Re: stirring chocolate pudding backwards: identifying coordinates

• To: mathgroup at smc.vnet.net
• Subject: [mg101569] Re: stirring chocolate pudding backwards: identifying coordinates
• From: Ray Koopman <koopman at sfu.ca>
• Date: Fri, 10 Jul 2009 06:46:12 -0400 (EDT)
• References: <h31umo\$bp1\$1@smc.vnet.net>

On Jul 8, 4:08 am, Andreas <aa... at ix.netcom.com> wrote:
> I think I need to better illustrate my problem. Let's try this:
>
> values = Array[v## &, {3, 3}];
> "values =" MatrixForm[values]
>
> The above generates: 3 time intervals in 3 dimensions.
> "values" is a times series. Time runs down the matrix.
>
> samples = Array[s## &, {4, 3}];
> "samples =" MatrixForm[samples]
>
> The above generates 4 sample points in 3 dimensions.
> "samples" is not a time series.
>
> Each of the numbers in each row of the "samples" matrix
> represents a percent as well as a coordinate. The total of
> each row in "samples" always equals 1. In the real world
> they could represent proportions of 3 fluids in a mixture
> or perhaps percentages of 3 assets in a portfolio.
>
> Each of the numbers in "values" represents an amount relative
> to some original value. They are independent measurements and
> must be >=0.
>
> The number of columns in each of the two matrices needs to
> correspond. The number of rows does not.
>
> In its planned application, I might sample 1,000 or even 1,000,000
> coordinates to better get an approximation of the weighted average
> of the sample space.
>
> Next we generate a time series (corresponding to the time
> increments in "values") of weighted averages of the samples
> where the weights stand in proportion to the measurements
> in each column at each interval in the time series "values":
>
> values.Mean[samples]
>
> Still, so far so good (and much speeded up from my original code
> thanks to your values.Mean[samples]).
>
> Now, back to my original representation of the space from which
> I select samples:
>
> sampleSimplex[sampleSize_, dimensions_] := Normalize[#, Total]& /@
> RandomReal[ExponentialDistribution[2], {sampleSize, dimensions}]
>
> The following graphs the coordinates of the samples
> so you can see what the sampleSimplex[] function does:
>
> Graphics3D[Point /@ sampleSimplex[1000, 3, 0, 1],
> ImageSize -> 250, ViewPoint -> {Pi, Pi/2, 2}]
>
> Instead of trying to calculate from the entire simplex (the
> triangle defined by coordinates: {{1,0,0},{0,1,0},{0,0,1}}),
> I sample from a finite set of points. The entire simplex would
> represent all possible mixtures.
>
> An aside, I wonder if Mathematica has a way for me to do this
> continuously rather than messing around with sampling? I still
> have a lot to learn.
>
> Using my original post's example:
>
> samples = {{0.375153,0.412505,0.212342},
>            {0.470678,0.360788,0.168533},
>            {0.0510186,0.828575,0.120407},
>            {0.24481,0.00851303,0.746677}}
>
> values = {{0.518072,0.490701,0.813364},
>           {0.404083,0.356724,0.362498},
>           {0.507292,0.436016,0.247148}}
>
> values.Mean[samples]
>
> {0.599181, 0.372042, 0.397434}
>
> I want to identify the coordinates on the plane (from which I
> select "samples") for each of these three weighted averages.
> The plane has an infinite number of points so at least one must
> have the same individual value as the weighted average of the
> sample. More generally, this point should vary, over time, across
> the plane (especially as, in its projected application, I'll have
> values for thousands of time increments and they may have far
> greater variance than my simple example). Also, it seems like it
> wouldn't have to correspond to any specific point in the list of
> samples.
>
> I hope this makes the problem clearer.
>
> Thanks for your patience.

For any set of values, the weighting scheme you describe will always
give an average that would be observed at the mean (centroid) of
the simplex. If this is counter-intuitive then perhaps you should
reconsider the weighting.

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