On Feb 21, 2011, at 9:53 AM, Erich Neuwirth wrote:
We want to generate a distribution on the unit square with the
following
properties
* It is concentrated on a "reasonable" subset of the square,
and the restricted distribution is uniform on this subset.
* Both marginal distributions are uniform on the unit interval.
* All horizontal and all vertical cross sections are sets of lines
segments with the same total length
If we find a geometric figure with these properties, we have solved
the
problem.
So we define the distribution to be uniform on the following area:
(it is distorted but should give the idea)
x***/-----------------/***x
|**/-----------------/****|
|*/-----------------/*****|
|/-----------------/******|
|-----------------/******/|
|----------------/******/-|
|---------------/******/--|
|--------------/******/---|
|-------------/******/----|
|------------/******/-----|
|-----------/******/------|
|----------/******/-------|
|---------/******/--------|
|--------/******/---------|
|-------/******/----------|
|------/******/-----------|
|-----/******/------------|
|----/******/-------------|
|---/******/--------------|
|--/******/---------------|
|-/******/----------------|
|/******/-----------------|
|******/-----------------/|
|*****/-----------------/*|
|****/-----------------/**|
x***/-----------------/***x
There is the same number of stars in each horizontal row and each
vertical column.
I love it! I plotted the data that your method generated and got a
plot with points in the regions you displayed. After pondering its
curious pathology (three disjoint regions), I thought I could cure
that pathology by flipping quadrants. I proceeded to do so but
discovered that the pathology wasn't really cured but only
concentrated at the ends and middle. Using your ASCII art, my version
wuld look like this where the \\\ regions are "double dense".
y[x>0.5 & y <0.5] = 0.5 +abs(0.5-y[x>0.5 & y <0.5])
y[x<0.5 & y >0.5] = 0.5 -abs(0.5-y[x<0.5 & y >0.5])
plot(x,y)
x---------------------\\\\x
|--------------------/*\\\|
|-------------------/***\\|
|------------------/*****\|
|-----------------/******/|
|----------------/******/-|
|---------------/******/--|
|--------------/******/---|
|-------------/******/----|
|-------------|\\***/-----|
|-------------|\\\*/------|
|-------------|\\\\-------|
|-------------*-----------|
|--------/\\\\------------|
|-------/**\\\------------|
|------/****\\------------|
|-----/******\------------|
|----/******/-------------|
|---/******/--------------|
|--/******/---------------|
|-/******/----------------|
|/******/-----------------|
|\*****/------------------|
|\\***/-------------------|
|\\\\/--------------------|
\\\\\---------------------x
So it not only created a still slightly less pathological counterpart,
but the correlation jumped from 0.5 to 0.95. It looks to be a
promising basis for homework problems in probability courses, an
experience I have never has the ?pleasure? to experience except during
self-study to repair my (many) mathematical deficiencies.
> cor(x,y)
[1] 0.9449256
--
David.
So we define
g(x1,x2)= 1 abs(x1-x2) <= a or
abs(x1-x2+1) <= a or
abs(x1-x2-1) <= a
0 elsewhere
The total area of the shape is 2*a.
The admissible range for a is <0,1/2>
therefore
f(x1,x2)=g(x1,x2)/(2*a)
is a density functions.
This is where simple algebra comes in.
This distribution has
expected value 1/2 and variance 1/12 for both margins
(uniform distribution), and it has
covariance = (1-3*a+2*a2)/12
and correlation = 1 - 3*a + 2*a2
The inverse function of 1 - 3*2 + 2*a2 is
(3-sqrt(1+8*r))/4
Therefore we can compute that our distribution with
a=(3-sqrt(1+8*r))/4
will produce a given r.
Ho do we create random numbers from this distribution?
By using conditional densities.
x1 is sampled from the uniform distribution, and for a give x1
we produce x2 by a uniform distribution on the along the vertical
cross
cut of the geometrical shape (which is either 1 or 2 intervals).
And which is most easily implemented by using the modulo operator %%.
This mechanism is NOT a convolution. Applying module after the
addition
makes it a nonconvolution. Adding independent random variables
without doing anything further is a convolution, by applying a
trimming
operation, the convolution property gets lost.
--
David Winsemius, MD
West Hartford, CT
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