volatility is quoted:
1. quote volatility at an annual (not daily) rate;
2. quote volatility in percentage (not decimal) terms.
To change our daily volatility c = .0110875 into annual
terms, we note that there are about 256 trading days in
the year. The square root of 256 is 16, so to change daily
volatility into annual volatility, we simply multiply it by
16:
annual c = 16 (daily c) = 16 (.0110875) = .1774.
Then we convert this to percent (by multiplying by 100
and calling the result "percent"):
annual c = 17.74 percent.
The New York Stock Exchange Composite Index had a
historical volatility of 17.74 percent over the sample
period during May.
Note that an annual volatility of 16 percent corresponds
to a daily volatility of 1 percent. This is a useful
relationship to remember, because we can look at a price
or index, mentally divide by 100, and say the price
change will fall in the range of plus or minus that amount
with 2/3 probability (approximately). For example, if the
current gold volatility is 16 percent, and the price is
$260, we can say the coming day™s price change will fall
in the range of plus or minus $2.60 with about 2/3
probability.
Notice that 256 trading days give us a probability
interval that is only 16 times as large as the probability
interval for 1 day. This translates into a Hausdorff
dimension for time (in the probability calculation) as D =
log(16)/log(256) = ½ or 0.5, which is just the
Bachelier-Einstein square-root-of-T (T0.5) law.
The way we calculated the scale c is called "historical
volatility," because we used actual historical data to
estimate c. In the options markets, there is another
measure of volatility, called "implied volatility."
Implied volatility is found by back-solving an option
value (using a valuation formula) for the volatility, c,
that gives the current option price. Hence this volatility,


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which pertains to the future (specifically, to the future
life of the option) is implied by the price at which the
option is trading.
Fractal Sums of Random Variables
Now for the fun part. We have been looking at random
variables x(t) (representing changes in the log of price).
Under the assumption these random variables were
normal, we estimated a scale parameter c, which allows
us to do probability calculations.
In order to estimate c, we took the sums of random
variables (or, in this instance, the sums of squares of
x(t)).
Were our calculations proper and valid? Do they make
any sense? The answer to these questions depends on the
issue of the probability distribution of a sum of random
variables. How does the distribution of the sum relate to
the distributions of the individual random variables that
are added together?
In answering this question we want to focus on ways we
can come up with a location parameter m, and a scale
parameter c. For the normal distribution, m is the mean,
but for the Cauchy distribution the mean doesn™t exist
("is infinite"). For the normal distribution, the scale
parameter c is the standard deviation, but for the Cauchy
distribution the standard deviation doesn™t exist.
Nevertheless, a location m and a scale c exist for the
Cauchy distribution. The maximum likelihood estimator
for c will not be the same in the case of the Cauchy
distribution as it was for the normal. We can™t take
squares if the x(t) have a Cauchy distribution.
Suppose we have n random variables Xi, all with the
same distribution, and we calculate their sum X:
X = X1 + X2 + ¦ + Xn-1 + Xn.

Does the distribution of the sum X have a simple form?
In particular, can we relate the distribution of X to the
common distribution of the Xi? Let™s be even more
specific. We have looked at the normal (Gaussian) and
Cauchy distributions, both of which were parameterized
with a location m and scale c. If each of the Xi has a


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location m and scale c, whether normal or Cauchy, can
that information be translated into a location and a scale
for the sum X?
The answer to all these questions is yes, for a class of
distributions called stable distributions. (They are also
sometimes called "Levy stable", "Pareto-Levy", or
"stable Paretian" distributions.) Both the normal and the
Cauchy are stable distributions. But there are many
more.
We will use the notation "˜" as shorthand for "has the
same distribution as." For example,
X1 ˜ X2

means X1 and X2 have the same distribution. We now
use "˜" in the following definition of stable distributions:
Definition: A random variable X is said to have a stable
distribution if for any n >= 2 (greater than or equal to
2), there is a positive number Cn and a real number Dn
such that
X1 + X2 + ¦ + Xn-1 + Xn ˜ Cn X + Dn

where X1, X2, ¦, Xn are all independent copies of X.

Think of what this definition means. If their distribution
is stable, then the sum of n identically distributed
random variables has the same distribution as any one of
them, except by multiplication by a scale factor Cn and a
further adjustment by a location Dn .

Does this remind you of fractals? Fractals are
geometrical objects that look the same at different scales.
Here we have random variables whose probability
distributions look the same at different scales (except for
the add factor Dn).

Let™s define two more terms.[2]
Definition: A stable random variable X is strictly stable
if Dn = 0.

So strictly stable distributions are clearly fractal in
nature, because the sum of n independent copies of the
underlying distribution looks exactly the same as the


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underlying distribution itself, once adjust by the scale
factor Cn. One type of strictly stable distributions are
symmetric stable distributions.
Definition: A stable random variable X is symmetric
stable if its distribution is symmetric”that is, if X and
-X have the same distribution.
The scale parameter Cn necessarily has the form [3]:

Cn = n1/ ± , where 0< ± <=2.

So if we have n independent copies of a symmetric
stable distribution, their sum has the same distribution
with a scale that is n1/ ± times as large.
For the normal or Gaussian distribution, ± = 2. So for n
independent copies of a normal distribution, their sum
has a scale that is n1/ ± = n1/ 2 times as large.
For the Cauchy distribution, ± = 1. So for n independent
copies of a Cauchy distribution, their sum has a scale
that is n1/ ± = n1/ 1 = n times as large.
Thus if, for example, Brownian particles had a Cauchy
distribution, they would scale not according to a T0.5
law, but rather according to a T law!
Notice that we can also calculate a Hausdorff dimension
for symmetric stable distributions. If we divide a
symmetric stable random variable X by a scale factor of
c = n1/ ± , we get the probability equivalent [4] of N = n
copies of X/n1/ ± . So the Hausdorff dimension is

D = log N/ log c = log n/ log(n1/ ± ) = ± .
This gives us a simple interpretation of ± . The
parameter ± is simply the Hausdorff dimension of a
symmetric stable distribution. For the normal, the
Hausdorff dimension is equal to 2, equivalent to that of a
plane. For the Cauchy, the Hausdorff dimension is equal
to 1, equivalent to that of a line. In between is a full
range of values, including symmetric stable distributions
with Hausdorff dimensions equivalent to the Koch Curve
(log 4/log 3) and the Sierpinski Carpet (log 8/log3).
Some Fun with Logistic Art


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