ñòð. 4 |

y(n+1) = 2.8*y(n)/(x(n)*x(n)-2*x(n)+2+y(n)*y(n))

with probability q = Â½.

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 weâ€™ve learned to measure length, weâ€™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 = 1.8927â€¦ 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 letâ€™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, letâ€™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 â†’ âˆž (as n goes to infinity), we have (2/3)n â†’ 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 letâ€™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 â†’ âˆž (as n goes to infinity), we have (8/9)n â†’ 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.

Letâ€™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, letâ€™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 1.8927â€¦

Similarly, to create Cantor dust, we start with a line of

one dimension. Our working space is one dimension. So

letâ€™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 = .6309â€¦

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 canâ€™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".

"Weâ€™re being invaded by foreigners!" "How many of

them are there?" "Many!"

Itâ€™s not a very good number system, since it canâ€™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 doesnâ€™t tell us how Cantor

dust differs from a single point. The topological

dimension of one doesnâ€™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=2Ï€r

where Ï€ = 3.141592653â€¦ 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. (Iâ€™ve measured my Sierpinski

carpet; havenâ€™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 rodâ€”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 Spainâ€™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, weâ€™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 isnâ€™t.

To see the principles involved, letâ€™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, letâ€™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 = 1.2618â€¦

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 triangleâ€”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, thereâ€™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? Canâ€™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. Letâ€™s try something. Instead of

L=mN,

letâ€™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. Weâ€™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, letâ€™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! Weâ€™ve just destroyed the blob again! We have a

finite length. Itâ€™s not zero and itâ€™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 = 91.8927â€¦ = 64. A

Sierpinski carpet with 10 on each side has an "area" of N

= 101.8927â€¦ = 78.12. And so on.

The Hausdorff dimension, D = 1.8927â€¦, 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, theyâ€™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 â€˜manyâ€™ ".

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 timeâ€”and from now on,

without further discussion, we will pretend that thatâ€™s

what Bachelier said also.

The issue we want to consider is whether the appropriate

dimension for time is D = Â½. In order to calculate

probability should we use T1/2, or TD, where D may take

values different from Â½?

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 = Â½ 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 Bohmâ€™s hidden-variable interpretation of the

quantum pilot wave (which obeys the rules of quantum

probability) is discussed in John Gribbin, Schrodingerâ€™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. Henryâ€™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 Henryâ€™s net winnings, then

x(n) is determined by the dynamical system:

x(n) = x(n) + 1, with probability p = Â½

x(n) = x(n) â€“ 1, with probability q = Â½.

The graph of 10,000 coin tosses in Part 3 simply shows the fluctuations in Henryâ€™s

wealth (starting from 0) over the course of the 10,000 coin tosses.

Letâ€™s do this in real time, although we will restrict ourselves to 3200 coin tosses.

Letâ€™s plot Henryâ€™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 "gamblerâ€™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 Henryâ€™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. Itâ€™s called the

Gamblerâ€™s Ruin Problem. You donâ€™t need nonnormal distributions to encounter

gambler's ruin. Normal ones will do just fine.

Futures Trading and the Gamblerâ€™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. Itâ€™s not

necessarily because they are smarter than you. Itâ€™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 partnerâ€”the person on the

other side of your bet who wins when you lose and loses when you winâ€”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 â‰ q

Ruin probability = â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”,

(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 gamblerâ€™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 theyâ€™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 casinoâ€™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 gamblerâ€™s ruin problem. So you if you sit at the blackjack

table at Harrahâ€™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-Ã -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. Thatâ€™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. Thatâ€™s lousy odds, no matter the game. Itâ€™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/(2Ï€ )0.5] exp(-x2/2) , - âˆž < x < âˆž

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 pointâ€™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 â€“ âˆž and +

âˆž . 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 - âˆž to + âˆž 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 - âˆž to + âˆž ) is 1, we have

I(- âˆž ,âˆž ) 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 -âˆž < x < b. So this value is given by the area under the curve from -âˆž to b. We

will write this as F(b):

F(b) = I(-âˆž ,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(-âˆž ) = 0, while

F(+âˆž ) = 1.

The probability between a and b, a <x < b, then, can be written simply as

F(b) â€“ F(a).

The probability x > b can be written as:

1 â€“ F(b).

Now. Here is a different function for spreading probability, called the Cauchy

density:

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g(x) = 1/[Ï€ (1 + x2)], - âˆž < x < âˆž

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 â€“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 marketsâ€”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, -

âˆž < x < âˆž , 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/(2Ï€ )0.5] exp(-(x-m)2/2) , - âˆž < x < âˆž .

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/[Ï€ (1 + (x-m)2)], - âˆž < x < âˆž

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 doesnâ€™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) doesnâ€™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 donâ€™t exist

("are infinite").

Here is the normal density written with the addition of a scale parameter c:

f(x) = [1/(c (2Ï€ )0.5)] exp(-((x-m)/c)2/2) , - âˆž < x < âˆž .

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/[cÏ€ (1 + ((x-m)/c)2)], - âˆž < x < âˆž

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. "Heâ€™s a terrorist." Pete didnâ€™t

bother to mention the blocks had been there for years. He

knew Bachelier wouldnâ€™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 weâ€™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 â€˜Goldsteinâ€™ and the daily scheduled â€˜Two Minutes

Hateâ€™ in George Orwellâ€™s book 1984."

"1984? Letâ€™s see, that was fifteen years ago," Bachelier

said. "A historical work?"

"Actually, itâ€™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

daysâ€”except in the stock market."

Bachelier didnâ€™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.

Bachelierâ€™s Scale for Stock Prices

Here is Bachelierâ€™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) â€“ 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) â€“ 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 dayâ€™s m and c are the same as

any other dayâ€™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 todayâ€™s price is P, Bachelier

modeled the probability interval around the log of the

price change by

(log P â€“ 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 â€“ 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 Bachelierâ€™s changing probability interval for log S

becomes:

(log S â€“ .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 =

2.718281â€¦), and get

( S exp(â€“ .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(â€“ .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 wasnâ€™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(â€“ (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 dayâ€™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) = Â½ 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 doesnâ€™t exist

("is infinite"). For the normal distribution, the scale

parameter c is the standard deviation, but for the Cauchy

distribution the standard deviation doesnâ€™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 canâ€™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 + â€¦ + 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? Letâ€™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 + â€¦ + Xn-1 + Xn ˜ Cn X + Dn

where X1, X2, â€¦, 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).

Letâ€™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 symmetricâ€”that is, if X and

-X have the same distribution.

The scale parameter Cn necessarily has the form [3]:

Cn = n1/ Î± , where 0< Î± <=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/ Î± times as large.

For the normal or Gaussian distribution, Î± = 2. So for n

independent copies of a normal distribution, their sum

has a scale that is n1/ Î± = n1/ 2 times as large.

For the Cauchy distribution, Î± = 1. So for n independent

copies of a Cauchy distribution, their sum has a scale

that is n1/ Î± = 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/ Î± , we get the probability equivalent [4] of N = n

copies of X/n1/ Î± . So the Hausdorff dimension is

D = log N/ log c = log n/ log(n1/ Î± ) = Î± .

This gives us a simple interpretation of Î± . The

parameter Î± 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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