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| Science Forum Index » Fractals Science Forum » Box-counting dimension of random and Gaussian data points? |
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| Ross Evans |
 Posted: Tue Dec 07, 2004 8:45 pm |
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Guest
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I need help understanding something.
The box-counting dimension of a set of random data points -- random in the
sense that they are white noise, equally likely to occur anywhere within an
interval -- is theoretically 2. That is computed by adding 1 to the slope
of the log/log plot of the single-variable data points, counted at various
sizes of box measurent..
That works out empirically for me. My measured slope of such a set of data
points is very close to 1.
But what about data points that are distributed not as such white noise, but
as Gaussian-distributed data centered around a mean. These data points
visually do not fill the space as completely, because by definition they
tend to be clustered around the mean. And when I measure the slope of such
a set of points, it is only about .92.
But the literature generally seems to expect a box-counting dimension of 2
for Gaussian data, not just white noise. I don't get it. |
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| Roger Bagula |
| Posted: Wed Dec 08, 2004 11:19 am |
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Guest
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Dear Ross Evans,
Are you measuring a time series as:
s[n]->{s[1],s[2], s[3],...,s[n]}
Or
{r[n],s[n]},-> {{r[1],s[1]},{r[2],s[2]},{r[3],s[3]},...,{r[n],s[n]}}
The first will be one dimensional and the second should be two dimensional.
How are you producing your white noise/ normal variates?
Different randomization method of producing your distribution will
produce different
results.
Ross Evans wrote:
[quote:6a2d3d889f]I need help understanding something.
The box-counting dimension of a set of random data points -- random in the
sense that they are white noise, equally likely to occur anywhere within an
interval -- is theoretically 2. That is computed by adding 1 to the slope
of the log/log plot of the single-variable data points, counted at various
sizes of box measurent..
That works out empirically for me. My measured slope of such a set of data
points is very close to 1.
But what about data points that are distributed not as such white noise, but
as Gaussian-distributed data centered around a mean. These data points
visually do not fill the space as completely, because by definition they
tend to be clustered around the mean. And when I measure the slope of such
a set of points, it is only about .92.
But the literature generally seems to expect a box-counting dimension of 2
for Gaussian data, not just white noise. I don't get it.
[/quote:6a2d3d889f]
--
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
alternative email: rlbtftn@netscape.net
URL : http://home.earthlink.net/~tftn |
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| Roger Bagula |
| Posted: Thu Dec 09, 2004 12:23 pm |
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Guest
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Let me say that this is a good quest
and the answer in not obvious
and may well be not all that certain, but
it is what the Einstein theory of Brownian motion tells us.
A French economist actually discovered this diffusion law before
Einstein, but
he is the one who applied it best to physics.
It all ties in with his ideas about curvature of space time as well.
To give you a counter idea: suppose that light got dimmer as the cube of
the distance
instead of as the square; that is the "measure" for space was:
m3(x,y,z)=(x^3+y^3+z^3)^(1/3)
instead of
m2(x,y,z)=(x^2+y^2+z^2)^(1/2)
The result that the diffusion of Brownian motion would
be much slower and be governed by a different set of differential
equations altogether.
The limiting Femat type nth measure is the Hausdorff space measure:
Mhd=Max[x]+Max[y]+Max[z]
Which is a 3d space filling measure.
If we add in an non-Euclidean time in a negative hyperbolic sense:
Mhd4d=Max[x]+Max[y]+Max[z]-c*Max[t]
You see the time probability and the curvature of space is "intrinsic"
in the diffusion measures that Brownian motion is based on.
Ross Evans wrote:
[quote:3fd78b4953]I need help understanding something.
The box-counting dimension of a set of random data points -- random in the
sense that they are white noise, equally likely to occur anywhere within an
interval -- is theoretically 2. That is computed by adding 1 to the slope
of the log/log plot of the single-variable data points, counted at various
sizes of box measurent..
That works out empirically for me. My measured slope of such a set of data
points is very close to 1.
But what about data points that are distributed not as such white noise, but
as Gaussian-distributed data centered around a mean. These data points
visually do not fill the space as completely, because by definition they
tend to be clustered around the mean. And when I measure the slope of such
a set of points, it is only about .92.
But the literature generally seems to expect a box-counting dimension of 2
for Gaussian data, not just white noise. I don't get it.
[/quote:3fd78b4953]
--
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
alternative email: rlbtftn@netscape.net
URL : http://home.earthlink.net/~tftn |
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| Roger Bagula |
| Posted: Thu Dec 09, 2004 12:41 pm |
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Guest
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Looking it up in a very old noise theory book the
diffusion of a particle in a uniform field ( gravitation) is:
dPhi/dt=Sum[beta(i)*dPhi/dx(i)+D(i)*d^2Phi/dx(i)^2,{i,1,3}]
You notice that the derivative level is the second derivative.
Page 299 of Selected Papers Noise and Stochastic Processes, edited by
Nelson Wax, Dover books, 1954
Norbert Weiner and Benoit Mandelbrot and a professor name Ito and his
calculus have
modified the general theory some since then, but not
the basis of curvature, diffusion and time.
Roger Bagula wrote:
[quote:5700e8ad88]Let me say that this is a good quest
and the answer in not obvious
and may well be not all that certain, but
it is what the Einstein theory of Brownian motion tells us.
A French economist actually discovered this diffusion law before
Einstein, but
he is the one who applied it best to physics.
It all ties in with his ideas about curvature of space time as well.
To give you a counter idea: suppose that light got dimmer as the cube
of the distance
instead of as the square; that is the "measure" for space was:
m3(x,y,z)=(x^3+y^3+z^3)^(1/3)
instead of
m2(x,y,z)=(x^2+y^2+z^2)^(1/2)
The result that the diffusion of Brownian motion would
be much slower and be governed by a different set of differential
equations altogether.
The limiting Femat type nth measure is the Hausdorff space measure:
Mhd=Max[x]+Max[y]+Max[z]
Which is a 3d space filling measure.
If we add in an non-Euclidean time in a negative hyperbolic sense:
Mhd4d=Max[x]+Max[y]+Max[z]-c*Max[t]
You see the time probability and the curvature of space is "intrinsic"
in the diffusion measures that Brownian motion is based on.
Ross Evans wrote:
I need help understanding something.
The box-counting dimension of a set of random data points -- random
in the
sense that they are white noise, equally likely to occur anywhere
within an
interval -- is theoretically 2. That is computed by adding 1 to the
slope
of the log/log plot of the single-variable data points, counted at
various
sizes of box measurent..
That works out empirically for me. My measured slope of such a set
of data
points is very close to 1.
But what about data points that are distributed not as such white
noise, but
as Gaussian-distributed data centered around a mean. These data points
visually do not fill the space as completely, because by definition they
tend to be clustered around the mean. And when I measure the slope
of such
a set of points, it is only about .92.
But the literature generally seems to expect a box-counting dimension
of 2
for Gaussian data, not just white noise. I don't get it.
[/quote:5700e8ad88]
--
Respectfully, Roger L. Bagula
tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
alternative email: rlbtftn@netscape.net
URL : http://home.earthlink.net/~tftn |
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| Tony Roberts |
| Posted: Thu Dec 09, 2004 6:38 pm |
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Guest
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You have not commented on my reply,
Tony
Tony Roberts wrote:
[quote:8aadbb0d82]Firstly, I presume you are distributing the points in the plane.
Generically, uniformly random points in n dimensions should have
dimension n.
Secondly, yes indeed a Gaussian cloud of points in n dimensions should
have dimension n.
Thirdly, well the previous statement only holds for some scales. You
must understand that a fractal dimension ONLY applies over a finite
(usually) range of length scales. Suppose your Gaussian cloud of k
points is distrubuted with standard deviation 1:
* on large lengths scales (say >10), the entire cloud will look like
just one point as they are all located at near ly the same pplace, thus
the cloud looks like it has dimension 1
* on medium length scales (say <2 and > k^(1/n)), the cloud of points
looks as if it fills up the space in a fuzz of points and som has
dimension n
* on really small length scales (say <k^(/n)) you will perceive the
points as individuals and the set of points will again have dimension 0.
The right dimension depends upon the length scale at which you want to
interact with the object.
Fourthly, the box counting dimension is a poor methodology which mixes
up the dimension of the middle regime with that of the surrounding
regimes. Thus the box counting dimension (in this example) will be a
little low.
Tony
Ross Evans wrote:
I need help understanding something.
The box-counting dimension of a set of random data points -- random in
the
sense that they are white noise, equally likely to occur anywhere
within an
interval -- is theoretically 2. That is computed by adding 1 to the
slope
of the log/log plot of the single-variable data points, counted at
various
sizes of box measurent..
That works out empirically for me. My measured slope of such a set of
data
points is very close to 1.
But what about data points that are distributed not as such white
noise, but
as Gaussian-distributed data centered around a mean. These data points
visually do not fill the space as completely, because by definition they
tend to be clustered around the mean. And when I measure the slope of
such
a set of points, it is only about .92.
But the literature generally seems to expect a box-counting dimension
of 2
for Gaussian data, not just white noise. I don't get it.
[/quote:8aadbb0d82] |
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